Question 1

In the previous problem, assume that $f ( x ) = \sin ( \pi x )$ . The solution for $U ( x , s )$ is

Accepted Answer

A)  $U = \sinh ( s x ) + \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$ 
B)  $U = \cosh ( s x ) + \frac { s } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$ 
C)  $U = \frac { 1 } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$ 
D)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$ 
E)  $U = \frac { 1 } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$ 
A)  $U = \sinh ( s x ) + \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$ 
B)  $U = \cosh ( s x ) + \frac { s } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$ 
C)  $U = \frac { 1 } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$ 
D)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$ 
E)  $U = \frac { 1 } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$ D

Question 2

In the three previous problems, the solution for the temperature $u ( x , t )$ is

Accepted Answer

A)  $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$ 
B)  $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$ 
C)  $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$ 
D)  $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$ 
E)  $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$ 
A)  $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$ 
B)  $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$ 
C)  $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$ 
D)  $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$ 
E)  $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$ E

Question 3

Let $f ( x ) = \left\{ \begin{array} { c c c } 
0 & \text { if } & x  1
\end{array} \right\}$ . The Fourier integral representation of f is $f ( x ) = \int _ { 0 } ^ { \infty } [ A ( \alpha ) \cos ( \alpha x ) + B ( \alpha ) \sin ( \alpha x ) ] d \alpha / \pi$ , where

Accepted Answer

A)  $A ( \alpha ) = \sin \alpha / \alpha \text { and } B ( \alpha ) = \cos \alpha / \alpha$ 
B)  $$A ( \alpha ) = 2 \cos \alpha / \alpha \text { and } B ( \alpha ) = 2 \sin \alpha / \alpha$$ 
C)  $A ( \alpha ) = 2 \sin \alpha / \alpha \text { and } B ( \alpha ) = 2 \cos \alpha / \alpha$ 
D)  $A ( \alpha ) = 0 \text { and } B ( \alpha ) = 2 \sin \alpha / \alpha$ 
E)  $A ( \alpha ) = 2 \sin \alpha / \alpha \text { and } B ( \alpha ) = 0$ 
A)  $A ( \alpha ) = \sin \alpha / \alpha \text { and } B ( \alpha ) = \cos \alpha / \alpha$ 
B)  $$A ( \alpha ) = 2 \cos \alpha / \alpha \text { and } B ( \alpha ) = 2 \sin \alpha / \alpha$$ 
C)  $A ( \alpha ) = 2 \sin \alpha / \alpha \text { and } B ( \alpha ) = 2 \cos \alpha / \alpha$ 
D)  $A ( \alpha ) = 0 \text { and } B ( \alpha ) = 2 \sin \alpha / \alpha$ 
E)  $A ( \alpha ) = 2 \sin \alpha / \alpha \text { and } B ( \alpha ) = 0$ E

Question 4

Let $f ( x ) = \left\{ \begin{array} { c c c } 
0 & \text { if } & x  3
\end{array} \right\}$ . The Fourier integral representation of f is

Accepted Answer

A)  $f ( x ) = \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$ 
B)  $f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha$ 
C)  $f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$ 
D)  $$f ( x ) = \int _ { 0 } ^ { \infty } \{ [ \sin ( a x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha$$ 
E)  none of the above 
A)  $f ( x ) = \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$ 
B)  $f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha$ 
C)  $f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi$ 
D)  $$f ( x ) = \int _ { 0 } ^ { \infty } \{ [ \sin ( a x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha$$ 
E)  none of the above

Question 5

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is

Accepted Answer

A)  $1 / ( \sqrt { s } ( s + 1 ) )$ 
B)  $1 / ( \sqrt { s } ( s - 1 ) )$ 
C)  $1 / ( s ( s + 1 ) )$ 
D)  $1 / ( s ( \sqrt { s } - 1 ) )$ 
E)  $1 / ( s ( \sqrt { s } + 1 ) )$ 
A)  $1 / ( \sqrt { s } ( s + 1 ) )$ 
B)  $1 / ( \sqrt { s } ( s - 1 ) )$ 
C)  $1 / ( s ( s + 1 ) )$ 
D)  $1 / ( s ( \sqrt { s } - 1 ) )$ 
E)  $1 / ( s ( \sqrt { s } + 1 ) )$

Question 6

In the previous problem, the integral representation converges at $x = 1$ to the value

Accepted Answer

A)  0 
B)  1 
C)  $1 / 2$ 
D)  -1 
E)  none of the above 
A)  0 
B)  1 
C)  $1 / 2$ 
D)  -1 
E)  none of the above

Question 7

In the previous problem, the solution is

Accepted Answer

A)  $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right)$ 
B)  $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right)$ 
C)  $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$ 
D)  $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$ 
E)  $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha ^ { 2 }$ 
A)  $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right)$ 
B)  $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right)$ 
C)  $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$ 
D)  $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$ 
E)  $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha ^ { 2 }$

Question 8

In the three previous problems, the solution for $u ( x , t )$ is

Accepted Answer

A)  $u = \sin ( \pi ) u ( t - x / a )$ 
B)  $u = \sin ( \pi ( t - x / a ) ) u ( t - x / a )$ 
C)  $u = \sin ( \pi x ) \cosh ( \pi t )$ 
D)  $u = \cos ( \pi x ) \sinh ( \pi t )$ 
E)  $u = \cos ( \pi ( t - x / a ) ) u ( t - x / a )$ 
A)  $u = \sin ( \pi ) u ( t - x / a )$ 
B)  $u = \sin ( \pi ( t - x / a ) ) u ( t - x / a )$ 
C)  $u = \sin ( \pi x ) \cosh ( \pi t )$ 
D)  $u = \cos ( \pi x ) \sinh ( \pi t )$ 
E)  $u = \cos ( \pi ( t - x / a ) ) u ( t - x / a )$

Question 9

Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$ . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

Accepted Answer

A)  $\int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$ 
B)  $\int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$ 
C)  $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$ 
D)  $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$ 
E)  none of the above 
A)  $\int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$ 
B)  $\int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$ 
C)  $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$ 
D)  $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$ 
E)  none of the above

Question 10

In the previous problem, assume that $f ( x ) = \sin ( \pi t )$ . The solution for $U ( x , s )$ is

Accepted Answer

A)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { s x / \alpha }$ 
B)  $$U = \frac { s } { s ^ { 2 } - \pi ^ { 2 } } e ^ { - s x / \alpha }$$ 
C)  $U = \frac { \pi } { s ^ { 2 } - \pi ^ { 2 } } e ^ { s x / \alpha }$ 
D)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { - s x / \alpha }$ 
E)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sinh ( - s x / \alpha )$ 
A)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { s x / \alpha }$ 
B)  $$U = \frac { s } { s ^ { 2 } - \pi ^ { 2 } } e ^ { - s x / \alpha }$$ 
C)  $U = \frac { \pi } { s ^ { 2 } - \pi ^ { 2 } } e ^ { s x / \alpha }$ 
D)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { - s x / \alpha }$ 
E)  $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sinh ( - s x / \alpha )$

Question 11

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} , \text { then } \left\{ \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } \right\}$ is

Accepted Answer

A)  $U _ { s } ( x , s )$ 
B)  $s U ( x , s )$ 
C)  $s ^ { 2 } U ( x , s ) - u ( x , 0 )$ 
D)  $s ^ { 2 } U ( x , s ) - u ( x , 0 ) - s u _ { t } ( x , 0 )$ 
E)  $s ^ { 2 } U ( x , s ) - s u ( x , 0 ) - u _ { t } ( x , 0 )$ 
A)  $U _ { s } ( x , s )$ 
B)  $s U ( x , s )$ 
C)  $s ^ { 2 } U ( x , s ) - u ( x , 0 )$ 
D)  $s ^ { 2 } U ( x , s ) - u ( x , 0 ) - s u _ { t } ( x , 0 )$ 
E)  $s ^ { 2 } U ( x , s ) - s u ( x , 0 ) - u _ { t } ( x , 0 )$

Question 12

Consider a semi-infinite, elastic, vibrating string, with zero initial position and velocity, driven by a vertical force at $x = 0$ , so that $u ( 0 , t ) = f ( t )$ . Assume that $\lim _ { x \rightarrow \infty } u ( x , t ) = 0$ . The mathematical model for the deflection, $u ( x , t )$ , is

Accepted Answer

A)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$ 
B)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = f ( t ) , u _ { t } ( x , 0 ) = 0$ 
C)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$ 
D)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$ 
E)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = f ( t ) , u _ { t } ( x , 0 ) = 0$ 
A)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$ 
B)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = f ( t ) , u _ { t } ( x , 0 ) = 0$ 
C)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$ 
D)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = 0$ 
E)  $\alpha ^ { 2 } \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( 0 , t ) = f ( t ) , u ( x , 0 ) = f ( t ) , u _ { t } ( x , 0 ) = 0$

Question 13

In the previous problem, the integral converges at $x = 0$ to the value

Accepted Answer

A)  0 
B)  $1 / 2$ 
C)  1 
D)  2 
E)  $e$ 
A)  0 
B)  $1 / 2$ 
C)  1 
D)  2 
E)  $e$

Question 14

Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is

Accepted Answer

A)  $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$ 
B)  $k \alpha U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$ 
C)  $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$ 
D)  $- k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$ 
E)  $- k \alpha U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$ 
A)  $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$ 
B)  $k \alpha U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$ 
C)  $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$ 
D)  $- k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$ 
E)  $- k \alpha U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$

Question 15

The Fourier cosine integral of a function f defined on $[ 0 , \infty ]$ is

Accepted Answer

A)  $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$ 
B)  $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$ 
C)  $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha$ 
D)  $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x$ 
E)  $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x / \pi$ 
A)  $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$ 
B)  $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$ 
C)  $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha$ 
D)  $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x$ 
E)  $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x / \pi$

Question 16

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$ is

Accepted Answer

A)  $s U ( x , s ) - u ( x , 0 )$ 
B)  $s U ( x , s ) - u _ { x } ( x , 0 )$ 
C)  $s U ( x , s ) - s u ( x , 0 )$ 
D)  $U _ { x } ( x , s )$ 
E)  $$U _ { s } ( x , s )$$ 
A)  $s U ( x , s ) - u ( x , 0 )$ 
B)  $s U ( x , s ) - u _ { x } ( x , 0 )$ 
C)  $s U ( x , s ) - s u ( x , 0 )$ 
D)  $U _ { x } ( x , s )$ 
E)  $$U _ { s } ( x , s )$$

Question 17

The error function is defined as

Accepted Answer

A)  $\operatorname { erf } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$ 
B)  $\operatorname { erf } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$ 
C)  $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$ 
D)  $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$ 
E)  $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$ 
A)  $\operatorname { erf } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$ 
B)  $\operatorname { erf } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$ 
C)  $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$ 
D)  $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$ 
E)  $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$

Question 18

The value of $\mathcal { L } \{ \operatorname { erfc } ( \sqrt { t } ) \}$ is

Accepted Answer

A)  $( 1 + 1 / \sqrt { s + 1 } ) / s$ 
B)  $( 1 + 1 / \sqrt { s - 1 } ) / s$ 
C)  $( 1 - 1 / \sqrt { s + 1 } ) / s$ 
D)  $( 1 - 1 / \sqrt { s - 1 } ) / s$ 
E)  none of the above 
A)  $( 1 + 1 / \sqrt { s + 1 } ) / s$ 
B)  $( 1 + 1 / \sqrt { s - 1 } ) / s$ 
C)  $( 1 - 1 / \sqrt { s + 1 } ) / s$ 
D)  $( 1 - 1 / \sqrt { s - 1 } ) / s$ 
E)  none of the above

Question 19

The Laplace transform of a function f is

Accepted Answer

A)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { t } f ( t ) d t$ 
B)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - t } f ( t ) d t$ 
C)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { s t } f ( t ) d t$ 
D)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - s t } f ( t ) d t$ 
E)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { s } e ^ { s t } f ( t ) d t$ 
A)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { t } f ( t ) d t$ 
B)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - t } f ( t ) d t$ 
C)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { s t } f ( t ) d t$ 
D)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - s t } f ( t ) d t$ 
E)  $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { s } e ^ { s t } f ( t ) d t$

Question 20

The complementary error function is defined as

Accepted Answer

A)  $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$ 
B)  $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$ 
C)  $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$ 
D)  $\operatorname { erfc } ( x ) = 2 \int _ { x } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$ 
E)  $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$ 
A)  $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$ 
B)  $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$ 
C)  $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$ 
D)  $\operatorname { erfc } ( x ) = 2 \int _ { x } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$ 
E)  $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$

In the previous problem, assume that $f ( x ) = \sin ( \pi x )$ . The solution for $U ( x , s )$ is

In the three previous problems, the solution for the temperature $u ( x , t )$ is

Let $f ( x ) = \left\{ \begin{array} { c c c } 0 & \text { if } & x < 0 \\2 & \text { if } & 0 \leq x \leq 3 \\0 & \text { if } & x > 3\end{array} \right\}$ . The Fourier integral representation of f is

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is

In the previous problem, the integral representation converges at $x = 1$ to the value

In the previous problem, the solution is

In the three previous problems, the solution for $u ( x , t )$ is

Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$ . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

In the previous problem, assume that $f ( x ) = \sin ( \pi t )$ . The solution for $U ( x , s )$ is

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} , \text { then } \left\{ \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } \right\}$ is

Consider a semi-infinite, elastic, vibrating string, with zero initial position and velocity, driven by a vertical force at $x = 0$ , so that $u ( 0 , t ) = f ( t )$ . Assume that $\lim _ { x \rightarrow \infty } u ( x , t ) = 0$ . The mathematical model for the deflection, $u ( x , t )$ , is

In the previous problem, the integral converges at $x = 0$ to the value

Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is

The Fourier cosine integral of a function f defined on $[ 0 , \infty ]$ is

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$ is

The error function is defined as

The value of $\mathcal { L } \{ \operatorname { erfc } ( \sqrt { t } ) \}$ is

The Laplace transform of a function f is

The complementary error function is defined as

Introduction to Differential Equations

First-Order Differential Equations

Modeling With First-Order Differential Equations

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Laplace Transform

Systems of Linear First-Order Differential Equations

Numerical Solutions of Ordinary Differential Equations

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Exam 14: Integral Transform Method

In the previous problem, assume that f(x)=sin⁡(πx)f ( x ) = \sin ( \pi x )f(x)=sin(πx) . The solution for U(x,s)U ( x , s )U(x,s) is

In the three previous problems, the solution for the temperature u(x,t)u ( x , t )u(x,t) is

The value of L{eterf⁡(t)}\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}L{eterf(t​)} is

In the previous problem, the integral representation converges at x=1x = 1x=1 to the value

In the previous problem, the solution is

In the three previous problems, the solution for u(x,t)u ( x , t )u(x,t) is

Suppose F{f(x)}=F(α),F{g(x)}=G(α)\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )F{f(x)}=F(α),F{g(x)}=G(α) . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

In the previous problem, assume that f(x)=sin⁡(πt)f ( x ) = \sin ( \pi t )f(x)=sin(πt) . The solution for U(x,s)U ( x , s )U(x,s) is

If U(x,s)=C{u(x,t)}, then {∂2u∂t2}U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} , \text { then } \left\{ \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } \right\}U(x,s)=C{u(x,t)}, then {∂t2∂2u​} is

In the previous problem, the integral converges at x=0x = 0x=0 to the value

Apply a Fourier transform in xxx in the previous problem. The resulting equation for U(α,t)=F{u(x,t)}U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}U(α,t)=F{u(x,t)} is

The Fourier cosine integral of a function f defined on [0,∞][ 0 , \infty ][0,∞] is

If U(x,s)=C{u(x,t)}, then L{∂u∂x}U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}U(x,s)=C{u(x,t)}, then L{∂x∂u​} is

The error function is defined as

The value of L{erfc⁡(t)}\mathcal { L } \{ \operatorname { erfc } ( \sqrt { t } ) \}L{erfc(t​)} is

The Laplace transform of a function f is

The complementary error function is defined as

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

Numerical Solutions of Ordinary Differential Equations

Plane Autonomous Systems

Orthogonal Functions and Fourier Series

Boundary-Value Problems in Rectangular Coordinates

Boundary-Value Problems in Other Coordinate Systems

Numerical Solutions of Partial Differential Equations

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In the previous problem, assume that $f ( x ) = \sin ( \pi x )$ . The solution for $U ( x , s )$ is

In the three previous problems, the solution for the temperature $u ( x , t )$ is

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is

In the previous problem, the integral representation converges at $x = 1$ to the value

In the three previous problems, the solution for $u ( x , t )$ is

Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$ . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

In the previous problem, assume that $f ( x ) = \sin ( \pi t )$ . The solution for $U ( x , s )$ is

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} , \text { then } \left\{ \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } \right\}$ is

In the previous problem, the integral converges at $x = 0$ to the value

Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is

The Fourier cosine integral of a function f defined on $[ 0 , \infty ]$ is

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$ is

The value of $\mathcal { L } \{ \operatorname { erfc } ( \sqrt { t } ) \}$ is