Deck 9: Numerical Solutions of Ordinary Differential Equations

Question 1

When entering the number $1 / 7$ into a three digit base ten calculator, the round-off error is

A) 0.00143
B) 0.000143
C) $1 / 70$
D) $1 / 700$
E) $1 / 7000$

Answer 1

$1 / 7000$

Question 2

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

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

Answer 2

$0.005 y ^ { \prime \prime } ( c ) \text {, where } x _ { n } < c < x _ { n + 1 }$

Question 3

The improved Euler's method is what type of Runge-Kutta method?

A) first order
B) second order
C) third order
D) fourth order
E) It is not a Runge-Kutta method

Answer 3

B

Question 4

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

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 5

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

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

Question 6

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

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

Question 7

Using the Adams-Bashforth-Moulton method from the previous three problems, the solution of $y ^ { \prime } = y , y ( 0 ) = 1$ for $y ( 0.4 )$ with $h = 0.1$ is

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

Question 8

Using the Adams-Bashforth method from the previous problem, and using the values $y _ { 0 } = 1 , y _ { 1 } = 1.1052 , y _ { 2 } = 1.2214 , y _ { 3 } = 1.3499$ the solution $y ^ { \prime } = y , y ( 0 ) = 1$ for $y _ { n + 1 } ^ { * } = y ( 0.4 )$ with $h = 0.1$ is

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

Question 9

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

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 10

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

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

Question 11

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

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 12

Which of the following are second order Runge-Kutta methods for the solution of $y ^ { t } = f ( x , y ) , y \left( x _ { 0 } \right) = y _ { 0 }$ ? Select all that apply.

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

Question 13

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

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

Question 14

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

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 15

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

A) 1.2055
B) 1.21625
C) 1.24205
D) 1.226525
E) 1.235625

Question 16

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

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

Question 17

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

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 18

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

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

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

Question 20

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

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

Question 21

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

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

Question 22

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

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

Question 23

The Euler formula for solving the system $y ^ { t } = u , u ^ { t } = f ( x , y , u ) , y \left( x _ { 0 } \right) = y _ { 0 } , u \left( x _ { 0 } \right) = u _ { 0 }$ is

A) $y _ { n + 1 } = y _ { n } + h u _ { n } , u _ { n + 1 } = u _ { n } + h f \left( x _ { n } , y _ { n } , u _ { n } \right)$
B) $y _ { n + 1 } = y _ { n } - h u _ { n } , u _ { n + 1 } = u _ { n } + h f \left( x _ { n } , y _ { n } , u _ { n } \right)$
C) $y _ { n + 1 } = y _ { n } + h u _ { n } , u _ { n + 1 } = u _ { n } - h f \left( x _ { n } , y _ { n } , u _ { n } \right)$
D) $y _ { n + 1 } = y _ { n } + h f \left( x _ { n } , y _ { n } , u _ { n } \right) , u _ { n + 1 } = u _ { n } + h u _ { n }$
E) none of the above

Question 24

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

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

Question 25

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

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

Question 26

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

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

Question 27

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

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

Question 28

When entering the number $1 / 3$ into a three digit base ten calculator, the actual value entered is

A) $1 / 3$
B) $.333$
C) $.334$
D) $.300$
E) $.330$

Question 29

The fourth order Runge-Kutta method for solving $y ^ { \prime \prime } = f \left( x , y , y ^ { \prime } \right) , y \left( x _ { 0 } \right) = y _ { 0 } , y ^ { \prime } \left( x _ { 0 } \right) = u _ { 0 }$ is

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

Question 30

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.

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 31

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

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 32

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

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 33

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

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 34

When entering the number $1 / 3$ into a three digit base ten calculator, the round-off error is

A) $1 / 30$
B) $1 / 300$
C) $1 / 3000$
D) $0.003$
E) $0.0003$

Question 35

Euler's method is what type of Runge-Kutta method?

A) first order
B) second order
C) third order
D) fourth order
E) It is not a Runge-Kutta method

Question 36

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

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

Question 37

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

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

Question 38

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

A) 1.2
B) 1.21
C) 1.214
D) 1.22
E) 1.24

Question 39

Using Euler's method on the previous problem and using a value of $h = 0.1$ , the solution for $y ( 0.2 )$ is

A) 0.11
B) 0.2
C) 0.21
D) 0.22
E) 0.221

Question 40

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

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