Consider the equation $u _ { xx } - u _ { t } = 0$ with conditions $u ( 0 , t ) = 0 , u _ { x } = ( L , t ) = 0 , u ( x , 0 ) = f ( x )$ . When separating variables with $u ( x , t ) = X ( x ) T ( t )$ , the resulting problems for $X , T$ are
A) $X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } - \lambda T = 0 , T ( 0 ) = f ( x )$
B) $X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0 , T ( 0 ) = f ( x )$
C) $X ^ { \prime \prime } - \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0$
D) $X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } - \lambda T = 0$
E) $X ^ { \prime \prime } + \lambda X = 0 , X ( 0 ) = 0 , X ^ { \prime } ( L ) = 0 , T ^ { \prime } + \lambda T = 0$

Question

Quizplus · Accepted Answer

Substituting $u(x,t) = X(x)T(t)$ into the given PDE $u_{xx} - u_t = 0$ and dividing by $XT$ leads to $\frac{X''}{X} = \frac{T'}{T}$, implying both sides equal a constant $\lambda$. The boundary conditions for $X(x)$ are directly given as $X(0) = 0$ and $X'(L) = 0$. The condition $T'(0) = f(x)$ is not applicable since $T$ is a function of $t$ only, not $x$. Thus, the correct separation leads to $X'' + \lambda X = 0$ with the given boundary conditions for $X$, and $T' + \lambda T = 0$ for $T$, without an initial condition involving $f(x)$ directly in the separated equations.

Consider the Equation uxx−ut=0u _ { xx } - u _ { t } = 0uxx​−ut​=0

Consider the Equation $u _ { xx } - u _ { t } = 0$