Deck 10: Differential Equations

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Question
Consider the differential equation y' = y - <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true. <div style=padding-top: 35px> . Which of the following statements is/are true?

A) The function f(t) = <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true. <div style=padding-top: 35px> is a solution to this differential equation with initial condition <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true. <div style=padding-top: 35px>
B) This differential equation has infinitely many solutions.
C) The constant function f(t) = 1 is a solution to this differential equation.
D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true. <div style=padding-top: 35px>
E) All of these statements are true.
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Question
Consider the differential equation <strong>Consider the differential equation   =   (y + 3). Which of the following statements is/are true? (I) f(t) = -3 is a constant solution to this differential equation. (II) f(t) = 0 is a constant solution to this differential equation. (III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then   (1) = 3.</strong> A) I and III B) II only C) III only D) I only E) I, II, and III <div style=padding-top: 35px> = <strong>Consider the differential equation   =   (y + 3). Which of the following statements is/are true? (I) f(t) = -3 is a constant solution to this differential equation. (II) f(t) = 0 is a constant solution to this differential equation. (III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then   (1) = 3.</strong> A) I and III B) II only C) III only D) I only E) I, II, and III <div style=padding-top: 35px> (y + 3). Which of the following statements is/are true?
(I) f(t) = -3 is a constant solution to this differential equation.
(II) f(t) = 0 is a constant solution to this differential equation.
(III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then <strong>Consider the differential equation   =   (y + 3). Which of the following statements is/are true? (I) f(t) = -3 is a constant solution to this differential equation. (II) f(t) = 0 is a constant solution to this differential equation. (III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then   (1) = 3.</strong> A) I and III B) II only C) III only D) I only E) I, II, and III <div style=padding-top: 35px> (1) = 3.

A) I and III
B) II only
C) III only
D) I only
E) I, II, and III
Question
Solve the differential equation: y' = <strong>Solve the differential equation: y' =   sin t.</strong> A) y = cos(ln t) + C B) y =   + C C) y = ln(-sin t) + C D) y = -ln(cos t + C <div style=padding-top: 35px> sin t.

A) y = cos(ln t) + C
B) y = <strong>Solve the differential equation: y' =   sin t.</strong> A) y = cos(ln t) + C B) y =   + C C) y = ln(-sin t) + C D) y = -ln(cos t + C <div style=padding-top: 35px> + C
C) y = ln(-sin t) + C
D) y = -ln(cos t + C
Question
Given the differential equation: Given the differential equation:   , is this the solution  <div style=padding-top: 35px> , is this the solution Given the differential equation:   , is this the solution  <div style=padding-top: 35px>
Question
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <div style=padding-top: 35px>

A) y = - <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <div style=padding-top: 35px>
B) y = - <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <div style=padding-top: 35px>
C) y = ln <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <div style=padding-top: 35px>
D) y = <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <div style=padding-top: 35px> <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <div style=padding-top: 35px>
Question
Find f'(1) if f(t) is a solution to the initial value problem: Find f'(1) if f(t) is a solution to the initial value problem:   Enter just a real number (no approximations).<div style=padding-top: 35px> Enter just a real number (no approximations).
Question
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7   <div style=padding-top: 35px> ?

A) y = <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7   <div style=padding-top: 35px>
B) y = <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7   <div style=padding-top: 35px>
C) y = 7 <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7   <div style=padding-top: 35px>
D) y = 7 <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7   <div style=padding-top: 35px>
Question
Find a constant solution of Find a constant solution of   Enter just a reduced fraction of form   .<div style=padding-top: 35px> Enter just a reduced fraction of form Find a constant solution of   Enter just a reduced fraction of form   .<div style=padding-top: 35px> .
Question
Given the differential equation: Given the differential equation:   , is this the solution   <div style=padding-top: 35px> , is this the solution Given the differential equation:   , is this the solution   <div style=padding-top: 35px>
Question
Given the differential equation: Given the differential equation:   , is this the solution   <div style=padding-top: 35px> , is this the solution Given the differential equation:   , is this the solution   <div style=padding-top: 35px>
Question
Find f'(1) if f(t) is a solution to the initial value problem: Find f'(1) if f(t) is a solution to the initial value problem:   Enter just an integer.<div style=padding-top: 35px> Enter just an integer.
Question
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <div style=padding-top: 35px>

A) y = <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <div style=padding-top: 35px> + 3
B) y = <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <div style=padding-top: 35px> <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <div style=padding-top: 35px> + 3
C) y = - <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <div style=padding-top: 35px> <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <div style=padding-top: 35px> + 3x
D) none of these
Question
Given the differential equation: y' is this the solution Given the differential equation: y' is this the solution   <div style=padding-top: 35px>
Question
Find f'(0) if f(t) is a solution to the initial value problem: Find f'(0) if f(t) is a solution to the initial value problem:   Enter just an integer.<div style=padding-top: 35px> Enter just an integer.
Question
Given the differential equation: Given the differential equation:   is this the solution  <div style=padding-top: 35px> is this the solution Given the differential equation:   is this the solution  <div style=padding-top: 35px>
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Given the differential equation: Given the differential equation:   , is this the solution:   <div style=padding-top: 35px> , is this the solution: Given the differential equation:   , is this the solution:   <div style=padding-top: 35px>
Question
Find a constant solution of Find a constant solution of   Enter just an integer.<div style=padding-top: 35px> Enter just an integer.
Question
Given the differential equation: Given the differential equation:   , is this the solution   <div style=padding-top: 35px> , is this the solution Given the differential equation:   , is this the solution   <div style=padding-top: 35px>
Question
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y = -   C) y = ln 4t D) none of these <div style=padding-top: 35px> ?

A) y = <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y = -   C) y = ln 4t D) none of these <div style=padding-top: 35px>
B) y = - <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y = -   C) y = ln 4t D) none of these <div style=padding-top: 35px>
C) y = ln 4t
D) none of these
Question
Write a differential equation that expresses the following description of a rate: When ice cream is removed from the freezer, it warms up at a rate proportional to the difference between the temperature of the ice cream and the room temperature of 76°. (Use y for the temperature of the ice cream, t for the time, and k for an unknown constant.)

A) y' = 76 - ky
B) y' = k(76 - y)
C) y' = k(76 - t)
D) y' = 76t - ky
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 <div style=padding-top: 35px>

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 <div style=padding-top: 35px> - 2
B) y = 2 <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 <div style=padding-top: 35px> -1
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 <div style=padding-top: 35px> + 1
D) y = t + 1
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these <div style=padding-top: 35px>

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these <div style=padding-top: 35px> - 2t + 1
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these <div style=padding-top: 35px> - 2t
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these <div style=padding-top: 35px>
D) y = 0
E) none of these
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px>
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these <div style=padding-top: 35px> , y(0) = 0

A) y = -ln <strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these <div style=padding-top: 35px>
B) y = 0
C) y = <strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these <div style=padding-top: 35px>
D) y = <strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these <div style=padding-top: 35px>
E) none of these
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px>
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px>
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   <div style=padding-top: 35px>

A) y = 4 - <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   <div style=padding-top: 35px>
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   <div style=padding-top: 35px>
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   <div style=padding-top: 35px> + <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   <div style=padding-top: 35px>
D) y = 4 + <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   <div style=padding-top: 35px>
Question
Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.

A) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 <div style=padding-top: 35px> = ky; There is not enough information given to determine initial conditions.
B) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 <div style=padding-top: 35px> = ky; y(0) = 0
C) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 <div style=padding-top: 35px> = k(1 - y); There is not enough information given to determine initial conditions.
D) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 <div style=padding-top: 35px> = k(1 - y); y(0) = 0
E) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 <div style=padding-top: 35px> = (1 + y); y(0) =0
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution   <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution   <div style=padding-top: 35px>
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) t + 1 B)   C) t - 1 D)   <div style=padding-top: 35px>

A) t + 1
B) <strong>Solve the differential equation with the given initial condition.  </strong> A) t + 1 B)   C) t - 1 D)   <div style=padding-top: 35px>
C) t - 1
D) <strong>Solve the differential equation with the given initial condition.  </strong> A) t + 1 B)   C) t - 1 D)   <div style=padding-top: 35px>
Question
Find the constant solutions to the differential equation: Find the constant solutions to the differential equation:   . Enter just one integer or two separated by a comma (no label).<div style=padding-top: 35px> .
Enter just one integer or two separated by a comma (no label).
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t <div style=padding-top: 35px>

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t <div style=padding-top: 35px> - <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t <div style=padding-top: 35px>
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t <div style=padding-top: 35px>
C) y = 16 + <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t <div style=padding-top: 35px>
D) y = 64 - <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t <div style=padding-top: 35px> t
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution   <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution   <div style=padding-top: 35px>
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px>
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   <div style=padding-top: 35px>

A) y = ln <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   <div style=padding-top: 35px> + 1
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   <div style=padding-top: 35px> + 1
C) y = tan t + 1
D) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   <div style=padding-top: 35px> + <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   <div style=padding-top: 35px>
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =   <div style=padding-top: 35px>

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =   <div style=padding-top: 35px>
B) y = ± <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =   <div style=padding-top: 35px>
C) y = ± <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =   <div style=padding-top: 35px>
D) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =   <div style=padding-top: 35px>
Question
Given the differential equation: Given the differential equation:   is this the solution   <div style=padding-top: 35px> is this the solution Given the differential equation:   is this the solution   <div style=padding-top: 35px>
Question
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px> is this the solution Given the differential equation with the given initial condition:   is this the solution    <div style=padding-top: 35px>
Question
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +   <div style=padding-top: 35px>

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +   <div style=padding-top: 35px>
B) y = - <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +   <div style=padding-top: 35px>
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +   <div style=padding-top: 35px>
D) y = -1 + <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +   <div style=padding-top: 35px>
Question
Given the differential equation: Given the differential equation:   is this the solution   <div style=padding-top: 35px> is this the solution Given the differential equation:   is this the solution   <div style=padding-top: 35px>
Question
Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> - 4y = -2 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> ; y(0) = -1

A) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
General solution: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> - 2 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
B) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
General solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
Particular solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> - 2 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
C) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
General solution: y = -2t + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
Particular solution: y = -2t - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
D) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
General solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
Particular solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px> <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <div style=padding-top: 35px>
Solve the equation using an integrating factor.
Question
Solve the initial value problem using an integrating factor.
t <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px> + 3y = 5t; <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px> , t > 0

A) y = <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px> t - <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px>
B) y = <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px> + 1 - <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px>
C) y = 5 <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px> - 4t
D) y = 5 <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px> - 4 <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   <div style=padding-top: 35px>
Question
The annual sales y (in millions of dollars) of a company satisfy the differential equation <strong>The annual sales y (in millions of dollars) of a company satisfy the differential equation   . Which of the following is a verbal description of the rate of change of annual sales ?</strong> A) The annual sales are increasing at a rate proportional to the annual sales. B) The annual sales are increasing at $0.2 million ($200,000) per year. C) The annual sales are decreasing at a rate proportional to the annual sales. D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year. <div style=padding-top: 35px> . Which of the following is a verbal description of the rate of change of annual sales ?

A) The annual sales are increasing at a rate proportional to the annual sales.
B) The annual sales are increasing at $0.2 million ($200,000) per year.
C) The annual sales are decreasing at a rate proportional to the annual sales.
D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year.
Question
Combine the terms y and Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   <div style=padding-top: 35px> into the derivative of a product, then solve the equation. Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   <div style=padding-top: 35px> Is this the solution: Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   <div style=padding-top: 35px>
Question
Solve the problem.
An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?

A) A = 60,000 - 36,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years <div style=padding-top: 35px> = 10)017 years
B) A = 80,000 - 56,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years <div style=padding-top: 35px> = 8)352 years
C) A = 80,000 - 56,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years <div style=padding-top: 35px> = 7)134 years
D) A = 60,000 - 36,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years <div style=padding-top: 35px> = 8)352 years
Question
Combine the terms y and Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? <div style=padding-top: 35px> into the derivative of a product: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? <div style=padding-top: 35px> .
Is this derivative correct: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? <div style=padding-top: 35px> ?
Question
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px> + <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px> y = 4 <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px> , t > 0

A) y = 4x + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px>
B) y = 4x + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px>
C) y = 4 + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px>
D) y = 4 + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   <div style=padding-top: 35px>
Question
A cool object is to be heated to a maximum temperature M = M°C. At any time t, the rate at which the temperature rises is proportional to the difference between the actual temperature and the maximal temperature. If the object is originally 0°C, find and solve a differential equation describing this situation. Is this the solution: A cool object is to be heated to a maximum temperature M = M°C. At any time t, the rate at which the temperature rises is proportional to the difference between the actual temperature and the maximal temperature. If the object is originally 0°C, find and solve a differential equation describing this situation. Is this the solution:    <div style=padding-top: 35px>
Question
Solve the initial value problem using an integrating factor.
<strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t <div style=padding-top: 35px> + y = 3; y(0) = 0.

A) y = <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t <div style=padding-top: 35px> (3t - <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t <div style=padding-top: 35px> )
B) y = <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t <div style=padding-top: 35px>
C) y = 3 - 3 <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t <div style=padding-top: 35px>
D) y = 3t
Question
Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.

A) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> , k < 0
B) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> , k > 0
C) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> , k < 0; y(0) =0
D) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these <div style=padding-top: 35px> , k > 0; y(0) = 0
E) none of these
Question
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C   <div style=padding-top: 35px> , t > 0

A) y = t + 1/2 + C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C   <div style=padding-top: 35px>
B) y = t - C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C   <div style=padding-top: 35px>
C) y = Ct + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C   <div style=padding-top: 35px>
D) y = 1/2 + C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C   <div style=padding-top: 35px>
Question
Combine the terms y and Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   <div style=padding-top: 35px> into the derivative of a product, then solve the equation. Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   <div style=padding-top: 35px> Is this the solution: Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   <div style=padding-top: 35px>
Question
The annual sales y (in millions of dollars) of a company satisfy the differential equation <strong>The annual sales y (in millions of dollars) of a company satisfy the differential equation   . Which of the following is a verbal description of the rate of change of annual sales.</strong> A) The annual sales are decreasing at a rate proportional to the annual sales. B) The annual sales are increasing at $0.2 million ($200,000) per year to an upper limit of $5 million. C) The annual sales are increasing at a rate proportional to the difference between the annual sales and an upper limit of $10 million. D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year. <div style=padding-top: 35px> . Which of the following is a verbal description of the rate of change of annual sales.

A) The annual sales are decreasing at a rate proportional to the annual sales.
B) The annual sales are increasing at $0.2 million ($200,000) per year to an upper limit of $5 million.
C) The annual sales are increasing at a rate proportional to the difference between the annual sales and an upper limit of $10 million.
D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year.
Question
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   <div style=padding-top: 35px> , t > 0

A) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   <div style=padding-top: 35px> + C
B) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   <div style=padding-top: 35px>
C) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   <div style=padding-top: 35px> - 2
D) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   <div style=padding-top: 35px>
Question
Suppose the relationship between the price p, of a product and the weekly sales, s, of the product is given by the differential equation <strong>Suppose the relationship between the price p, of a product and the weekly sales, s, of the product is given by the differential equation   Then</strong> A) as sales increase, the price increases. B) as the price increases the rate of change of the price also increases. C) the rate of the decrease of the price is proportional to the sales. D) s = 0 is a constant solution to this differential equation. E) all of these <div style=padding-top: 35px> Then

A) as sales increase, the price increases.
B) as the price increases the rate of change of the price also increases.
C) the rate of the decrease of the price is proportional to the sales.
D) s = 0 is a constant solution to this differential equation.
E) all of these
Question
v
An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?

A) A = 60,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 <div style=padding-top: 35px> - 52,000= $31,041.84
B) A = 52,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 <div style=padding-top: 35px> - 44,000= $27,969.59
C) A = 48,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 <div style=padding-top: 35px> - 40,000= $26,433.47
D) A = 44,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 <div style=padding-top: 35px> - 36,000= $24,897.34
Question
Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> - y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> ; y(1) = -1, t > 0

A) integrating factor:\ <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
General solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
B) integrating factor: 2t
General solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
C) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
General solution: y = -4 + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
Particular solution: y = -4 + 3 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
D) integrating factor: 2t
General solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px> + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   <div style=padding-top: 35px>
Question
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px> , t > 0

A) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px> + C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px>
B) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px>
C) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px> + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px>
D) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C <div style=padding-top: 35px> + C
Question
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> , t > 0

A) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px>
B) y = - <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> + C
C) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px>
D) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px> <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <div style=padding-top: 35px>
Question
Combine the terms y and Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? <div style=padding-top: 35px> into the derivative of a product: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? <div style=padding-top: 35px> .
Is this derivative correct: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? <div style=padding-top: 35px> ?
Question
A jug of milk at 50° is placed outdoors at a temperature of 100°. If after 5 minutes the temperature of the milk is 60°, write the equation giving the temperature of the milk as a function of time. Enter your answer exactly as:
T = a A jug of milk at 50° is placed outdoors at a temperature of 100°. If after 5 minutes the temperature of the milk is 60°, write the equation giving the temperature of the milk as a function of time. Enter your answer exactly as: T = a   + c<div style=padding-top: 35px> + c
Question
A tank contains 2000 L of a solution consisting of 50 kg of salt dissolved in water. Pure water is pumped into the tank at the rate of 10L/s, and the mixture (kept uniform by stirring) is pumped out at the same rate. How long will it be until only 5 kg of salt remain in the tank?

A) approximately 689 seconds
B) approximately 703 seconds
C) approximately 460 seconds
D) approximately 276 seconds
Question
A fly population increases at a rate proportional to the amount present. After two years the population has doubled. After three years it is 20,000. Find the number of flies initially present. Enter just an integer.
Question
How much would you need to invest per month - in effect, continuously - in an investment account that pays an annual interest rate of 9%, compounded continuously, in order for the account to be worth $100,000 after 20 years? Enter just an integer representing dollars to the nearest dollar (no units)
Question
One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. Do these graphs represent: One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. Do these graphs represent:     <div style=padding-top: 35px> One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. Do these graphs represent:     <div style=padding-top: 35px>
Question
Consider the differential equation <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these <div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these <div style=padding-top: 35px>
Indicate whether the following statements are true or false.
Let <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these <div style=padding-top: 35px> = <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these <div style=padding-top: 35px> . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing.
(II) It has an inflection point.
(III) It is always concave down.

A) I only
B) I and II
C) II only
D) III only
E) none of these
Question
Suppose that $1000 is deposited in a savings account that pays 6% annual interest compounded continuously. At what rate (in dollars per year) is it earning interest after 5 years? Enter just an integer representing the amount to the nearest dollar (no units).
Question
Consider the differential equation <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these <div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these <div style=padding-top: 35px>
Indicate whether the following statements are true or false.
For what y value(s) does a solution of <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these <div style=padding-top: 35px> = <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these <div style=padding-top: 35px> - 3y + 2 have inflection points?

A) y = 2
B) y = 0
C) y = 2 and y = 1
D) y = <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these <div style=padding-top: 35px>
E) none of these
Question
Suppose the following is a graph of z = g(y). <strong>Suppose the following is a graph of z = g(y).   Which of the following can then be said about the solution y = f(t) to the initial value problem   ? (I) f(t) is an increasing function (II) f(t) is always positive (III) f(t) has an inflection point when y = 2.</strong> A) III only B) I, II, and III C) I only D) II only E) I and III <div style=padding-top: 35px> Which of the following can then be said about the solution y = f(t) to the initial value problem <strong>Suppose the following is a graph of z = g(y).   Which of the following can then be said about the solution y = f(t) to the initial value problem   ? (I) f(t) is an increasing function (II) f(t) is always positive (III) f(t) has an inflection point when y = 2.</strong> A) III only B) I, II, and III C) I only D) II only E) I and III <div style=padding-top: 35px> ?
(I) f(t) is an increasing function
(II) f(t) is always positive
(III) f(t) has an inflection point when y = 2.

A) III only
B) I, II, and III
C) I only
D) II only
E) I and III
Question
A nutritionist proposes the following model for weight loss on a program she is developing: <strong>A nutritionist proposes the following model for weight loss on a program she is developing:   + 0.006w =   C where w(t) is a person's weight (in pounds) after t days of consuming exactly C calories per day. A person weighing 180 pounds goes on this diet program consuming 2400 calories per day. Use the above model to predict how long will it take this person to lose 15 pounds.</strong> A) 41 days B) 37 days C) 39 days D) 35 days <div style=padding-top: 35px> + 0.006w = <strong>A nutritionist proposes the following model for weight loss on a program she is developing:   + 0.006w =   C where w(t) is a person's weight (in pounds) after t days of consuming exactly C calories per day. A person weighing 180 pounds goes on this diet program consuming 2400 calories per day. Use the above model to predict how long will it take this person to lose 15 pounds.</strong> A) 41 days B) 37 days C) 39 days D) 35 days <div style=padding-top: 35px> C where w(t) is a person's weight (in pounds) after t days of consuming exactly C calories per day. A person weighing 180 pounds goes on this diet program consuming 2400 calories per day. Use the above model to predict how long will it take this person to lose 15 pounds.

A) 41 days
B) 37 days
C) 39 days
D) 35 days
Question
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = -3, y = 1, and y = 5 are the constant solutions to   = g(y).<div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = -3, y = 1, and y = 5 are the constant solutions to   = g(y).<div style=padding-top: 35px>
Indicate whether the following statements are true or false.
y = -3, y = 1, and y = 5 are the constant solutions to Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = -3, y = 1, and y = 5 are the constant solutions to   = g(y).<div style=padding-top: 35px> = g(y).
Question
Consider the differential equation <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down. (II) It is a constant solution. (III) It is always decreasing.</strong> A) I only B) III only C) I and III D) II only E) none of these <div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down. (II) It is a constant solution. (III) It is always decreasing.</strong> A) I only B) III only C) I and III D) II only E) none of these <div style=padding-top: 35px>
Indicate whether the following statements are true or false.
Let <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down. (II) It is a constant solution. (III) It is always decreasing.</strong> A) I only B) III only C) I and III D) II only E) none of these <div style=padding-top: 35px> = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down.
(II) It is a constant solution.
(III) It is always decreasing.

A) I only
B) III only
C) I and III
D) II only
E) none of these
Question
One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph.   Do these graphs represent the situation?   <div style=padding-top: 35px> Do these graphs represent the situation? One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph.   Do these graphs represent the situation?   <div style=padding-top: 35px>
Question
There is a differential equation that is a mathematical model of the situation in which the time rate of change in the population of a certain organism is proportional to the product of the current population and the difference between the current population and the limiting factor of 100,000. Is this the equation There is a differential equation that is a mathematical model of the situation in which the time rate of change in the population of a certain organism is proportional to the product of the current population and the difference between the current population and the limiting factor of 100,000. Is this the equation   <div style=padding-top: 35px>
Question
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = 2 is the only constant solution of   = g(y) .<div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = 2 is the only constant solution of   = g(y) .<div style=padding-top: 35px>
Indicate whether the following statements are true or false.
y = 2 is the only constant solution of Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = 2 is the only constant solution of   = g(y) .<div style=padding-top: 35px> = g(y) .
Question
Suppose <strong>Suppose   = ky + b and the graphs of several solutions of the differential equation are as below:   Then (I) k is negative. (II) k is positive. (III) b is positive. (IV) b is negative.</strong> A) II and IV B) I and IV C) II and III D) I and III E) not enough information given <div style=padding-top: 35px> = ky + b and the graphs of several solutions of the differential equation are as below: <strong>Suppose   = ky + b and the graphs of several solutions of the differential equation are as below:   Then (I) k is negative. (II) k is positive. (III) b is positive. (IV) b is negative.</strong> A) II and IV B) I and IV C) II and III D) I and III E) not enough information given <div style=padding-top: 35px> Then (I) k is negative.
(II) k is positive.
(III) b is positive.
(IV) b is negative.

A) II and IV
B) I and IV
C) II and III
D) I and III
E) not enough information given
Question
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is greater than 6, then the corresponding solution will be an increasing function.<div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is greater than 6, then the corresponding solution will be an increasing function.<div style=padding-top: 35px>
Indicate whether the following statements are true or false.
If the initial value of y(0) is greater than 6, then the corresponding solution will be an increasing function.
Question
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 3, then the corresponding solution has an inflection point.<div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 3, then the corresponding solution has an inflection point.<div style=padding-top: 35px>
Indicate whether the following statements are true or false.
If the initial value of y(0) is 3, then the corresponding solution has an inflection point.
Question
Suppose the graph below gives a solution to the differential equation <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV <div style=padding-top: 35px> = g(P) where P is the price of a product and s is the weekly sales. <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV <div style=padding-top: 35px> Which of the following statements is/are true?
(I) g(M) = 0
(II) <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV <div style=padding-top: 35px>
(III) <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV <div style=padding-top: 35px>
(IV) g( <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV <div style=padding-top: 35px> ) > 0

A) I only
B) IV only
C) I, III, and IV
D) I and IV
E) I, II, and IV
Question
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 2, then the corresponding solution has an inflection point.<div style=padding-top: 35px> = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 2, then the corresponding solution has an inflection point.<div style=padding-top: 35px>
Indicate whether the following statements are true or false.
If the initial value of y(0) is 2, then the corresponding solution has an inflection point.
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Deck 10: Differential Equations
1
Consider the differential equation y' = y - <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true. . Which of the following statements is/are true?

A) The function f(t) = <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true. is a solution to this differential equation with initial condition <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true.
B) This differential equation has infinitely many solutions.
C) The constant function f(t) = 1 is a solution to this differential equation.
D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then <strong>Consider the differential equation y' = y -   . Which of the following statements is/are true?</strong> A) The function f(t) =   is a solution to this differential equation with initial condition   B) This differential equation has infinitely many solutions. C) The constant function f(t) = 1 is a solution to this differential equation. D) If f(t) is a solution to the differential equation satisfying the initial condition y(0) = 0, then   E) All of these statements are true.
E) All of these statements are true.
All of these statements are true.
2
Consider the differential equation <strong>Consider the differential equation   =   (y + 3). Which of the following statements is/are true? (I) f(t) = -3 is a constant solution to this differential equation. (II) f(t) = 0 is a constant solution to this differential equation. (III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then   (1) = 3.</strong> A) I and III B) II only C) III only D) I only E) I, II, and III = <strong>Consider the differential equation   =   (y + 3). Which of the following statements is/are true? (I) f(t) = -3 is a constant solution to this differential equation. (II) f(t) = 0 is a constant solution to this differential equation. (III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then   (1) = 3.</strong> A) I and III B) II only C) III only D) I only E) I, II, and III (y + 3). Which of the following statements is/are true?
(I) f(t) = -3 is a constant solution to this differential equation.
(II) f(t) = 0 is a constant solution to this differential equation.
(III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then <strong>Consider the differential equation   =   (y + 3). Which of the following statements is/are true? (I) f(t) = -3 is a constant solution to this differential equation. (II) f(t) = 0 is a constant solution to this differential equation. (III) If f(t) is a solution to the differential equation with initial conditions y(1) = 0, then   (1) = 3.</strong> A) I and III B) II only C) III only D) I only E) I, II, and III (1) = 3.

A) I and III
B) II only
C) III only
D) I only
E) I, II, and III
I and III
3
Solve the differential equation: y' = <strong>Solve the differential equation: y' =   sin t.</strong> A) y = cos(ln t) + C B) y =   + C C) y = ln(-sin t) + C D) y = -ln(cos t + C sin t.

A) y = cos(ln t) + C
B) y = <strong>Solve the differential equation: y' =   sin t.</strong> A) y = cos(ln t) + C B) y =   + C C) y = ln(-sin t) + C D) y = -ln(cos t + C + C
C) y = ln(-sin t) + C
D) y = -ln(cos t + C
y = -ln(cos t + C
4
Given the differential equation: Given the differential equation:   , is this the solution  , is this the solution Given the differential equation:   , is this the solution
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5
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =

A) y = - <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =
B) y = - <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =
C) y = ln <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =
D) y = <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =     <strong>Which of the following functions solves the differential equation:  </strong> A) y = -   B) y = -   C) y = ln   D) y =
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6
Find f'(1) if f(t) is a solution to the initial value problem: Find f'(1) if f(t) is a solution to the initial value problem:   Enter just a real number (no approximations). Enter just a real number (no approximations).
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7
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7   ?

A) y = <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7
B) y = <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7
C) y = 7 <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7
D) y = 7 <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y =   C) y = 7   D) y = 7
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8
Find a constant solution of Find a constant solution of   Enter just a reduced fraction of form   . Enter just a reduced fraction of form Find a constant solution of   Enter just a reduced fraction of form   . .
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9
Given the differential equation: Given the differential equation:   , is this the solution   , is this the solution Given the differential equation:   , is this the solution
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10
Given the differential equation: Given the differential equation:   , is this the solution   , is this the solution Given the differential equation:   , is this the solution
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11
Find f'(1) if f(t) is a solution to the initial value problem: Find f'(1) if f(t) is a solution to the initial value problem:   Enter just an integer. Enter just an integer.
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12
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these

A) y = <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these + 3
B) y = <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these + 3
C) y = - <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these <strong>Which of the following functions solves the differential equation:  </strong> A) y =   + 3 B) y =     + 3 C) y = -     + 3x D) none of these + 3x
D) none of these
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13
Given the differential equation: y' is this the solution Given the differential equation: y' is this the solution
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14
Find f'(0) if f(t) is a solution to the initial value problem: Find f'(0) if f(t) is a solution to the initial value problem:   Enter just an integer. Enter just an integer.
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15
Given the differential equation: Given the differential equation:   is this the solution  is this the solution Given the differential equation:   is this the solution
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16
Given the differential equation: Given the differential equation:   , is this the solution:   , is this the solution: Given the differential equation:   , is this the solution:
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17
Find a constant solution of Find a constant solution of   Enter just an integer. Enter just an integer.
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18
Given the differential equation: Given the differential equation:   , is this the solution   , is this the solution Given the differential equation:   , is this the solution
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19
Which of the following functions solves the differential equation: <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y = -   C) y = ln 4t D) none of these ?

A) y = <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y = -   C) y = ln 4t D) none of these
B) y = - <strong>Which of the following functions solves the differential equation:   ?</strong> A) y =   B) y = -   C) y = ln 4t D) none of these
C) y = ln 4t
D) none of these
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20
Write a differential equation that expresses the following description of a rate: When ice cream is removed from the freezer, it warms up at a rate proportional to the difference between the temperature of the ice cream and the room temperature of 76°. (Use y for the temperature of the ice cream, t for the time, and k for an unknown constant.)

A) y' = 76 - ky
B) y' = k(76 - y)
C) y' = k(76 - t)
D) y' = 76t - ky
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21
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 - 2
B) y = 2 <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 -1
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2 B) y = 2   -1 C) y =   + 1 D) y = t + 1 + 1
D) y = t + 1
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22
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these - 2t + 1
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these - 2t
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   - 2t + 1 B) y =   - 2t C) y =   D) y = 0 E) none of these
D) y = 0
E) none of these
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23
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    is this the solution Given the differential equation with the given initial condition:   is this the solution
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24
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these , y(0) = 0

A) y = -ln <strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these
B) y = 0
C) y = <strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these
D) y = <strong>Solve the differential equation with the given initial condition.   , y(0) = 0</strong> A) y = -ln   B) y = 0 C) y =   D) y =   E) none of these
E) none of these
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25
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    is this the solution Given the differential equation with the given initial condition:   is this the solution
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26
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    is this the solution Given the differential equation with the given initial condition:   is this the solution
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27
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +

A) y = 4 - <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +   + <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +
D) y = 4 + <strong>Solve the differential equation with the given initial condition.  </strong> A) y = 4 -   B) y =   C) y =   +   D) y = 4 +
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28
Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.

A) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 = ky; There is not enough information given to determine initial conditions.
B) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 = ky; y(0) = 0
C) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 = k(1 - y); There is not enough information given to determine initial conditions.
D) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 = k(1 - y); y(0) = 0
E) <strong>Let t represent the number of hours that a packing machine is operated and y(t) represent the probability that the machine breaks down at least once during the t hours of operation. It has been observed that the rate of increase of the probability of a breakdown is proportional to the probability of not having a breakdown. Find a differential equation describing this situation.</strong> A)   = ky; There is not enough information given to determine initial conditions. B)   = ky; y(0) = 0 C)   = k(1 - y); There is not enough information given to determine initial conditions. D)   = k(1 - y); y(0) = 0 E)   = (1 + y); y(0) =0 = (1 + y); y(0) =0
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29
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution   is this the solution Given the differential equation with the given initial condition:   is this the solution
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30
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) t + 1 B)   C) t - 1 D)

A) t + 1
B) <strong>Solve the differential equation with the given initial condition.  </strong> A) t + 1 B)   C) t - 1 D)
C) t - 1
D) <strong>Solve the differential equation with the given initial condition.  </strong> A) t + 1 B)   C) t - 1 D)
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31
Find the constant solutions to the differential equation: Find the constant solutions to the differential equation:   . Enter just one integer or two separated by a comma (no label). .
Enter just one integer or two separated by a comma (no label).
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32
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t - <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t
C) y = 16 + <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t
D) y = 64 - <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   -   B) y =   C) y = 16 +   D) y = 64 -   t t
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33
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution   is this the solution Given the differential equation with the given initial condition:   is this the solution
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34
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    is this the solution Given the differential equation with the given initial condition:   is this the solution
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35
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +

A) y = ln <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   + 1
B) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   + 1
C) y = tan t + 1
D) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +   + <strong>Solve the differential equation with the given initial condition.  </strong> A) y = ln   + 1 B) y =   + 1 C) y = tan t + 1 D) y =   +
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36
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =
B) y = ± <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =
C) y = ± <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =
D) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = ±   C) y = ±   D) y =
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37
Given the differential equation: Given the differential equation:   is this the solution   is this the solution Given the differential equation:   is this the solution
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38
Given the differential equation with the given initial condition: Given the differential equation with the given initial condition:   is this the solution    is this the solution Given the differential equation with the given initial condition:   is this the solution
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39
Solve the differential equation with the given initial condition.
<strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +

A) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +
B) y = - <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +
C) y = <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +
D) y = -1 + <strong>Solve the differential equation with the given initial condition.  </strong> A) y =   B) y = -   C) y =   D) y = -1 +
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40
Given the differential equation: Given the differential equation:   is this the solution   is this the solution Given the differential equation:   is this the solution
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41
Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. - 4y = -2 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. ; y(0) = -1

A) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
General solution: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. - 2 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
B) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
General solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
Particular solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. - 2 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
C) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
General solution: y = -2t + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
Particular solution: y = -2t - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
D) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
General solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
Particular solution: y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor. <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition.   - 4y = -2   ; y(0) = -1</strong> A) integrating factor:   General solution:   Particular solution: y = -   - 2   B) integrating factor:   General solution: y =   + C   Particular solution: y =   - 2   C) integrating factor:   General solution: y = -2t + C   Particular solution: y = -2t -   D) integrating factor:   General solution: y =     + C   Particular solution: y =     -     Solve the equation using an integrating factor.
Solve the equation using an integrating factor.
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42
Solve the initial value problem using an integrating factor.
t <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   + 3y = 5t; <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   , t > 0

A) y = <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   t - <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4
B) y = <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   + 1 - <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4
C) y = 5 <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   - 4t
D) y = 5 <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4   - 4 <strong>Solve the initial value problem using an integrating factor. t   + 3y = 5t;   , t > 0</strong> A) y =   t -   B) y =   + 1 -   C) y = 5   - 4t D) y = 5   - 4
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43
The annual sales y (in millions of dollars) of a company satisfy the differential equation <strong>The annual sales y (in millions of dollars) of a company satisfy the differential equation   . Which of the following is a verbal description of the rate of change of annual sales ?</strong> A) The annual sales are increasing at a rate proportional to the annual sales. B) The annual sales are increasing at $0.2 million ($200,000) per year. C) The annual sales are decreasing at a rate proportional to the annual sales. D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year. . Which of the following is a verbal description of the rate of change of annual sales ?

A) The annual sales are increasing at a rate proportional to the annual sales.
B) The annual sales are increasing at $0.2 million ($200,000) per year.
C) The annual sales are decreasing at a rate proportional to the annual sales.
D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year.
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44
Combine the terms y and Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   into the derivative of a product, then solve the equation. Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   Is this the solution: Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:
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45
Solve the problem.
An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?

A) A = 60,000 - 36,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years = 10)017 years
B) A = 80,000 - 56,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years = 8)352 years
C) A = 80,000 - 56,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years = 7)134 years
D) A = 60,000 - 36,000 <strong>Solve the problem. An initial deposit of $24,000 is made into an account that earns 5% compounded continuously. Money is then withdrawn at a constant rate of $4000 a year until the amount in the account is 0. Find the equation for the amount in the account at any time t. When is the amount 0?</strong> A) A = 60,000 - 36,000   = 10)017 years B) A = 80,000 - 56,000  = 8)352 years C) A = 80,000 - 56,000  = 7)134 years D) A = 60,000 - 36,000  = 8)352 years = 8)352 years
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46
Combine the terms y and Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? into the derivative of a product: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? .
Is this derivative correct: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? ?
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47
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   + <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   y = 4 <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C   , t > 0

A) y = 4x + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C
B) y = 4x + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C
C) y = 4 + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C
D) y = 4 + C <strong>Solve the equation using an integrating factor.   +   y = 4   , t > 0</strong> A) y = 4x + C   B) y = 4x + C   C) y = 4 + C   D) y = 4 + C
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48
A cool object is to be heated to a maximum temperature M = M°C. At any time t, the rate at which the temperature rises is proportional to the difference between the actual temperature and the maximal temperature. If the object is originally 0°C, find and solve a differential equation describing this situation. Is this the solution: A cool object is to be heated to a maximum temperature M = M°C. At any time t, the rate at which the temperature rises is proportional to the difference between the actual temperature and the maximal temperature. If the object is originally 0°C, find and solve a differential equation describing this situation. Is this the solution:
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49
Solve the initial value problem using an integrating factor.
<strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t + y = 3; y(0) = 0.

A) y = <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t (3t - <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t )
B) y = <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t
C) y = 3 - 3 <strong>Solve the initial value problem using an integrating factor.   + y = 3; y(0) = 0.</strong> A) y =   (3t -   ) B) y =   C) y = 3 - 3   D) y = 3t
D) y = 3t
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50
Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.

A) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these , k < 0
B) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these , k > 0
C) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these , k < 0; y(0) =0
D) <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these = <strong>Suppose water is seeping from an underground storage facility at a rate that is proportional to the square amount of water present. If f(t) = y is the amount of water present at time t, find a differential equation describing the situation.</strong> A)   =   , k < 0 B)   =   , k > 0 C)   =   , k < 0; y(0) =0 D)   =   , k > 0; y(0) = 0 E) none of these , k > 0; y(0) = 0
E) none of these
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51
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C   , t > 0

A) y = t + 1/2 + C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C
B) y = t - C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C
C) y = Ct + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C
D) y = 1/2 + C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = t + 1/2 + C   B) y = t - C   C) y = Ct +   D) y = 1/2 + C
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52
Combine the terms y and Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   into the derivative of a product, then solve the equation. Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:   Is this the solution: Combine the terms y and   into the derivative of a product, then solve the equation.   Is this the solution:
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53
The annual sales y (in millions of dollars) of a company satisfy the differential equation <strong>The annual sales y (in millions of dollars) of a company satisfy the differential equation   . Which of the following is a verbal description of the rate of change of annual sales.</strong> A) The annual sales are decreasing at a rate proportional to the annual sales. B) The annual sales are increasing at $0.2 million ($200,000) per year to an upper limit of $5 million. C) The annual sales are increasing at a rate proportional to the difference between the annual sales and an upper limit of $10 million. D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year. . Which of the following is a verbal description of the rate of change of annual sales.

A) The annual sales are decreasing at a rate proportional to the annual sales.
B) The annual sales are increasing at $0.2 million ($200,000) per year to an upper limit of $5 million.
C) The annual sales are increasing at a rate proportional to the difference between the annual sales and an upper limit of $10 million.
D) The annual sales are increasing at a rate proportional to $0.2 million ($200,000) per year.
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54
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   , t > 0

A) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   + C
B) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C
C) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C   - 2
D) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   + C B) y = C   C) y = C   - 2 D) y = C
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55
Suppose the relationship between the price p, of a product and the weekly sales, s, of the product is given by the differential equation <strong>Suppose the relationship between the price p, of a product and the weekly sales, s, of the product is given by the differential equation   Then</strong> A) as sales increase, the price increases. B) as the price increases the rate of change of the price also increases. C) the rate of the decrease of the price is proportional to the sales. D) s = 0 is a constant solution to this differential equation. E) all of these Then

A) as sales increase, the price increases.
B) as the price increases the rate of change of the price also increases.
C) the rate of the decrease of the price is proportional to the sales.
D) s = 0 is a constant solution to this differential equation.
E) all of these
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56
v
An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?

A) A = 60,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 - 52,000= $31,041.84
B) A = 52,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 - 44,000= $27,969.59
C) A = 48,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 - 40,000= $26,433.47
D) A = 44,000 <strong>v An initial deposit of $8,000 is made into an account earning 6.5% compounded continuously. Thereafter, money is deposited into the account at a constant rate of $2600 per year. Find the amount in this account at any time t. How much is in this account after 5 years?</strong> A) A = 60,000   - 52,000= $31,041.84 B) A = 52,000   - 44,000= $27,969.59 C) A = 48,000   - 40,000= $26,433.47 D) A = 44,000   - 36,000= $24,897.34 - 36,000= $24,897.34
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57
Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   - y = <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   ; y(1) = -1, t > 0

A) integrating factor:\ <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
General solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
B) integrating factor: 2t
General solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
C) integrating factor: <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
General solution: y = -4 + C <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
Particular solution: y = -4 + 3 <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
D) integrating factor: 2t
General solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
Particular solution: y = - <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   + <strong>Find the integrating factor, the general solution, and the particular solution satisfying the initial condition. 2t   - y =   ; y(1) = -1, t > 0</strong> A) integrating factor:\   General solution: y = -   + C   Particular solution: y = -   +   B) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +   C) integrating factor:   General solution: y = -4 + C   Particular solution: y = -4 + 3   D) integrating factor: 2t General solution: y = -   +   Particular solution: y = -   +
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58
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C , t > 0

A) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C + C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C
B) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C
C) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C
D) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y =   + C   B) y = C   C) y =   +   D) y =   + C + C
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59
Solve the equation using an integrating factor.
<strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     , t > 0

A) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +
B) y = - <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     + C
C) y = <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +
D) y = C <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     + <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +     <strong>Solve the equation using an integrating factor.   , t > 0</strong> A) y = C   +   B) y = -   +   + C C) y =   +   D) y = C   +
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60
Combine the terms y and Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? into the derivative of a product: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? .
Is this derivative correct: Combine the terms y and   into the derivative of a product:   . Is this derivative correct:   ? ?
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61
A jug of milk at 50° is placed outdoors at a temperature of 100°. If after 5 minutes the temperature of the milk is 60°, write the equation giving the temperature of the milk as a function of time. Enter your answer exactly as:
T = a A jug of milk at 50° is placed outdoors at a temperature of 100°. If after 5 minutes the temperature of the milk is 60°, write the equation giving the temperature of the milk as a function of time. Enter your answer exactly as: T = a   + c + c
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62
A tank contains 2000 L of a solution consisting of 50 kg of salt dissolved in water. Pure water is pumped into the tank at the rate of 10L/s, and the mixture (kept uniform by stirring) is pumped out at the same rate. How long will it be until only 5 kg of salt remain in the tank?

A) approximately 689 seconds
B) approximately 703 seconds
C) approximately 460 seconds
D) approximately 276 seconds
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63
A fly population increases at a rate proportional to the amount present. After two years the population has doubled. After three years it is 20,000. Find the number of flies initially present. Enter just an integer.
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64
How much would you need to invest per month - in effect, continuously - in an investment account that pays an annual interest rate of 9%, compounded continuously, in order for the account to be worth $100,000 after 20 years? Enter just an integer representing dollars to the nearest dollar (no units)
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65
One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. Do these graphs represent: One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. Do these graphs represent:     One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. Do these graphs represent:
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66
Consider the differential equation <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these = g(y) where g(y) is the function whose graph is shown below: <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these
Indicate whether the following statements are true or false.
Let <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these = <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   =   . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing. (II) It has an inflection point. (III) It is always concave down.</strong> A) I only B) I and II C) II only D) III only E) none of these . Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = -2? (I) It is always increasing.
(II) It has an inflection point.
(III) It is always concave down.

A) I only
B) I and II
C) II only
D) III only
E) none of these
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67
Suppose that $1000 is deposited in a savings account that pays 6% annual interest compounded continuously. At what rate (in dollars per year) is it earning interest after 5 years? Enter just an integer representing the amount to the nearest dollar (no units).
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68
Consider the differential equation <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these = g(y) where g(y) is the function whose graph is shown below: <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these
Indicate whether the following statements are true or false.
For what y value(s) does a solution of <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these = <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these - 3y + 2 have inflection points?

A) y = 2
B) y = 0
C) y = 2 and y = 1
D) y = <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. For what y value(s) does a solution of   =   - 3y + 2 have inflection points?</strong> A) y = 2 B) y = 0 C) y = 2 and y = 1 D) y =   E) none of these
E) none of these
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69
Suppose the following is a graph of z = g(y). <strong>Suppose the following is a graph of z = g(y).   Which of the following can then be said about the solution y = f(t) to the initial value problem   ? (I) f(t) is an increasing function (II) f(t) is always positive (III) f(t) has an inflection point when y = 2.</strong> A) III only B) I, II, and III C) I only D) II only E) I and III Which of the following can then be said about the solution y = f(t) to the initial value problem <strong>Suppose the following is a graph of z = g(y).   Which of the following can then be said about the solution y = f(t) to the initial value problem   ? (I) f(t) is an increasing function (II) f(t) is always positive (III) f(t) has an inflection point when y = 2.</strong> A) III only B) I, II, and III C) I only D) II only E) I and III ?
(I) f(t) is an increasing function
(II) f(t) is always positive
(III) f(t) has an inflection point when y = 2.

A) III only
B) I, II, and III
C) I only
D) II only
E) I and III
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70
A nutritionist proposes the following model for weight loss on a program she is developing: <strong>A nutritionist proposes the following model for weight loss on a program she is developing:   + 0.006w =   C where w(t) is a person's weight (in pounds) after t days of consuming exactly C calories per day. A person weighing 180 pounds goes on this diet program consuming 2400 calories per day. Use the above model to predict how long will it take this person to lose 15 pounds.</strong> A) 41 days B) 37 days C) 39 days D) 35 days + 0.006w = <strong>A nutritionist proposes the following model for weight loss on a program she is developing:   + 0.006w =   C where w(t) is a person's weight (in pounds) after t days of consuming exactly C calories per day. A person weighing 180 pounds goes on this diet program consuming 2400 calories per day. Use the above model to predict how long will it take this person to lose 15 pounds.</strong> A) 41 days B) 37 days C) 39 days D) 35 days C where w(t) is a person's weight (in pounds) after t days of consuming exactly C calories per day. A person weighing 180 pounds goes on this diet program consuming 2400 calories per day. Use the above model to predict how long will it take this person to lose 15 pounds.

A) 41 days
B) 37 days
C) 39 days
D) 35 days
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71
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = -3, y = 1, and y = 5 are the constant solutions to   = g(y). = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = -3, y = 1, and y = 5 are the constant solutions to   = g(y).
Indicate whether the following statements are true or false.
y = -3, y = 1, and y = 5 are the constant solutions to Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = -3, y = 1, and y = 5 are the constant solutions to   = g(y). = g(y).
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72
Consider the differential equation <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down. (II) It is a constant solution. (III) It is always decreasing.</strong> A) I only B) III only C) I and III D) II only E) none of these = g(y) where g(y) is the function whose graph is shown below: <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down. (II) It is a constant solution. (III) It is always decreasing.</strong> A) I only B) III only C) I and III D) II only E) none of these
Indicate whether the following statements are true or false.
Let <strong>Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. Let   = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down. (II) It is a constant solution. (III) It is always decreasing.</strong> A) I only B) III only C) I and III D) II only E) none of these = 2 - y. Which of the following properties hold for the solution y = f(t) determined by the initial condition y(0) = 1? (I) It is always concave down.
(II) It is a constant solution.
(III) It is always decreasing.

A) I only
B) III only
C) I and III
D) II only
E) none of these
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73
One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph. One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph.   Do these graphs represent the situation?   Do these graphs represent the situation? One or more initial conditions are given for the differential equation. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solution. Include a yz-graph as well as a ty-graph.   Do these graphs represent the situation?
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74
There is a differential equation that is a mathematical model of the situation in which the time rate of change in the population of a certain organism is proportional to the product of the current population and the difference between the current population and the limiting factor of 100,000. Is this the equation There is a differential equation that is a mathematical model of the situation in which the time rate of change in the population of a certain organism is proportional to the product of the current population and the difference between the current population and the limiting factor of 100,000. Is this the equation
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75
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = 2 is the only constant solution of   = g(y) . = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = 2 is the only constant solution of   = g(y) .
Indicate whether the following statements are true or false.
y = 2 is the only constant solution of Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. y = 2 is the only constant solution of   = g(y) . = g(y) .
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76
Suppose <strong>Suppose   = ky + b and the graphs of several solutions of the differential equation are as below:   Then (I) k is negative. (II) k is positive. (III) b is positive. (IV) b is negative.</strong> A) II and IV B) I and IV C) II and III D) I and III E) not enough information given = ky + b and the graphs of several solutions of the differential equation are as below: <strong>Suppose   = ky + b and the graphs of several solutions of the differential equation are as below:   Then (I) k is negative. (II) k is positive. (III) b is positive. (IV) b is negative.</strong> A) II and IV B) I and IV C) II and III D) I and III E) not enough information given Then (I) k is negative.
(II) k is positive.
(III) b is positive.
(IV) b is negative.

A) II and IV
B) I and IV
C) II and III
D) I and III
E) not enough information given
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77
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is greater than 6, then the corresponding solution will be an increasing function. = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is greater than 6, then the corresponding solution will be an increasing function.
Indicate whether the following statements are true or false.
If the initial value of y(0) is greater than 6, then the corresponding solution will be an increasing function.
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78
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 3, then the corresponding solution has an inflection point. = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 3, then the corresponding solution has an inflection point.
Indicate whether the following statements are true or false.
If the initial value of y(0) is 3, then the corresponding solution has an inflection point.
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79
Suppose the graph below gives a solution to the differential equation <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV = g(P) where P is the price of a product and s is the weekly sales. <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV Which of the following statements is/are true?
(I) g(M) = 0
(II) <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV
(III) <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV
(IV) g( <strong>Suppose the graph below gives a solution to the differential equation   = g(P) where P is the price of a product and s is the weekly sales.   Which of the following statements is/are true? (I) g(M) = 0 (II)   (III)   (IV) g(   ) > 0</strong> A) I only B) IV only C) I, III, and IV D) I and IV E) I, II, and IV ) > 0

A) I only
B) IV only
C) I, III, and IV
D) I and IV
E) I, II, and IV
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80
Consider the differential equation Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 2, then the corresponding solution has an inflection point. = g(y) where g(y) is the function whose graph is shown below: Consider the differential equation   = g(y) where g(y) is the function whose graph is shown below:   Indicate whether the following statements are true or false. If the initial value of y(0) is 2, then the corresponding solution has an inflection point.
Indicate whether the following statements are true or false.
If the initial value of y(0) is 2, then the corresponding solution has an inflection point.
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