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

Question

Quizplus · Accepted Answer

The correct solution involves an integral over all positive frequencies, accounting for the initial condition $u_0$, and includes a normalization factor of $1/\pi$. This is a common form for solutions to partial differential equations using Fourier transforms, where the $1/\pi$ factor comes from the normalization of the sine function in the integral representation.

In the Previous Two Problem, the Solution For u(x,t)u ( x , t )u(x,t)

In the Previous Two Problem, the Solution For $u ( x , t )$