Question 1

The solution of the integral equation $\int _ { 0 } ^ { - \infty } f ( x ) \cos ( a x ) d x = F ( \alpha )$ is&#10;A) $f ( x ) = \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha / \pi$&#10;B) $f ( x ) = 2 \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha / \pi$&#10;C) $f ( x ) = 2 \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha$&#10;D) $f ( x ) = \int _ { 0 } ^ { \infty } F ( \alpha ) \cos ( \alpha x ) d \alpha$&#10;E) none of the above

Accepted Answer

The correct solution involves the use of the Fourier cosine transform and its inverse. The given integral equation resembles the Fourier cosine transform, and its solution is obtained by applying the inverse Fourier cosine transform, which includes a factor of $2/\pi$.

Question 2

In the three previous problems, the solution for $u ( x , t )$ is&#10;A) $u = \sin ( \pi ) u ( t - x / a )$&#10;B) $u = \sin ( \pi ( t - x / a ) ) u ( t - x / a )$&#10;C) $u = \sin ( \pi x ) \cosh ( \pi t )$&#10;D) $u = \cos ( \pi x ) \sinh ( \pi t )$&#10;E) $u = \cos ( \pi ( t - x / a ) ) u ( t - x / a )$

Accepted Answer

$u = \sin ( \pi ( t - x / a ) ) u ( t - x / a )$&#10;

Question 3

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

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

f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi

Accepted Answer

The correct Fourier integral representation of $f(x)$ for a piecewise function like the one given, where $f(x) = 2$ for $0 \leq x \leq 3$ and $f(x) = 0$ otherwise, involves integrating over the interval where $f(x)$ is non-zero. The factor of $2/\pi$ is typical for the normalization in the Fourier sine integral representation for functions defined on a finite interval. The expression inside the integral correctly represents the Fourier sine components for the given function $f(x)$, and the factor of $2$ outside the integral accounts for the amplitude of $f(x)$ over its non-zero interval.

Question 4

In the previous problem, the solution for $U ( \alpha , t )$ is&#10;A) $e ^ { k a t } / \left( 1 + \alpha ^ { 2 } \right)$&#10;B) $e ^ { k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right)$&#10;C) $2 e ^ { k c s } / \left( 1 + \alpha ^ { 2 } \right)$&#10;D) $2 e ^ { - k c e } / \left( 1 + \alpha ^ { 2 } \right)$&#10;E) $2 e ^ { - k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right)$

Accepted Answer

The correct solution for $U(\alpha, t)$ typically involves an exponential decay term due to diffusion processes, which is represented by $e^{-k\alpha^2t}$, and a normalization factor, which is represented by $1/(1+\alpha^2)$. This matches the form of option E.

Question 5

In the previous problem, the integral converges at $x = 0$ to the value&#10;A) 0&#10;B) $1 / 2$&#10;C) 1&#10;D) 2&#10;E) $e$

Accepted Answer

Without the specific integral provided in the "previous problem," it's impossible to accurately determine the answer. However, based on common integrals that converge to specific values, one might guess or infer certain outcomes. The correct answer, in this case, would depend entirely on the details of the integral in question, which are not provided. The choice of "C" as the correct answer is arbitrary without the context of the integral.

Question 6

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is&#10;A) $1 / ( \sqrt { s } ( \sqrt { s } + 1 ) )$&#10;B) $1 / ( \sqrt { s } ( \sqrt { s } - 1 ) )$&#10;C) $1 / ( s ( \sqrt { s } + 1 ) )$&#10;D) $1 / ( s ( \sqrt { s } - 1 ) )$&#10;E) none of the above

Accepted Answer

The Laplace transform of $e^t \operatorname{erf}(\sqrt{t})$ is $1 / ( \sqrt { s } ( \sqrt { s } + 1 ) )$, where $\operatorname{erf}(\cdot)$ is the error function. This result can be derived using the convolution theorem or properties of the Laplace transform and the error function.

Question 7

The value of $\mathcal { L } \{ \operatorname { erfc } ( \sqrt { t } ) \}$ is&#10;A) $( 1 + 1 / \sqrt { s + 1 } ) / s$&#10;B) $( 1 + 1 / \sqrt { s - 1 } ) / s$&#10;C) $( 1 - 1 / \sqrt { s + 1 } ) / s$&#10;D) $( 1 - 1 / \sqrt { s - 1 } ) / s$&#10;E) none of the above

Accepted Answer

The Laplace transform of the complementary error function $\operatorname{erfc}(\sqrt{t})$ is $(1 - 1/\sqrt{s+1})/s$.

Question 8

In the previous two problems, the integral converges for $x < 0$ to the function&#10;A) $e ^ { - \mathrm { x } }$&#10;B) $e ^ { x }$&#10;C) $x$&#10;D) 1&#10;E) none of the above

Accepted Answer

The answer of In the previous two problems, the integral...

Question 9

In the previous three problems, the solution for $u ( x , t )$ is&#10;A) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha x } e ^ { k at } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / ( 2 \pi )$&#10;B) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha x } e ^ { - k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$&#10;C) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha \alpha } e ^ { k \alpha t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$&#10;D) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha x } e ^ { - k \alpha t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$&#10;E) $\int _ { - \infty } ^ { \infty } \left[ e ^ { - i \alpha \alpha } e ^ { k \alpha ^ { 2 } t } / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / ( 2 \pi )$

Accepted Answer

Option B is correct because it represents a solution to a differential equation that models diffusion or heat conduction, where $e^{-k\alpha^2t}$ is a typical term for the time-dependent part of the solution in the Fourier transform space, indicating diffusion or heat dissipation over time. The $e^{-i\alpha x}$ term represents the spatial component in the Fourier transform, and the denominator $(1+\alpha^2)$ could be part of the initial or boundary conditions transformed into the Fourier space. The division by $\pi$ is related to the normalization in the inverse Fourier transform.

Question 10

The complementary error function is defined as&#10;A) $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$&#10;B) $\operatorname { erfc } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$&#10;C) $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$&#10;D) $\operatorname { erfc } ( x ) = 2 \int _ { x } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$&#10;E) $\operatorname { erfc } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$

Accepted Answer

The complementary error function, $\operatorname{erfc}(x)$, is defined as $1 - \operatorname{erf}(x)$, where $\operatorname{erf}(x)$ is the error function. The correct definition involves an integral from $x$ to $\infty$ of $e^{-u^2}$ scaled by $\frac{2}{\sqrt{\pi}}$, which matches option D.

Question 11

In the previous problem, assume that $f ( x ) = \sin ( \pi t )$ . The solution for $U ( x , s )$ is&#10;A) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { s x / \alpha }$&#10;B) $$U = \frac { s } { s ^ { 2 } - \pi ^ { 2 } } e ^ { - s x / \alpha }$$&#10;C) $U = \frac { \pi } { s ^ { 2 } - \pi ^ { 2 } } e ^ { s x / \alpha }$&#10;D) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } e ^ { - s x / \alpha }$&#10;E) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sinh ( - s x / \alpha )$

Accepted Answer

The answer of In the previous problem, assume that \(f...

Question 12

Apply a Fourier transform in $x$ in the previous problem. The resulting equation for $U ( \alpha , t ) = \mathcal { F } \{ u ( x , t ) \}$ is&#10;A) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$&#10;B) $k \alpha U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$&#10;C) $k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 1 / \left( 1 + \alpha ^ { 2 } \right)$&#10;D) $- k \alpha ^ { 2 } U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$&#10;E) $- k \alpha U = U _ { t } , U ( \alpha , 0 ) = 2 / \left( 1 + \alpha ^ { 2 } \right)$

Accepted Answer

The answer of Apply a Fourier transform in $x$ in...

Question 13

The Fourier cosine integral of a function f defined on $[ 0 , \infty ]$ is&#10;A) $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$&#10;B) $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha / \pi$&#10;C) $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d x \right] \cos ( \alpha x ) d \alpha$&#10;D) $\int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x$&#10;E) $2 \int _ { 0 } ^ { \infty } \left[ \int _ { 0 } ^ { \infty } f ( x ) \cos ( \alpha x ) d \alpha \right] \cos ( \alpha x ) d x / \pi$

Accepted Answer

The answer of The Fourier cosine integral of a function...

Question 14

In the previous problem, the integral converges at $x = 0$ to the value&#10;A) 0&#10;B) 1&#10;C) 2&#10;D) 3&#10;E) 4

Accepted Answer

The answer of In the previous problem, the integral converges...

Question 15

In the previous problem, if $F ( \alpha ) = e ^ { - \alpha } \text {, then } f ( x )$ is&#10;A) $x$&#10;B) $1$&#10;C) $1 / \left( \pi \left( 1 + x ^ { 2 } \right) \right)$&#10;D) $2 / \left( \pi \left( 1 + x ^ { 2 } \right) \right)$&#10;E) 0

Accepted Answer

The answer of In the previous problem, if \(F (...

Question 16

Consider the temperature, $u ( x , t )$ , in an infinite rod $( - \infty < x < \infty )$ , with an initial temperature of $f ( x ) = e ^ { - | x | }$ . The mathematical model for this is&#10;A) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( x , 0 ) = e ^ { - | x | }$&#10;B) $k \frac { \partial u } { \partial x } = \frac { \partial u } { \partial t } , u ( x , 0 ) = e ^ { - | x | }$&#10;C) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = e ^ { - | x | }$&#10;D) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( x , 0 ) = e ^ { - | x | }$&#10;E) $k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = e ^ { - | x | }$

Accepted Answer

The answer of Consider the temperature, \(u ( x ,...

Question 17

In the previous problem, apply the Laplace transform. The resulting equation for $U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}$ is&#10;A) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = - s f ( x )$&#10;B) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = 0$&#10;C) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - f ( x )$&#10;D) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = s f ( x )$&#10;E) $a ^ { 2 } \frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = 0$

Accepted Answer

The answer of In the previous problem, apply the Laplace...

Question 18

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

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

Accepted Answer

The answer of Consider a semi-infinite, elastic, vibrating string, with...

Question 19

The value of $\lim _ { x \rightarrow \infty } \operatorname { erfc } ( x )$ is&#10;A) 0&#10;B) 1&#10;C) $\pi / 2$&#10;D) $\sqrt { \pi / 2 }$&#10;E) $\sqrt { \pi } / 2$

Accepted Answer

The answer of The value of \(\lim _ { x...

Question 20

The Fourier cosine integral of $f ( x ) = e ^ { - x } , x \geq 0$ is&#10;A) $f ( x ) = 2 \int _ { 0 } ^ { \infty } [ \alpha \cos ( \alpha x ) / ( 1 + \alpha ) ] d \alpha$&#10;B) $f ( x ) = \int _ { 0 } ^ { \infty } \left[ \alpha \cos ( \alpha x ) / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$&#10;C) $f ( x ) = \int _ { 0 } ^ { \infty } \left[ \alpha \cos ( \alpha x ) / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha$&#10;D) $f ( x ) = 2 \int _ { 0 } ^ { \infty } \left[ \cos ( \alpha x ) / \left( 1 + \alpha ^ { 2 } \right) \right] d \alpha / \pi$&#10;E) $f ( x ) = 2 \int _ { 0 } ^ { \infty } [ \cos ( \alpha x ) / ( 1 + \alpha ) ] d \alpha / \pi$

Accepted Answer

The answer of The Fourier cosine integral of \(f (...

Question 21

In the previous problem, assume that $f ( x ) = \sin ( \pi x )$ . The solution for $U ( x , s )$ is&#10;A) $U = \sinh ( s x ) + \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$&#10;B) $U = \cosh ( s x ) + \frac { s } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$&#10;C) $U = \frac { 1 } { s ^ { 2 } - \pi ^ { 2 } } \sin ( \pi x )$&#10;D) $U = \frac { s } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$&#10;E) $U = \frac { 1 } { s ^ { 2 } + \pi ^ { 2 } } \sin ( \pi x )$

Accepted Answer

The answer of In the previous problem, assume that \(f...

Question 22

The value of $\mathcal { L } \{ \operatorname { erf } ( \sqrt { t } ) \}$ is&#10;A) $1 / ( s \sqrt { s + 1 } )$&#10;B) $1 / ( s \sqrt { s - 1 } )$&#10;C) $1 / ( ( s + 1 ) \sqrt { s } )$&#10;D) $1 / ( ( s - 1 ) \sqrt { s } )$&#10;E) none of the above

Accepted Answer

The answer of The value of \(\mathcal { L }...

Question 23

The error function is defined as&#10;A) $\operatorname { erf } ( x ) = \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u$&#10;B) $\operatorname { erf } ( x ) = \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u$&#10;C) $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \pi$&#10;D) $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { \infty } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$&#10;E) $\operatorname { erf } ( x ) = 2 \int _ { 0 } ^ { x } e ^ { - u ^ { 2 } } d u / \sqrt { \pi }$

Accepted Answer

The answer of The error function is defined as&#10;A) \(\operatorname...

Question 24

Consider the problem of finding the temperature in a semi-infinite rod with zero initial temperature and a fixed constant temperature, $u _ { 0 }$ , at $x = 0$ . The mathematical model for this problem is

A)

k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }

B)

k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }

C)

k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = 0 , u ( 0 , t ) = u _ { 0 }

D)

k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( x , 0 ) = u _ { 0 } , u ( 0 , t ) = 0

E)

k \frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = u _ { 0 } , u ( 0 , t ) = 0

Accepted Answer

The answer of Consider the problem of finding the temperature...

Question 25

The value of $\mathcal { L } \left\{ \mathrm { e } ^ { t } \operatorname { erf } ( \sqrt { t } ) \right\}$ is&#10;A) $1 / ( \sqrt { s } ( s + 1 ) )$&#10;B) $1 / ( \sqrt { s } ( s - 1 ) )$&#10;C) $1 / ( s ( s + 1 ) )$&#10;D) $1 / ( s ( \sqrt { s } - 1 ) )$&#10;E) $1 / ( s ( \sqrt { s } + 1 ) )$

Accepted Answer

The answer of The value of \(\mathcal { L }...

Question 26

The value of $\lim _ { x \rightarrow \infty } \operatorname { erf } ( x )$ is&#10;A) 0&#10;B) 1&#10;C) $\pi / 2$&#10;D) $\sqrt { \pi / 2 }$&#10;E) $\sqrt { \pi } / 2$

Accepted Answer

The answer of The value of \(\lim _ { x...

Question 27

The Laplace transform of a function f is&#10;A) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { t } f ( t ) d t$&#10;B) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - t } f ( t ) d t$&#10;C) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { s t } f ( t ) d t$&#10;D) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { \infty } e ^ { - s t } f ( t ) d t$&#10;E) $\mathcal { L } \{ f ( x ) \} = \int _ { 0 } ^ { s } e ^ { s t } f ( t ) d t$

Accepted Answer

The answer of The Laplace transform of a function f...

Question 28

The Fourier integral representation of a function f is given by&#10;A) $\int _ { - \infty } ^ { \infty } f ( x ) \cos ( \alpha x ) d x$&#10;B) $\int _ { - \infty } ^ { \infty } f ( x ) \sin ( \alpha x ) d x$&#10;C) $\int _ { 0 } ^ { \infty } \left[ \int _ { - \infty } ^ { \infty } f ( t ) \cos ( \alpha t ) d t \cos ( \alpha x ) + \int _ { - \infty } ^ { \infty } f ( t ) \sin ( \alpha t ) d t \sin ( \alpha x ) \right] d \alpha$&#10;D) $\int _ { 0 } ^ { \infty } \left[ \int _ { - \infty } ^ { \infty } f ( t ) \cos ( \alpha t ) d t \cos ( \alpha x ) + \int _ { - \infty } ^ { \infty } f ( t ) \sin ( \alpha t ) d t \sin ( \alpha x ) \right] d \alpha / \pi$&#10;E) $\int _ { 0 } ^ { \infty } \left[ \int _ { - \infty } ^ { \infty } f ( t ) \cos ( \alpha t ) d t \cos ( \alpha x ) + \int _ { - \infty } ^ { \infty } f ( t ) \sin ( \alpha t ) d t \sin ( \alpha x ) \right] d \alpha / ( 2 \pi )$

Accepted Answer

The answer of The Fourier integral representation of a function...

Question 29

In the three previous problems, the solution for the temperature $u ( x , t )$ is&#10;A) $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$&#10;B) $u = u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$&#10;C) $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha$&#10;D) $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$&#10;E) $u = 2 u _ { 0 } \int _ { 0 } ^ { \infty } \left[ \sin ( \alpha x ) \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha \right] d \alpha / \pi$

Accepted Answer

The answer of In the three previous problems, the solution...

Question 30

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} , \text { then } \left\{ \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } \right\}$ is&#10;A) $U _ { s } ( x , s )$&#10;B) $s U ( x , s )$&#10;C) $s ^ { 2 } U ( x , s ) - u ( x , 0 )$&#10;D) $s ^ { 2 } U ( x , s ) - u ( x , 0 ) - s u _ { t } ( x , 0 )$&#10;E) $s ^ { 2 } U ( x , s ) - s u ( x , 0 ) - u _ { t } ( x , 0 )$

Accepted Answer

The answer of If \(U ( x , s )...

Question 31

In the previous problem, apply the Laplace transform. The resulting equation for $U ( x , s ) = \mathcal { L } \{ u ( x , t ) \}$ is&#10;A) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = - s f ( x )$&#10;B) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } + s ^ { 2 } U = f ( x )$&#10;C) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - f ( x )$&#10;D) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = s f ( x )$&#10;E) $\frac { \partial ^ { 2 } U } { \partial x ^ { 2 } } - s ^ { 2 } U = - s f ( x )$

Accepted Answer

The answer of In the previous problem, apply the Laplace...

Question 32

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.&#10;A) $\int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$&#10;B) $\int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}$&#10;C) $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$&#10;D) $\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )$&#10;E) none of the above

Accepted Answer

The answer of Suppose \(\mathcal { F } \{ f...

Question 33

The complex form of the Fourier integral of a function f is&#10;A) $\int _ { - \infty } ^ { \infty } f ( t ) e ^ { i \alpha t} d t$&#10;B) $\int _ { - \infty } ^ { \infty } \left( \int _ { - \infty } ^ { \infty } f ( t ) e ^ { i \alpha t } d t \right) e ^ { - i \alpha \alpha } d \alpha / ( 2 \pi )$&#10;C) $\int _ { - \infty } ^ { \infty } f ( t ) e ^ { - i \alpha t } d t$&#10;D) $\int _ { - \infty } ^ { \infty } \left( \int _ { - \infty } ^ { \infty } f ( t ) e ^ { - i \alpha t } d t \right) e ^ { i \alpha x } d \alpha / \pi$&#10;E) $\int _ { - \infty } ^ { \infty } \left( \int _ { - \infty } ^ { \infty } f ( t ) e ^ { \alpha t } d t \right) e ^ { - i \alpha x } d \alpha / \pi$

Accepted Answer

The answer of The complex form of the Fourier integral...

Question 34

In the previous problem, the integral representation converges at $x = 1$ to the value&#10;A) 0&#10;B) 1&#10;C) $1 / 2$&#10;D) -1&#10;E) none of the above

Accepted Answer

The answer of In the previous problem, the integral representation...

Question 35

In the three previous problems, the solution for $u ( x , t )$ is&#10;A) $u = \sin ( \pi x ) \sin ( \pi t )$&#10;B) $u = \sin ( \pi x ) \cos ( \pi t )$&#10;C) $u = \sin ( \pi x ) \cosh ( \pi t )$&#10;D) $u = \cos ( \pi x ) \sinh ( \pi t )$&#10;E) $u = \cos ( \pi x ) \cos ( \pi i )$

Accepted Answer

The answer of In the three previous problems, the solution...

Question 36

If $U ( x , s ) = \mathcal { C } \{ u ( x , t ) \} \text {, then } \mathcal { L } \left\{ \frac { \partial u } { \partial x } \right\}$ is&#10;A) $s U ( x , s ) - u ( x , 0 )$&#10;B) $s U ( x , s ) - u _ { x } ( x , 0 )$&#10;C) $s U ( x , s ) - s u ( x , 0 )$&#10;D) $U _ { x } ( x , s )$&#10;E) $$U _ { s } ( x , s )$$

Accepted Answer

The answer of If \(U ( x , s )...

Question 37

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

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

f ( x ) = 2 \int _ { 0 } ^ { \infty } \{ [ \sin ( \alpha x ) + \sin ( ( 3 - x ) \alpha ) ] / \alpha \} d \alpha / \pi

Accepted Answer

The answer of Let \(f ( x ) = \left\{...

Question 38

Consider the problem of a vibrating string, tightly-stretched between $x = 0$ and $x = 1$ , with a fixed initial position, $f ( x )$ , and zero initial velocity. The mathematical problem for the deflection, $u ( x , t )$ , is $( \text { with } u ( 0 , t ) = 0 , u ( 1 , t ) = 0 )$

A)

\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial u } { \partial t } , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0

B)

\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial u } { \partial t } = 0 , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0

C)

\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0

D)

\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } = 0 , u ( x , 0 ) = f ( x ) , u _ { t } ( x , 0 ) = 0

E)

\frac { \partial ^ { 2 } u } { \partial x ^ { 2 } } = \frac { \partial ^ { 2 } u } { \partial t ^ { 2 } } , u ( x , 0 ) = 0 , u _ { t } ( x , 0 ) = f ( x )

Accepted Answer

The answer of Consider the problem of a vibrating string,...

Question 39

In the previous problem, apply a Fourier sine transform in x. The new problem for the transform $U ( a , t )$ is&#10;A) $k \frac { \partial U } { \partial t } + \alpha ^ { 2 } U = k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$&#10;B) $k \frac { \partial U } { \partial t } - \alpha ^ { 2 } U = k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$&#10;C) $\frac { \partial U } { \partial t } + k \alpha ^ { 2 } U = k \alpha \psi _ { 0 } , U ( \alpha , 0 ) = 0$&#10;D) $\frac { \partial U } { \partial t } - k \alpha ^ { 2 } U = k \alpha \psi _ { 0 } , U ( \alpha , 0 ) = 0$&#10;E) $\frac { \partial U } { \partial t } + \alpha ^ { 2 } U = k \alpha u _ { 0 } , U ( \alpha , 0 ) = 0$

Accepted Answer

The answer of In the previous problem, apply a Fourier...

Question 40

In the previous problem, the solution is&#10;A) $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right)$&#10;B) $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right)$&#10;C) $U = u _ { 0 } \left( 1 + e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$&#10;D) $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha$&#10;E) $U = u _ { 0 } \left( 1 - e ^ { - k \alpha ^ { 2 } t } \right) / \alpha ^ { 2 }$

Accepted Answer

The answer of In the previous problem, the solution is&#10;A)...

Deck 14: Integral Transform Method