Question 1

The solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using Euler's method with $h = 0.1$ is

Accepted Answer

A)  1.01 
B)  1.11 
C)  1.21 
D)  1.22 
E)  1.23 
A)  1.01 
B)  1.11 
C)  1.21 
D)  1.22 
E)  1.23

Question 2

The Euler's method solution for $y ( 0.2 )$ of $y ^ { \prime \prime } + y = 0 , y ^ { \prime } ( 0 ) = 1$ using $h = 0.1$ is

Accepted Answer

A)  0.14 
B)  0.2 
C)  0.21 
D)  0.11 
E)  0.12 
A)  0.14 
B)  0.2 
C)  0.21 
D)  0.11 
E)  0.12

Question 3

The solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using the improved Euler's method with $h = 0.1$ is

Accepted Answer

A)  1.22125 
B)  1.210625 
C)  1.226525 
D)  1.21525 
E)  1.221025 
A)  1.22125 
B)  1.210625 
C)  1.226525 
D)  1.21525 
E)  1.221025

Question 4

Using the method from the previous two problems, using the values $y _ { 0 } = 1 , y _ { 1 } = 1.1052 , y _ { 2 } = 1.2214 , y _ { 3 } = 1.3499$ the solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.4 )$ with $h = 0.1$ is

Accepted Answer

A)  1.4919 
B)  1.4967 
C)  1.4978 
D)  1.5003 
E)  none of the above 
A)  1.4919 
B)  1.4967 
C)  1.4978 
D)  1.5003 
E)  none of the above

Question 5

The solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using Euler's method with $h = 0.1$ is

Accepted Answer

A)  1.01 
B)  1.1 
C)  1.11 
D)  1.21 
E)  1.22 
A)  1.01 
B)  1.1 
C)  1.11 
D)  1.21 
E)  1.22

Question 6

Using the method from the previous problem, the solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

Accepted Answer

A)  1.24 
B)  1.241 
C)  1.214 
D)  1.2214 
E)  1.224 
A)  1.24 
B)  1.241 
C)  1.214 
D)  1.2214 
E)  1.224

Question 7

The standard backward difference approximation of $y ^ { \prime } ( x )$ is

Accepted Answer

A)  $( y ( x + h ) - y ( x ) ) / h$ 
B)  $( y ( x ) - y ( x - h ) ) / h$ 
C)  $( y ( x + h ) - y ( x ) ) / h ^ { 2 }$ 
D)  $y ( x + h ) - y ( x )$ 
E)  $y ( x ) - y ( x + h )$ 
A)  $( y ( x + h ) - y ( x ) ) / h$ 
B)  $( y ( x ) - y ( x - h ) ) / h$ 
C)  $( y ( x + h ) - y ( x ) ) / h ^ { 2 }$ 
D)  $y ( x + h ) - y ( x )$ 
E)  $y ( x ) - y ( x + h )$

Question 8

In the previous problem, the local truncation error in $y _ { n + 1 }$ is

Accepted Answer

A)  $0.005 e ^ { c } \text {, where } x _ { n - 1 } < c < x _ { n }$ 
B)  $0.05 e ^ { c } \text {, where } x _ { n - 1 } < c < x _ { n }$ 
C)  $0.005 e ^ { c } \text {, where } x _ { n } < c < x _ { n + 1 }$ 
D)  $0.05 e ^ { c } \text {, where } x _ { n } < c < x _ { n + 1 }$ 
E)  unknown 
A)  $0.005 e ^ { c } \text {, where } x _ { n - 1 } < c < x _ { n }$ 
B)  $0.05 e ^ { c } \text {, where } x _ { n - 1 } < c < x _ { n }$ 
C)  $0.005 e ^ { c } \text {, where } x _ { n } < c < x _ { n + 1 }$ 
D)  $0.05 e ^ { c } \text {, where } x _ { n } < c < x _ { n + 1 }$ 
E)  unknown

Question 9

The Adams-Bashforth formula for finding the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

Accepted Answer

A)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } - 59 y _ { n - 1 } ^ { \prime } - 37 y _ { n - 2 } ^ { \prime } + 65 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
B)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 59 y _ { n } ^ { \prime } - 55 y _ { n - 1 } ^ { \prime } + 37 y _ { n - 2 } ^ { \prime } - 17 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
C)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } + 59 y _ { n - 1 } ^ { \prime } - 37 y _ { n - 2 } ^ { \prime } - 9 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
D)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } - 59 y _ { n - 1 } ^ { \prime } + 37 y _ { n - 2 } ^ { \prime } - 9 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
E)  none of the above 
A)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } - 59 y _ { n - 1 } ^ { \prime } - 37 y _ { n - 2 } ^ { \prime } + 65 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
B)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 59 y _ { n } ^ { \prime } - 55 y _ { n - 1 } ^ { \prime } + 37 y _ { n - 2 } ^ { \prime } - 17 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
C)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } + 59 y _ { n - 1 } ^ { \prime } - 37 y _ { n - 2 } ^ { \prime } - 9 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
D)  $\begin{array} { l } 
y _ { n + 1 } ^ { * } = y _ { n } + h \left( 55 y _ { n } ^ { \prime } - 59 y _ { n - 1 } ^ { \prime } + 37 y _ { n - 2 } ^ { \prime } - 9 y _ { n - 3 } ^ { \prime } \right) / 24 , \text { where } y _ { n } ^ { \prime } = f \left( x _ { n } , y _ { n } \right) \
y _ { n - 1 } ^ { \prime } = f \left( x _ { n - 1 } , y _ { n - 1 } \right) , y _ { n - 2 } ^ { \prime } = f \left( x _ { n - 2 } , y _ { n - 2 } \right) , y _ { n - 3 } ^ { \prime } = f \left( x _ { n - 3 } , y _ { n - 3 } \right)
\end{array}$ 
E)  none of the above

Question 10

The problem $y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1$ can be written as a system of two equations as follows.

Accepted Answer

A)  $y ^ { \prime } = u , u ^ { \prime } = x y u , y ( 0 ) = 0 , u ( 0 ) = 0$ 
B)  $y ^ { \prime } = u , u ^ { \prime } = - x y u , y ( 0 ) = 1 , u ( 0 ) = 0$ 
C)  $y ^ { \prime } = u , u ^ { \prime } = x y u , y ( 0 ) = 1 , u ( 0 ) = 0$ 
D)  $y ^ { \prime } = u , u ^ { \prime } = x y u , y ( 0 ) = 0 , u ( 0 ) = 1$ 
E)  $y ^ { \prime } = u , u ^ { \prime } = - x y u , y ( 0 ) = 0 , u ( 0 ) = 1$ 
A)  $y ^ { \prime } = u , u ^ { \prime } = x y u , y ( 0 ) = 0 , u ( 0 ) = 0$ 
B)  $y ^ { \prime } = u , u ^ { \prime } = - x y u , y ( 0 ) = 1 , u ( 0 ) = 0$ 
C)  $y ^ { \prime } = u , u ^ { \prime } = x y u , y ( 0 ) = 1 , u ( 0 ) = 0$ 
D)  $y ^ { \prime } = u , u ^ { \prime } = x y u , y ( 0 ) = 0 , u ( 0 ) = 1$ 
E)  $y ^ { \prime } = u , u ^ { \prime } = - x y u , y ( 0 ) = 0 , u ( 0 ) = 1$

Question 11

Using the value of $y _ { n + 1 } ^ { * }$ from the previous problem, the Adams-Moulton corrector value for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

Accepted Answer

A)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } - 19 y _ { n } ^ { \prime } + 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
B)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } + 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 34 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
C)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } - 5 y _ { n - 1 } ^ { \prime } - y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
D)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } - 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
E)  none of the above 
A)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } - 19 y _ { n } ^ { \prime } + 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
B)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } + 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 34 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
C)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } - 5 y _ { n - 1 } ^ { \prime } - y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
D)  $y _ { n + 1 } = y _ { n } + h \left( 9 y _ { n + 1 } ^ { \prime } + 19 y _ { n } ^ { \prime } - 5 y _ { n - 1 } ^ { \prime } + y _ { n - 2 } ^ { \prime } \right) / 24 , \text { where } y _ { n + 1 } ^ { \prime } = f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right)$ 
E)  none of the above

Question 12

The local truncation error for the improved Euler's method is

Accepted Answer

A)  $O ( h )$ 
B)  $O \left( h ^ { 2 } \right)$ 
C)  $O \left( h ^ { 3 } \right)$ 
D)  $O \left( h ^ { 4 } \right)$ 
E)  unknown 
A)  $O ( h )$ 
B)  $O \left( h ^ { 2 } \right)$ 
C)  $O \left( h ^ { 3 } \right)$ 
D)  $O \left( h ^ { 4 } \right)$ 
E)  unknown

Question 13

Using the method from the previous problem, the solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

Accepted Answer

A)  1.241 
B)  1.242 
C)  1.2422 
D)  1.2426 
E)  1.2428 
A)  1.241 
B)  1.242 
C)  1.2422 
D)  1.2426 
E)  1.2428

Question 14

When entering the number $1 / 7$ into a three digit base ten calculator, the actual value entered is

Accepted Answer

A)  $1 / 7$ 
B)  $.143$ 
C)  $.142$ 
D)  $.140$ 
E)  $.150$ 
A)  $1 / 7$ 
B)  $.143$ 
C)  $.142$ 
D)  $.140$ 
E)  $.150$

Question 15

Euler's formula for solving $y ^ { \prime } = f ( x , y ) , y ( \bar { x } ) = \bar { y }$ is

Accepted Answer

A)  $y _ { n + 1 } = y _ { n } - f \left( x _ { n } , y _ { n } \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots$ 
B)  $y _ { n + 1 } = y _ { n } - h f \left( x _ { n } , y _ { n } \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots$ 
C)  $y _ { n + 1 } = y _ { n } + h f \left( x _ { n } , y _ { n } \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots$ 
D)  $y _ { n + 1 } = y _ { n } + h \left( f \left( x _ { n } , y _ { n } \right) + f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right) / 2 \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots \text { where } y _ { n + 1 } ^ { * }$
is predicted from Euler's formula 
E)  $y _ { n + 1 } = y _ { n } + \left( f \left( x _ { n } , y _ { n } \right) + f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right) / 2 \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots \text { where } y _ { n + 1 } ^ { * }$
is predicted from Euler's formula 
A)  $y _ { n + 1 } = y _ { n } - f \left( x _ { n } , y _ { n } \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots$ 
B)  $y _ { n + 1 } = y _ { n } - h f \left( x _ { n } , y _ { n } \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots$ 
C)  $y _ { n + 1 } = y _ { n } + h f \left( x _ { n } , y _ { n } \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots$ 
D)  $y _ { n + 1 } = y _ { n } + h \left( f \left( x _ { n } , y _ { n } \right) + f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right) / 2 \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots \text { where } y _ { n + 1 } ^ { * }$
is predicted from Euler's formula 
E)  $y _ { n + 1 } = y _ { n } + \left( f \left( x _ { n } , y _ { n } \right) + f \left( x _ { n + 1 } , y _ { n + 1 } ^ { * } \right) / 2 \right) , y _ { 0 } = \bar { y } , n = 0,1,2 , \ldots \text { where } y _ { n + 1 } ^ { * }$
is predicted from Euler's formula

Question 16

Using the notation from the text, the finite difference equation for solving the boundary value problem $y ^ { \prime \prime } + P ( x ) y ^ { \prime } + Q ( x ) y = f ( x ) , y ( a ) = \alpha , y ( b ) = \beta$ is

Accepted Answer

A)  $\left( 1 - h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 + h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
B)  $\left( 1 - h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
C)  $\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 + h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
D)  $\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( 2 - h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
E)  $\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
A)  $\left( 1 - h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 + h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
B)  $\left( 1 - h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
C)  $\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 + h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
D)  $\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( 2 - h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$ 
E)  $\left( 1 + h P _ { i } / 2 \right) y _ { i + 1 } + \left( - 2 + h ^ { 2 } Q _ { i } \right) y _ { i } + \left( 1 - h P _ { i } / 2 \right) y _ { i - 1 } = h ^ { 2 } f _ { i }$

Question 17

The most popular fourth order Runge-Kutta method for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

Accepted Answer

A)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( 2 k _ { 1 } + k _ { 2 } + k _ { 3 } + 2 k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } / 2 \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
B)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
C)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } / 2 \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
D)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( 2 k _ { 1 } + k _ { 2 } + k _ { 3 } + 2 k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
E)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 3 , y _ { n } + h k _ { 1 } / 2 \right) , k _ { 3 } = f \left( x _ { n } + 2 h / 3 , y _ { n } + h k _ { 2 } / 2 \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
A)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( 2 k _ { 1 } + k _ { 2 } + k _ { 3 } + 2 k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } / 2 \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
B)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
C)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } / 2 \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
D)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( 2 k _ { 1 } + k _ { 2 } + k _ { 3 } + 2 k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right) , k _ { 3 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 2 } \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$ 
E)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + 2 k _ { 2 } + 2 k _ { 3 } + k _ { 4 } \right) / 6 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , \
k _ { 2 } = f \left( x _ { n } + h / 3 , y _ { n } + h k _ { 1 } / 2 \right) , k _ { 3 } = f \left( x _ { n } + 2 h / 3 , y _ { n } + h k _ { 2 } / 2 \right) , \
k _ { 4 } = f \left( x _ { n } + h , y _ { n } + h k _ { 3 } \right)
\end{array}$

Question 18

The solution of $y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1$ for $y ( 0.1 )$ , using the Runge-Kutta method of order four, and using $h = 0.1$ , is

Accepted Answer

A)  0.0909 
B)  0.09999 
C)  0.09099 
D)  0.09899 
E)  0.08899 
A)  0.0909 
B)  0.09999 
C)  0.09099 
D)  0.09899 
E)  0.08899

Question 19

Using the method from the previous problem, the solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

Accepted Answer

A)  1.222 
B)  1.22 
C)  1.2213 
D)  1.24 
E)  1.21 
A)  1.222 
B)  1.22 
C)  1.2213 
D)  1.24 
E)  1.21

Question 20

A popular second order Runge-Kutta method for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

Accepted Answer

A)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right)
\end{array}$ 
B)  $y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h , y _ { n } + h k _ { 1 } / 2 \right)$ 
C)  $y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right)$ 
D)  $y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h , y _ { n } + h k _ { 1 } \right)$ 
E)  none of the above 
A)  $\begin{array} { l } 
y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 , \text { where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) \
k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } / 2 \right)
\end{array}$ 
B)  $y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h , y _ { n } + h k _ { 1 } / 2 \right)$ 
C)  $y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h / 2 , y _ { n } + h k _ { 1 } \right)$ 
D)  $y _ { n + 1 } = y _ { n } + h \left( k _ { 1 } + k _ { 2 } \right) / 2 \text {, where } k _ { 1 } = f \left( x _ { n } , y _ { n } \right) , k _ { 2 } = f \left( x _ { n } + h , y _ { n } + h k _ { 1 } \right)$ 
E)  none of the above

The solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using Euler's method with $h = 0.1$ is

The Euler's method solution for $y ( 0.2 )$ of $y ^ { \prime \prime } + y = 0 , y ^ { \prime } ( 0 ) = 1$ using $h = 0.1$ is

The solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using the improved Euler's method with $h = 0.1$ is

Using the method from the previous two problems, using the values $y _ { 0 } = 1 , y _ { 1 } = 1.1052 , y _ { 2 } = 1.2214 , y _ { 3 } = 1.3499$ the solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.4 )$ with $h = 0.1$ is

The solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using Euler's method with $h = 0.1$ is

Using the method from the previous problem, the solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

The standard backward difference approximation of $y ^ { \prime } ( x )$ is

In the previous problem, the local truncation error in $y _ { n + 1 }$ is

The Adams-Bashforth formula for finding the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

The problem $y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1$ can be written as a system of two equations as follows.

Using the value of $y _ { n + 1 } ^ { * }$ from the previous problem, the Adams-Moulton corrector value for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

The local truncation error for the improved Euler's method is

Using the method from the previous problem, the solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

When entering the number $1 / 7$ into a three digit base ten calculator, the actual value entered is

Euler's formula for solving $y ^ { \prime } = f ( x , y ) , y ( \bar { x } ) = \bar { y }$ is

Using the notation from the text, the finite difference equation for solving the boundary value problem $y ^ { \prime \prime } + P ( x ) y ^ { \prime } + Q ( x ) y = f ( x ) , y ( a ) = \alpha , y ( b ) = \beta$ is

The most popular fourth order Runge-Kutta method for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

The solution of $y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1$ for $y ( 0.1 )$ , using the Runge-Kutta method of order four, and using $h = 0.1$ , is

Using the method from the previous problem, the solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

A popular second order Runge-Kutta method for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

Introduction to Differential Equations

First-Order Differential Equations

Modeling With First-Order Differential Equations

Higher-Order Differential Equations

Modeling With Higher-Order Differential Equations

Series Solutions of Linear Equations

Laplace Transform

Systems of Linear First-Order Differential Equations

Plane Autonomous Systems

Orthogonal Functions and Fourier Series

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Exam 9: Numerical Solutions of Ordinary Differential Equations

The solution of y′=x+y,y(0)=1 for y(0.2)y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )y′=x+y,y(0)=1 for y(0.2) , using Euler's method with h=0.1h = 0.1h=0.1 is

The Euler's method solution for y(0.2)y ( 0.2 )y(0.2) of y′′+y=0,y′(0)=1y ^ { \prime \prime } + y = 0 , y ^ { \prime } ( 0 ) = 1y′′+y=0,y′(0)=1 using h=0.1h = 0.1h=0.1 is

The solution of y′=y,y(0)=1 for y(0.2)y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )y′=y,y(0)=1 for y(0.2) , using the improved Euler's method with h=0.1h = 0.1h=0.1 is

The solution of y′=y,y(0)=1 for y(0.2)y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )y′=y,y(0)=1 for y(0.2) , using Euler's method with h=0.1h = 0.1h=0.1 is

Using the method from the previous problem, the solution of y′=y,y(0)=1 for y(0.2)y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )y′=y,y(0)=1 for y(0.2) with h=0.2h = 0.2h=0.2 is

The standard backward difference approximation of y′(x)y ^ { \prime } ( x )y′(x) is

In the previous problem, the local truncation error in yn+1y _ { n + 1 }yn+1​ is

The Adams-Bashforth formula for finding the solution of yt=f(x,y),y(x0)=y0y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }yt=f(x,y),y(x0​)=y0​ is

The problem y′′+xyy′=0,y(0)=0,y′(0)=1y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1y′′+xyy′=0,y(0)=0,y′(0)=1 can be written as a system of two equations as follows.

Using the value of yn+1∗y _ { n + 1 } ^ { * }yn+1∗​ from the previous problem, the Adams-Moulton corrector value for the solution of yt=f(x,y),y(x0)=y0y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }yt=f(x,y),y(x0​)=y0​ is

The local truncation error for the improved Euler's method is

Using the method from the previous problem, the solution of y′=x+y,y(0)=1 for y(0.2)y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )y′=x+y,y(0)=1 for y(0.2) with h=0.2h = 0.2h=0.2 is

When entering the number 1/71 / 71/7 into a three digit base ten calculator, the actual value entered is

Euler's formula for solving y′=f(x,y),y(xˉ)=yˉy ^ { \prime } = f ( x , y ) , y ( \bar { x } ) = \bar { y }y′=f(x,y),y(xˉ)=yˉ​ is

Using the notation from the text, the finite difference equation for solving the boundary value problem y′′+P(x)y′+Q(x)y=f(x),y(a)=α,y(b)=βy ^ { \prime \prime } + P ( x ) y ^ { \prime } + Q ( x ) y = f ( x ) , y ( a ) = \alpha , y ( b ) = \betay′′+P(x)y′+Q(x)y=f(x),y(a)=α,y(b)=β is

The most popular fourth order Runge-Kutta method for the solution of yt=f(x,y),y(x0)=y0y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }yt=f(x,y),y(x0​)=y0​ is

The solution of y′′+xyy′=0,y(0)=0,y′(0)=1y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1y′′+xyy′=0,y(0)=0,y′(0)=1 for y(0.1)y ( 0.1 )y(0.1) , using the Runge-Kutta method of order four, and using h=0.1h = 0.1h=0.1 , is

Using the method from the previous problem, the solution of y′=x+y,y(0)=1 for y(0.2)y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )y′=x+y,y(0)=1 for y(0.2) with h=0.2h = 0.2h=0.2 is

A popular second order Runge-Kutta method for the solution of yt=f(x,y),y(x0)=y0y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }yt=f(x,y),y(x0​)=y0​ is

Introduction to Differential Equations

First-Order Differential Equations

Modeling With First-Order Differential Equations

Higher-Order Differential Equations

Modeling With Higher-Order Differential Equations

Series Solutions of Linear Equations

Laplace Transform

Systems of Linear First-Order Differential Equations

Plane Autonomous Systems

Orthogonal Functions and Fourier Series

Boundary-Value Problems in Rectangular Coordinates

Boundary-Value Problems in Other Coordinate Systems

Integral Transform Method

Numerical Solutions of Partial Differential Equations

Mathematics Problems: Differential Equations and Linear Algebra

Mathematical Problems and Solutions

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The solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using Euler's method with $h = 0.1$ is

The Euler's method solution for $y ( 0.2 )$ of $y ^ { \prime \prime } + y = 0 , y ^ { \prime } ( 0 ) = 1$ using $h = 0.1$ is

The solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using the improved Euler's method with $h = 0.1$ is

The solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ , using Euler's method with $h = 0.1$ is

Using the method from the previous problem, the solution of $y ^ { \prime } = y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

The standard backward difference approximation of $y ^ { \prime } ( x )$ is

In the previous problem, the local truncation error in $y _ { n + 1 }$ is

The Adams-Bashforth formula for finding the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

The problem $y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1$ can be written as a system of two equations as follows.

Using the value of $y _ { n + 1 } ^ { * }$ from the previous problem, the Adams-Moulton corrector value for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

Using the method from the previous problem, the solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

When entering the number $1 / 7$ into a three digit base ten calculator, the actual value entered is

Euler's formula for solving $y ^ { \prime } = f ( x , y ) , y ( \bar { x } ) = \bar { y }$ is

Using the notation from the text, the finite difference equation for solving the boundary value problem $y ^ { \prime \prime } + P ( x ) y ^ { \prime } + Q ( x ) y = f ( x ) , y ( a ) = \alpha , y ( b ) = \beta$ is

The most popular fourth order Runge-Kutta method for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is

The solution of $y ^ { \prime \prime } + x y y ^ { \prime } = 0 , y ( 0 ) = 0 , y ^ { \prime } ( 0 ) = 1$ for $y ( 0.1 )$ , using the Runge-Kutta method of order four, and using $h = 0.1$ , is

Using the method from the previous problem, the solution of $y ^ { \prime } = x + y , y ( 0 ) = 1 \text { for } y ( 0.2 )$ with $h = 0.2$ is

A popular second order Runge-Kutta method for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ is