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

Question

Quizplus · Accepted Answer

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

In the Previous Problem, Apply a Fourier Sine Transform in X