Question 1

Solve the differential equation $y ^ {tt} - 6 y ^ { t } + 9 y = 8 x ^ { 2 } e ^ { 3 x } \text { by the method of variation of }$ parameters.

Accepted Answer

A)  $y = \left( C _ { 1 } + C _ { 2 } x + \frac { 2 } { 3 } x ^ { 4 } \right) e ^ { 3 x }$ 
B)  $y = \left( C _ { 1 } + C _ { 2 } x + \frac { 8 } { 15 } x ^ { 4 } \right) e ^ { 3 x }$ 
C) $y = \left( C _ { 1 } + C _ { 2 } x + \frac { 4 } { 9 } x ^ { 2 } \right) e ^ { 3 x }$ 
D) $y = \left( C _ { 1 } + C _ { 2 } x - \frac { 8 } { 17 } x ^ { 2 } \right) e ^ { 3 x }$ 
E) $y = \left( C _ { 1 } + C _ { 2 } x - \frac { 4 } { 5 } x ^ { 2 } \right) e ^ { 3 x }$ 
A)  $y = \left( C _ { 1 } + C _ { 2 } x + \frac { 2 } { 3 } x ^ { 4 } \right) e ^ { 3 x }$ 
B)  $y = \left( C _ { 1 } + C _ { 2 } x + \frac { 8 } { 15 } x ^ { 4 } \right) e ^ { 3 x }$ 
C) $y = \left( C _ { 1 } + C _ { 2 } x + \frac { 4 } { 9 } x ^ { 2 } \right) e ^ { 3 x }$ 
D) $y = \left( C _ { 1 } + C _ { 2 } x - \frac { 8 } { 17 } x ^ { 2 } \right) e ^ { 3 x }$ 
E) $y = \left( C _ { 1 } + C _ { 2 } x - \frac { 4 } { 5 } x ^ { 2 } \right) e ^ { 3 x }$

Question 2

Taylor's Theorem to find the first four terms of the series solution of $y ^ { tt } + x ^ { 2 } y ^ { t } - ( \cos x ) y = 0$ given the initial conditions $y ( 0 ) = 7$, and $y ^ {t } ( 0 ) = 4$

Accepted Answer

A)  $y = 7 + \frac { 4 } { 1 ! } x + \frac { 7 } { 2 ! } x ^ { 2 } - \frac { 8 } { 3 ! } x ^ { 3 } + \cdots$ 
B) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 7 } { 2 ! } x ^ { 2 } + \frac { 4 } { 3 ! } x ^ { 3 } + \cdots$ 
C) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 14 } { 2 ! } x ^ { 2 } + \frac { 8 } { 3 ! } x ^ { 3 } + \cdots$ 
D) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 4 } { 2 ! } x ^ { 2 } - \frac { 14 } { 3 ! } x ^ { 3 } + \cdots$ 
E) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 4 } { 2 ! } x ^ { 2 } + \frac { 7 } { 3 ! } x ^ { 3 } + \cdots$ 
A)  $y = 7 + \frac { 4 } { 1 ! } x + \frac { 7 } { 2 ! } x ^ { 2 } - \frac { 8 } { 3 ! } x ^ { 3 } + \cdots$ 
B) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 7 } { 2 ! } x ^ { 2 } + \frac { 4 } { 3 ! } x ^ { 3 } + \cdots$ 
C) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 14 } { 2 ! } x ^ { 2 } + \frac { 8 } { 3 ! } x ^ { 3 } + \cdots$ 
D) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 4 } { 2 ! } x ^ { 2 } - \frac { 14 } { 3 ! } x ^ { 3 } + \cdots$ 
E) $y = 7 + \frac { 4 } { 1 ! } x + \frac { 4 } { 2 ! } x ^ { 2 } + \frac { 7 } { 3 ! } x ^ { 3 } + \cdots$

Question 3

Solve the differential equation $y ^ {tt } + 5 y ^ { t } - 6 y = 7 \cos 2 x \text {, where }$ $y ( 0 ) = - 1$ and $y ^ {t } ( 0 ) = 6$ by the method of undetermined coefficients.

Accepted Answer

A)  $y = \frac { 1 } { 5 } e ^ { x } - \frac { 81 } { 70 } e ^ { - 6 x } - \frac { 7 } { 25 } \cos 2 x + \frac { 7 } { 20 } \sin 2 x$ 
B)  $y = \frac { 7 } { 33 } e ^ { x } - \frac { 23 } { 20 } e ^ { - 6 x } + \frac { 7 } { 20 } \cos 2 x - \frac { 7 } { 20 } \sin 2 x$ 
C) $y = - \frac { 53 } { 35 } e ^ { - x } + \frac { 121 } { 140 } e ^ { 6 x } - \frac { 7 } { 20 } \cos 2 x - \frac { 7 } { 20 } \sin 2 x$ 
D) $y = \frac { 1 } { 5 } e ^ { x } - \frac { 23 } { 20 } e ^ { - 6 x } + \frac { 7 } { 20 } \cos 2 x - \frac { 7 } { 20 } \sin 2 x$ 
E) $y = \frac { 1 } { 5 } e ^ { x } - \frac { 17 } { 20 } e ^ { - 6 x } - \frac { 7 } { 20 } \cos 2 x + \frac { 7 } { 20 } \sin 2 x$ 
A)  $y = \frac { 1 } { 5 } e ^ { x } - \frac { 81 } { 70 } e ^ { - 6 x } - \frac { 7 } { 25 } \cos 2 x + \frac { 7 } { 20 } \sin 2 x$ 
B)  $y = \frac { 7 } { 33 } e ^ { x } - \frac { 23 } { 20 } e ^ { - 6 x } + \frac { 7 } { 20 } \cos 2 x - \frac { 7 } { 20 } \sin 2 x$ 
C) $y = - \frac { 53 } { 35 } e ^ { - x } + \frac { 121 } { 140 } e ^ { 6 x } - \frac { 7 } { 20 } \cos 2 x - \frac { 7 } { 20 } \sin 2 x$ 
D) $y = \frac { 1 } { 5 } e ^ { x } - \frac { 23 } { 20 } e ^ { - 6 x } + \frac { 7 } { 20 } \cos 2 x - \frac { 7 } { 20 } \sin 2 x$ 
E) $y = \frac { 1 } { 5 } e ^ { x } - \frac { 17 } { 20 } e ^ { - 6 x } - \frac { 7 } { 20 } \cos 2 x + \frac { 7 } { 20 } \sin 2 x$

Question 4

Find the general solution of the differential equation

Accepted Answer

A)  $4 x ^ { 2 } - 3 x y = C$ 
B)  $4 x ^ { 2 } + 4 y ^ { 2 } = C$ 
C)  $4 x ^ { 2 } - 3 x y + 4 y ^ { 2 } = C$ 
D)  $8 x ^ { 2 } - 3 x y + 8 y ^ { 2 } = C$ 
E)  $8 x ^ { 2 } + 8 y ^ { 2 } = C$ 
A)  $4 x ^ { 2 } - 3 x y = C$ 
B)  $4 x ^ { 2 } + 4 y ^ { 2 } = C$ 
C)  $4 x ^ { 2 } - 3 x y + 4 y ^ { 2 } = C$ 
D)  $8 x ^ { 2 } - 3 x y + 8 y ^ { 2 } = C$ 
E)  $8 x ^ { 2 } + 8 y ^ { 2 } = C$

Question 5

Use Taylor's Theorem to find the first five terms of the series solution of $y ^ {t} + ( 2 x - 1 ) y = 0$ given the initial condition $y ( 0 ) = 2$

Accepted Answer

A) 
$$y = 2 - x + \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 3 } + \frac { x ^ { 4 } } { 4 } + \cdots$$ 
B) 
$$y = 2 - \frac { 2 } { 1 ! } x + \frac { 6 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \frac { 6 } { 4 ! } x ^ { 4 } + \cdots$$ 
C) 
$$y = \frac { 2 } { 1 ! } x - \frac { 2 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \frac { 2 } { 4 ! } x ^ { 4 } - \frac { 2 } { 5 ! } x ^ { 5 } + \cdots$$ 
D) 
$$y = 2 + \frac { 2 } { 1 ! } x - \frac { 2 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \frac { 2 } { 4 ! } x ^ { 4 } + \cdots$$ 
E) 
$$y = 2 + x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 3 } + \frac { x ^ { 4 } } { 4 } + \cdots$$ 
A) 
$$y = 2 - x + \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 3 } + \frac { x ^ { 4 } } { 4 } + \cdots$$ 
B) 
$$y = 2 - \frac { 2 } { 1 ! } x + \frac { 6 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \frac { 6 } { 4 ! } x ^ { 4 } + \cdots$$ 
C) 
$$y = \frac { 2 } { 1 ! } x - \frac { 2 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \frac { 2 } { 4 ! } x ^ { 4 } - \frac { 2 } { 5 ! } x ^ { 5 } + \cdots$$ 
D) 
$$y = 2 + \frac { 2 } { 1 ! } x - \frac { 2 } { 2 ! } x ^ { 2 } - \frac { 10 } { 3 ! } x ^ { 3 } + \frac { 2 } { 4 ! } x ^ { 4 } + \cdots$$ 
E) 
$$y = 2 + x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 3 } + \frac { x ^ { 4 } } { 4 } + \cdots$$

Question 6

$\text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. }$

Accepted Answer

A)  $y = C _ { 1 } - C _ { 2 } e ^ { - 4 x }$ 
B)  $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x }$ 
C)  $y = C _ { 1 } + 2 C _ { 2 } e ^ { 2 x }$ 
D)  $y = C _ { 1 } - 2 C _ { 2 } e ^ { 2 x }$ 
E)  $y = C _ { 1 } + 4 C _ { 2 } e ^ { - 4 x }$ 
A)  $y = C _ { 1 } - C _ { 2 } e ^ { - 4 x }$ 
B)  $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x }$ 
C)  $y = C _ { 1 } + 2 C _ { 2 } e ^ { 2 x }$ 
D)  $y = C _ { 1 } - 2 C _ { 2 } e ^ { 2 x }$ 
E)  $y = C _ { 1 } + 4 C _ { 2 } e ^ { - 4 x }$

Question 7

Find the particular solution of the differential equation $\frac { w } { g } y ^ { tt } ( t ) + b y ^ { t } ( t ) + k y ( t ) = \frac { w } { g } F ( t )$ for the oscillating motion of an object on the end of a spring. In the equation, $y$ is the displacement from equilibrium (positive direction is downward) measured in feet, and $t$ is the time in seconds (see figure). The constant $w = 8$ is the weight of the object, $g = 128$ is the acceleration due to gravity, $b = 1$ is the magnitude of the resistance to the motion, $k = 4$ is the spring constant from Hooke's Law, $F ( t ) = 4 \sin 8 t$ is the acceleration imposed on the system, $y ( 0 ) = \frac { 1 } { 7 }$ and $y ^ {t } ( 0 ) = - 9$

Accepted Answer

A)  $y = \left( \frac { 39 } { 224 } + \frac { 213 } { 28 } t \right) e ^ { - 8 t } - \frac { 1 } { 36 } \cos 8 t$ 
B) $y = \left( \frac { 39 } { 224 } - \frac { 213 } { 28 } t \right) e ^ { - 8 t } - \frac { 1 } { 32 } \cos 8 t$ 
C)  $y = \left( \frac { 43 } { 252 } - \frac { 272 } { 35 } t \right) e ^ { - 8 t } - \frac { 1 } { 34 } \cos 8 t$ 
D)  $y = \left( \frac { 43 } { 252 } + \frac { 272 } { 35 } t \right) e ^ { - 8 t } - \frac { 1 } { 30 } \cos 8 t$ 
E)  $y = \left( \frac { 39 } { 224 } - \frac { 272 } { 35 } t \right) e ^ { - 8 t } - \frac { 1 } { 40 } \cos 8 t$ 
A)  $y = \left( \frac { 39 } { 224 } + \frac { 213 } { 28 } t \right) e ^ { - 8 t } - \frac { 1 } { 36 } \cos 8 t$ 
B) $y = \left( \frac { 39 } { 224 } - \frac { 213 } { 28 } t \right) e ^ { - 8 t } - \frac { 1 } { 32 } \cos 8 t$ 
C)  $y = \left( \frac { 43 } { 252 } - \frac { 272 } { 35 } t \right) e ^ { - 8 t } - \frac { 1 } { 34 } \cos 8 t$ 
D)  $y = \left( \frac { 43 } { 252 } + \frac { 272 } { 35 } t \right) e ^ { - 8 t } - \frac { 1 } { 30 } \cos 8 t$ 
E)  $y = \left( \frac { 39 } { 224 } - \frac { 272 } { 35 } t \right) e ^ { - 8 t } - \frac { 1 } { 40 } \cos 8 t$

Question 8

Find the particular solution of the differential equation $\frac { w } { g } y ^ { tt } ( t ) + b y ^ { t } ( t ) + k y ( t ) = \frac { w } { g } F ( t )$ for the oscillating motion of an object on the end of a spring. In the equation, $y$ is the displacement from equilibrium (positive direction is downward) measured in feet, and $t$ is the time in seconds (see figure). The constant $w = 4$ is the weight of the object, $g = 32$ is the acceleration due to gravity, $b = \frac { 1 } { 2 }$ is the magnitude of the resistance to the motion, $k = \frac { 25 } { 2 }$ is the spring constant from Hooke's Law, $F ( t ) = 4 \sin 8 t$ is the acceleration imposed on the system, $y ( 0 ) = \frac { 1 } { 5 }$ and $y ^ { t } ( 0 ) = - 6$

Accepted Answer

A)  $y = \frac { 37 } { 145 } e ^ { - 2 t } \cos ( 4 \sqrt { 6 } t ) + \frac { 217 \sqrt { 6 } } { 870 } e ^ { - 2 t } \sin ( 4 \sqrt { 6 } t ) + \frac { 9 } { 145 } \sin 8 t - \frac { 8 } { 145 } \cos 8 t$ 
B)  $y = \frac { 1 } { 5 } e ^ { 2 t } \sin ( 3 \sqrt { 6 } t ) + \frac { 7 \sqrt { 6 } } { 30 } e ^ { 2 t } \cos ( 3 \sqrt { 6 } t ) + \frac { 7 } { 145 } \sin 8 t + \frac { 8 } { 145 } \cos 8 t$ 
C)  $y = \frac { 37 } { 145 } e ^ { - 2 t } \sin ( 2 \sqrt { 5 } t ) + \frac { 7 \sqrt { 3 } } { 30 } e ^ { - 2 t } \cos ( 2 \sqrt { 5 } t ) - \frac { 9 } { 145 } \sin 8 t - \frac { 7 } { 145 } \cos 8 t$ 
D)  $y = \frac { 1 } { 5 } e ^ { - 2 t } \cos ( 8 \sqrt { 2 } t ) - \frac { 217 \sqrt { 6 } } { 870 } e ^ { - 2 t } \sin ( 8 \sqrt { 2 } t ) + \frac { 8 } { 145 } \sin 8 t - \frac { 9 } { 145 } \cos 8 t$ 
E) $y = \frac { 1 } { 5 } e ^ { 2 t } \cos ( 7 \sqrt { 3 } t ) - \frac { 7 \sqrt { 3 } } { 30 } e ^ { 2 t } \sin ( 7 \sqrt { 3 } t ) + \frac { 9 } { 145 } \sin 8 t - \frac { 7 } { 145 } \cos 8 t$ 
A)  $y = \frac { 37 } { 145 } e ^ { - 2 t } \cos ( 4 \sqrt { 6 } t ) + \frac { 217 \sqrt { 6 } } { 870 } e ^ { - 2 t } \sin ( 4 \sqrt { 6 } t ) + \frac { 9 } { 145 } \sin 8 t - \frac { 8 } { 145 } \cos 8 t$ 
B)  $y = \frac { 1 } { 5 } e ^ { 2 t } \sin ( 3 \sqrt { 6 } t ) + \frac { 7 \sqrt { 6 } } { 30 } e ^ { 2 t } \cos ( 3 \sqrt { 6 } t ) + \frac { 7 } { 145 } \sin 8 t + \frac { 8 } { 145 } \cos 8 t$ 
C)  $y = \frac { 37 } { 145 } e ^ { - 2 t } \sin ( 2 \sqrt { 5 } t ) + \frac { 7 \sqrt { 3 } } { 30 } e ^ { - 2 t } \cos ( 2 \sqrt { 5 } t ) - \frac { 9 } { 145 } \sin 8 t - \frac { 7 } { 145 } \cos 8 t$ 
D)  $y = \frac { 1 } { 5 } e ^ { - 2 t } \cos ( 8 \sqrt { 2 } t ) - \frac { 217 \sqrt { 6 } } { 870 } e ^ { - 2 t } \sin ( 8 \sqrt { 2 } t ) + \frac { 8 } { 145 } \sin 8 t - \frac { 9 } { 145 } \cos 8 t$ 
E) $y = \frac { 1 } { 5 } e ^ { 2 t } \cos ( 7 \sqrt { 3 } t ) - \frac { 7 \sqrt { 3 } } { 30 } e ^ { 2 t } \sin ( 7 \sqrt { 3 } t ) + \frac { 9 } { 145 } \sin 8 t - \frac { 7 } { 145 } \cos 8 t$

Question 9

Solve the differential equation $y ^ { tt } + 2 y ^ { t } = 4 e ^ { x } \text { by the method of undetermined }$ coefficients.

Accepted Answer

A) $y = C _ { 1 } + C _ { 2 } e ^ { 3 x } + \frac { 4 } { 3 } e ^ { x }$ 
B) $y = C _ { 1 } + C _ { 2 } e ^ { - 3 x } - \frac { 4 } { 3 } e ^ { x }$ 
C) $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x } + \frac { 4 } { 3 } e ^ { x }$ 
D)  $y = C _ { 1 } + C _ { 2 } e ^ { 3 x } - 3 e ^ { x }$ 
E)  $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x } + 4 e ^ { x }$ 
A) $y = C _ { 1 } + C _ { 2 } e ^ { 3 x } + \frac { 4 } { 3 } e ^ { x }$ 
B) $y = C _ { 1 } + C _ { 2 } e ^ { - 3 x } - \frac { 4 } { 3 } e ^ { x }$ 
C) $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x } + \frac { 4 } { 3 } e ^ { x }$ 
D)  $y = C _ { 1 } + C _ { 2 } e ^ { 3 x } - 3 e ^ { x }$ 
E)  $y = C _ { 1 } + C _ { 2 } e ^ { - 2 x } + 4 e ^ { x }$

Question 10

Use Taylor's Theorem to find the first six terms of the series solution of $\text { of } y ^ { tt } - 2 x y = 0$ given the initial conditions $y ( 0 ) = 5$ and $y ^ {t } ( 0 ) = - 7$

Accepted Answer

A) 
$y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 3 ! } x ^ { 3 } - \frac { 28 } { 4 ! } x ^ { 4 } + \frac { 80 } { 6 ! } x ^ { 6 } - \frac { 280 } { 7 ! } x ^ { 7 } + \cdots$ 
B) $y = 7 - \frac { 5 } { 1 ! } x + \frac { 14 } { 3 ! } x ^ { 3 } - \frac { 20 } { 4 ! } x ^ { 4 } + \frac { 112 } { 6 ! } x ^ { 6 } - \frac { 200 } { 7 ! } x ^ { 7 } + \cdots$ 
C) $y = 7 - \frac { 5 } { 1 ! } x + \frac { 14 } { 3 ! } x ^ { 3 } - \frac { 20 } { 4 ! } x ^ { 4 } + \frac { 112 } { 7 ! } x ^ { 7 } - \frac { 175 } { 8 ! } x ^ { 8 } + \cdots$ 
D) $y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 3 ! } x ^ { 3 } - \frac { 28 } { 4 ! } x ^ { 4 } + \frac { 90 } { 7 ! } x ^ { 7 } - \frac { 245 } { 8 ! } x ^ { 8 } + \cdots$ 
E) $y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 2 ! } x ^ { 2 } - \frac { 28 } { 3 ! } x ^ { 3 } + \frac { 80 } { 4 ! } x ^ { 4 } - \frac { 280 } { 5 ! } x ^ { 5 } + \cdots$ 
A) 
$y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 3 ! } x ^ { 3 } - \frac { 28 } { 4 ! } x ^ { 4 } + \frac { 80 } { 6 ! } x ^ { 6 } - \frac { 280 } { 7 ! } x ^ { 7 } + \cdots$ 
B) $y = 7 - \frac { 5 } { 1 ! } x + \frac { 14 } { 3 ! } x ^ { 3 } - \frac { 20 } { 4 ! } x ^ { 4 } + \frac { 112 } { 6 ! } x ^ { 6 } - \frac { 200 } { 7 ! } x ^ { 7 } + \cdots$ 
C) $y = 7 - \frac { 5 } { 1 ! } x + \frac { 14 } { 3 ! } x ^ { 3 } - \frac { 20 } { 4 ! } x ^ { 4 } + \frac { 112 } { 7 ! } x ^ { 7 } - \frac { 175 } { 8 ! } x ^ { 8 } + \cdots$ 
D) $y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 3 ! } x ^ { 3 } - \frac { 28 } { 4 ! } x ^ { 4 } + \frac { 90 } { 7 ! } x ^ { 7 } - \frac { 245 } { 8 ! } x ^ { 8 } + \cdots$ 
E) $y = 5 - \frac { 7 } { 1 ! } x + \frac { 10 } { 2 ! } x ^ { 2 } - \frac { 28 } { 3 ! } x ^ { 3 } + \frac { 80 } { 4 ! } x ^ { 4 } - \frac { 280 } { 5 ! } x ^ { 5 } + \cdots$

Question 11

Using the method of undetermined coefficients, determine the most suitable choice for $y _ { p }$ given $y ^ {tt } - 3 y ^ { t } - 54 y = x ^ { 2 }$. (You do not need to solve the differential equation.)

Accepted Answer

A)  $y _ { p } = A x + B$ 
B)  $y _ { p } = A x ^ { 2 } + B x + C$ 
C)  $y _ { p } = A x ^ { 3 } + B x ^ { 2 } + C x + D$ 
D)  $y _ { p } = A x ^ { 3 } + C x$ 
E)  $y _ { n } = A x ^ { 2 }$ 
A)  $y _ { p } = A x + B$ 
B)  $y _ { p } = A x ^ { 2 } + B x + C$ 
C)  $y _ { p } = A x ^ { 3 } + B x ^ { 2 } + C x + D$ 
D)  $y _ { p } = A x ^ { 3 } + C x$ 
E)  $y _ { n } = A x ^ { 2 }$

Question 12

Suppose a 32-pound weight is suspended on a spring. The weight is pulled $\frac { 1 } { 3 } \text { foot }$ below the equilibrium position and released. The motion takes place in a Med that furnishes a damping force of magnitude $\frac { 1 } { 8 }$ speed at all times. Assume that the weight stretches the spring $\frac { 2 } { 3 }$ foot from its natural position. Find a formula for the position of the weight as a function of time $t$.

Accepted Answer

A)  $y ( t ) = 3 \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
B)  $y ( t ) = \frac { e ^ { - t / 16 } } { 12 } \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
C)  $y ( t ) = \frac { e ^ { - t / 16 } } { 3 } \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
D) $y ( t ) = 6 \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
E) $y ( t ) = \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
A)  $y ( t ) = 3 \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
B)  $y ( t ) = \frac { e ^ { - t / 16 } } { 12 } \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
C)  $y ( t ) = \frac { e ^ { - t / 16 } } { 3 } \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
D) $y ( t ) = 6 \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$ 
E) $y ( t ) = \left( \cos \frac { \sqrt { 12,287 } t } { 16 } + \frac { \sqrt { 12,287 } } { 12,287 } \sin \frac { \sqrt { 12,287 } t } { 16 } \right)$

Question 13

If  y = c ( x ) represents the cost of producing  x  units in a manufacturing process, the elasticity of cost is defined as $E ( x ) = \frac { \text { marginal cost } } { \text { average cost } } = \frac { C ^ { \prime } ( x ) } { C ( x ) / x } = \frac { x } { y } \frac { d y } { d x }$. Find the cost function if the elasticity function is $E ( x ) = \frac { 12 x - y } { 2 y - 6 x }$, where $C ( 90 ) = 272$ and $x \geq 90$.

Accepted Answer

A)  $y = \frac { 3 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 730,210 x } \right) } { x }$ 
B)  $y = \frac { 6 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 730,210 x } \right) } { x }$ 
C)  $y = \frac { 6 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 728,960 } \right) } { x }$ 
D)  $y = \frac { 3 \left( x ^ { 2 } + 2 \sqrt { x ^ { 4 } - 734,041 } \right) } { x }$ 
E) $y = \frac { 3 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 728,960 x } \right) } { x }$ 
A)  $y = \frac { 3 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 730,210 x } \right) } { x }$ 
B)  $y = \frac { 6 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 730,210 x } \right) } { x }$ 
C)  $y = \frac { 6 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 728,960 } \right) } { x }$ 
D)  $y = \frac { 3 \left( x ^ { 2 } + 2 \sqrt { x ^ { 4 } - 734,041 } \right) } { x }$ 
E) $y = \frac { 3 \left( x ^ { 2 } + \sqrt { x ^ { 4 } - 728,960 x } \right) } { x }$

Question 14

$\text { Find an equation for the curve with the slope } \frac { d y } { d x } = \frac { y - x ^ { 2 } } { 7 y ^ { 2 } - x } \text { passing through the }$ point $( 3,2 )$.

Accepted Answer

A)  $x ^ { 3 } - 3 x y + 7 y ^ { 3 } = 77$ 
B)  $x ^ { 2 } - 2 x y + 7 y ^ { 2 } = 6$ 
C)  $x ^ { 3 } - 3 x y + 7 y ^ { 3 } = 65$ 
D)  $x ^ { 3 } + 3 x y - 7 y ^ { 3 } = - 11$ 
E)  $x ^ { 2 } - 2 x y + 7 y ^ { 2 } = - 7$ 
A)  $x ^ { 3 } - 3 x y + 7 y ^ { 3 } = 77$ 
B)  $x ^ { 2 } - 2 x y + 7 y ^ { 2 } = 6$ 
C)  $x ^ { 3 } - 3 x y + 7 y ^ { 3 } = 65$ 
D)  $x ^ { 3 } + 3 x y - 7 y ^ { 3 } = - 11$ 
E)  $x ^ { 2 } - 2 x y + 7 y ^ { 2 } = - 7$

Question 15

Find the interval of convergence for the solution of the differential equation $$y ^ { tt } - 11 x y ^ { t } = 0$$

Accepted Answer

A)  $[ 0 , \infty )$ 
B)  $( - \infty , \infty )$ 
C)  $( - 11 , \infty )$ 
D)  $( 0 , \infty )$ 
E)  $[ 11 , \infty )$ 
A)  $[ 0 , \infty )$ 
B)  $( - \infty , \infty )$ 
C)  $( - 11 , \infty )$ 
D)  $( 0 , \infty )$ 
E)  $[ 11 , \infty )$

Question 16

Find the value of k such that the differential equation $\left( x y ^ { 2 } + k x ^ { 5 } y + x ^ { 4 } \right) d x + \left( x ^ { 6 } + x ^ { 2 } y + y ^ { 6 } \right) d y = 0$ is exact.

Accepted Answer

A)  5 
B)  4 
C)  8 
D)  7 
E)  6 
A)  5 
B)  4 
C)  8 
D)  7 
E)  6

Question 17

$\text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. }$

Accepted Answer

A)  $y = a _ { 0 } + e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant 
B)  $y = a _ { 0 } e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant 
C)  $y = a _ { 0 } e ^ { x }$, where $a _ { 0 }$ is an arbitrary constant 
D)  $y = 3 + a _ { 0 } e ^ { z }$, where $a _ { 0 }$ is an arbitrary constant 
E)  $y = a _ { 0 } + 3 e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant 
A)  $y = a _ { 0 } + e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant 
B)  $y = a _ { 0 } e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant 
C)  $y = a _ { 0 } e ^ { x }$, where $a _ { 0 }$ is an arbitrary constant 
D)  $y = 3 + a _ { 0 } e ^ { z }$, where $a _ { 0 }$ is an arbitrary constant 
E)  $y = a _ { 0 } + 3 e ^ { 3 x }$, where $a _ { 0 }$ is an arbitrary constant

Question 18

Solve the differential equation $y ^ { tt } + y = 6 \sec x \text { by the method of variation of }$ parameters.

Accepted Answer

A)  $y = \left( C _ { 1 } + 6 \cot x \right) \cos x + \left( C _ { 2 } + 6 x ^ { 2 } \right) \sin x$ 
B)  $y = \left( C _ { 1 } + 6 \ln | \sin x | \right) \cos x + \left( C _ { 2 } + 6 x \right) \sin x$ 
C)  $y = \left( C _ { 1 } + 6 \tan x \right) \cos x + \left( C _ { 2 } + 6 x \right) \sin x$ 
D)  $y = \left( C _ { 1 } + 6 \ln | \cos x | \right) \cos x + \left( C _ { 2 } + 6 x \right) \sin x$ 
E)  $y = \left( C _ { 1 } + 6 \ln | \cos x | \right) \cos x + \left( C _ { 2 } + 6 x ^ { 2 } \right) \sin x$ 
A)  $y = \left( C _ { 1 } + 6 \cot x \right) \cos x + \left( C _ { 2 } + 6 x ^ { 2 } \right) \sin x$ 
B)  $y = \left( C _ { 1 } + 6 \ln | \sin x | \right) \cos x + \left( C _ { 2 } + 6 x \right) \sin x$ 
C)  $y = \left( C _ { 1 } + 6 \tan x \right) \cos x + \left( C _ { 2 } + 6 x \right) \sin x$ 
D)  $y = \left( C _ { 1 } + 6 \ln | \cos x | \right) \cos x + \left( C _ { 2 } + 6 x \right) \sin x$ 
E)  $y = \left( C _ { 1 } + 6 \ln | \cos x | \right) \cos x + \left( C _ { 2 } + 6 x ^ { 2 } \right) \sin x$

Question 19

Find the particular solution of the differential equation $\left( 2 x y - 25 x ^ { 4 } \right) d x + \left( 2 y + x ^ { 2 } + 5 \right) d y = 0$ that satisfies the initial condition $y ( 0 ) = 4$

Accepted Answer

A)  $x ^ { 2 } - 25 x ^ { 4 } + 2 y + 5 = 13$ 
B)  $x ^ { 3 } y - 5 x ^ { 4 } + 1 y ^ { 2 } + 5 y = 36$ 
C)  $x ^ { 2 } y - 5 x ^ { 5 } + y ^ { 2 } + 5 y = 36$ 
D)  $x ^ { 2 } - 5 x ^ { 5 } + y ^ { 2 } + 5 = 21$ 
E)  $x ^ { 2 } y - 5 x ^ { 5 } + 2 y ^ { 2 } + 5 = 37$ 
A)  $x ^ { 2 } - 25 x ^ { 4 } + 2 y + 5 = 13$ 
B)  $x ^ { 3 } y - 5 x ^ { 4 } + 1 y ^ { 2 } + 5 y = 36$ 
C)  $x ^ { 2 } y - 5 x ^ { 5 } + y ^ { 2 } + 5 y = 36$ 
D)  $x ^ { 2 } - 5 x ^ { 5 } + y ^ { 2 } + 5 = 21$ 
E)  $x ^ { 2 } y - 5 x ^ { 5 } + 2 y ^ { 2 } + 5 = 37$

Question 20

Sketch a graph of the solution of the differential equation $y ^ { tt } + 6 y ^ {t } + 18 y = 0 \text { with the }$ initial condition $y ( 0 ) = 3 , y ^ { t } ( 0 ) = - 9$

Accepted Answer

A) 
  
B) 
  
C) 
  
D) 
  
E) 
  
A) 
  
B) 
  
C) 
  
D) 
  
E)

Solve the differential equation $y ^ {tt} - 6 y ^ { t } + 9 y = 8 x ^ { 2 } e ^ { 3 x } \text { by the method of variation of }$ parameters.

Taylor's Theorem to find the first four terms of the series solution of $y ^ { tt } + x ^ { 2 } y ^ { t } - ( \cos x ) y = 0$ given the initial conditions $y ( 0 ) = 7$ , and $y ^ {t } ( 0 ) = 4$

Solve the differential equation $y ^ {tt } + 5 y ^ { t } - 6 y = 7 \cos 2 x \text {, where }$ $y ( 0 ) = - 1$ and $y ^ {t } ( 0 ) = 6$ by the method of undetermined coefficients.

Find the general solution of the differential equation

Use Taylor's Theorem to find the first five terms of the series solution of $y ^ {t} + ( 2 x - 1 ) y = 0$ given the initial condition $y ( 0 ) = 2$

$\text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. }$

Solve the differential equation $y ^ { tt } + 2 y ^ { t } = 4 e ^ { x } \text { by the method of undetermined }$ coefficients.

Use Taylor's Theorem to find the first six terms of the series solution of $\text { of } y ^ { tt } - 2 x y = 0$ given the initial conditions $y ( 0 ) = 5$ and $y ^ {t } ( 0 ) = - 7$

Using the method of undetermined coefficients, determine the most suitable choice for $y _ { p }$ given $y ^ {tt } - 3 y ^ { t } - 54 y = x ^ { 2 }$ . (You do not need to solve the differential equation.)

$\text { Find an equation for the curve with the slope } \frac { d y } { d x } = \frac { y - x ^ { 2 } } { 7 y ^ { 2 } - x } \text { passing through the }$ point $( 3,2 )$ .

Find the interval of convergence for the solution of the differential equation $y ^ { tt } - 11 x y ^ { t } = 0$

Find the value of k such that the differential equation $\left( x y ^ { 2 } + k x ^ { 5 } y + x ^ { 4 } \right) d x + \left( x ^ { 6 } + x ^ { 2 } y + y ^ { 6 } \right) d y = 0$ is exact.

$\text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. }$

Solve the differential equation $y ^ { tt } + y = 6 \sec x \text { by the method of variation of }$ parameters.

Find the particular solution of the differential equation $\left( 2 x y - 25 x ^ { 4 } \right) d x + \left( 2 y + x ^ { 2 } + 5 \right) d y = 0$ that satisfies the initial condition $y ( 0 ) = 4$

Sketch a graph of the solution of the differential equation $y ^ { tt } + 6 y ^ {t } + 18 y = 0 \text { with the }$ initial condition $y ( 0 ) = 3 , y ^ { t } ( 0 ) = - 9$

Graphs and Models

A Preview of Calculus

The Derivative and the Tangent Line Problem

Extrema on an Interval

Antiderivatives and Indefinite Integration

Slope Fields and Eulers Method

Area of a Region Between Two Curves

Basic Integration Rules

Sequences

Conics and Calculus

Vectors in the Plane

Vector-Valued Functions

Introduction to Functions of Several Variables

Iterated Integrals and Area in the Plane

Vector Fields

Filters

Exam 16: Exact First-Order Equations

Solve the differential equation ytt−6yt+9y=8x2e3x by the method of variation of y ^ {tt} - 6 y ^ { t } + 9 y = 8 x ^ { 2 } e ^ { 3 x } \text { by the method of variation of }ytt−6yt+9y=8x2e3x by the method of variation of parameters.

Taylor's Theorem to find the first four terms of the series solution of ytt+x2yt−(cos⁡x)y=0y ^ { tt } + x ^ { 2 } y ^ { t } - ( \cos x ) y = 0ytt+x2yt−(cosx)y=0 given the initial conditions y(0)=7y ( 0 ) = 7y(0)=7 , and yt(0)=4y ^ {t } ( 0 ) = 4yt(0)=4

Solve the differential equation ytt+5yt−6y=7cos⁡2x, where y ^ {tt } + 5 y ^ { t } - 6 y = 7 \cos 2 x \text {, where }ytt+5yt−6y=7cos2x, where y(0)=−1y ( 0 ) = - 1y(0)=−1 and yt(0)=6y ^ {t } ( 0 ) = 6yt(0)=6 by the method of undetermined coefficients.

Find the general solution of the differential equation

Use Taylor's Theorem to find the first five terms of the series solution of yt+(2x−1)y=0y ^ {t} + ( 2 x - 1 ) y = 0yt+(2x−1)y=0 given the initial condition y(0)=2y ( 0 ) = 2y(0)=2

Find the general solution of the linear differential equation ytt+2yt=0. \text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. } Find the general solution of the linear differential equation ytt+2yt=0.

Solve the differential equation ytt+2yt=4ex by the method of undetermined y ^ { tt } + 2 y ^ { t } = 4 e ^ { x } \text { by the method of undetermined }ytt+2yt=4ex by the method of undetermined coefficients.

Use Taylor's Theorem to find the first six terms of the series solution of of ytt−2xy=0\text { of } y ^ { tt } - 2 x y = 0 of ytt−2xy=0 given the initial conditions y(0)=5y ( 0 ) = 5y(0)=5 and yt(0)=−7y ^ {t } ( 0 ) = - 7yt(0)=−7

Using the method of undetermined coefficients, determine the most suitable choice for ypy _ { p }yp​ given ytt−3yt−54y=x2y ^ {tt } - 3 y ^ { t } - 54 y = x ^ { 2 }ytt−3yt−54y=x2 . (You do not need to solve the differential equation.)

Find the interval of convergence for the solution of the differential equation ytt−11xyt=0y ^ { tt } - 11 x y ^ { t } = 0ytt−11xyt=0

Find the value of k such that the differential equation (xy2+kx5y+x4)dx+(x6+x2y+y6)dy=0\left( x y ^ { 2 } + k x ^ { 5 } y + x ^ { 4 } \right) d x + \left( x ^ { 6 } + x ^ { 2 } y + y ^ { 6 } \right) d y = 0(xy2+kx5y+x4)dx+(x6+x2y+y6)dy=0 is exact.

Use a power series to solve the differential equation yt−3y=0. \text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. } Use a power series to solve the differential equation yt−3y=0.

Solve the differential equation ytt+y=6sec⁡x by the method of variation of y ^ { tt } + y = 6 \sec x \text { by the method of variation of }ytt+y=6secx by the method of variation of parameters.

Find the particular solution of the differential equation (2xy−25x4)dx+(2y+x2+5)dy=0\left( 2 x y - 25 x ^ { 4 } \right) d x + \left( 2 y + x ^ { 2 } + 5 \right) d y = 0(2xy−25x4)dx+(2y+x2+5)dy=0 that satisfies the initial condition y(0)=4y ( 0 ) = 4y(0)=4

Sketch a graph of the solution of the differential equation ytt+6yt+18y=0 with the y ^ { tt } + 6 y ^ {t } + 18 y = 0 \text { with the }ytt+6yt+18y=0 with the initial condition y(0)=3,yt(0)=−9y ( 0 ) = 3 , y ^ { t } ( 0 ) = - 9y(0)=3,yt(0)=−9

Graphs and Models

A Preview of Calculus

The Derivative and the Tangent Line Problem

Extrema on an Interval

Antiderivatives and Indefinite Integration

Slope Fields and Eulers Method

Area of a Region Between Two Curves

Basic Integration Rules

Sequences

Conics and Calculus

Vectors in the Plane

Vector-Valued Functions

Introduction to Functions of Several Variables

Iterated Integrals and Area in the Plane

Vector Fields

Filters

Solve the differential equation $y ^ {tt} - 6 y ^ { t } + 9 y = 8 x ^ { 2 } e ^ { 3 x } \text { by the method of variation of }$ parameters.

Taylor's Theorem to find the first four terms of the series solution of $y ^ { tt } + x ^ { 2 } y ^ { t } - ( \cos x ) y = 0$ given the initial conditions $y ( 0 ) = 7$ , and $y ^ {t } ( 0 ) = 4$

Solve the differential equation $y ^ {tt } + 5 y ^ { t } - 6 y = 7 \cos 2 x \text {, where }$ $y ( 0 ) = - 1$ and $y ^ {t } ( 0 ) = 6$ by the method of undetermined coefficients.

Use Taylor's Theorem to find the first five terms of the series solution of $y ^ {t} + ( 2 x - 1 ) y = 0$ given the initial condition $y ( 0 ) = 2$

$\text { Find the general solution of the linear differential equation } y ^ { t t } + 2 y ^ { t } = 0 \text {. }$

Solve the differential equation $y ^ { tt } + 2 y ^ { t } = 4 e ^ { x } \text { by the method of undetermined }$ coefficients.

Use Taylor's Theorem to find the first six terms of the series solution of $\text { of } y ^ { tt } - 2 x y = 0$ given the initial conditions $y ( 0 ) = 5$ and $y ^ {t } ( 0 ) = - 7$

Using the method of undetermined coefficients, determine the most suitable choice for $y _ { p }$ given $y ^ {tt } - 3 y ^ { t } - 54 y = x ^ { 2 }$ . (You do not need to solve the differential equation.)

Find the interval of convergence for the solution of the differential equation $y ^ { tt } - 11 x y ^ { t } = 0$

Find the value of k such that the differential equation $\left( x y ^ { 2 } + k x ^ { 5 } y + x ^ { 4 } \right) d x + \left( x ^ { 6 } + x ^ { 2 } y + y ^ { 6 } \right) d y = 0$ is exact.

$\text { Use a power series to solve the differential equation } y ^ { t } - 3 y = 0 \text {. }$

Solve the differential equation $y ^ { tt } + y = 6 \sec x \text { by the method of variation of }$ parameters.

Find the particular solution of the differential equation $\left( 2 x y - 25 x ^ { 4 } \right) d x + \left( 2 y + x ^ { 2 } + 5 \right) d y = 0$ that satisfies the initial condition $y ( 0 ) = 4$

Sketch a graph of the solution of the differential equation $y ^ { tt } + 6 y ^ {t } + 18 y = 0 \text { with the }$ initial condition $y ( 0 ) = 3 , y ^ { t } ( 0 ) = - 9$