I have this problem:
I know how to solve this problems with separation of variables, but I'm in trouble with this specific equation because of the initials conditions. I'm expecting to have something like that: $$X(x)=a\cos(ax)+b\sin(ax)$$Where $\lambda=\alpha^{2}$ a costant. And for the temporal part $$T(t)=ce^{-\lambda t}$$ I don't know how to use the initial condition specified for the first derivative of the themporal part. Could someone help me? Thank you, as always!!!.
If you try separated solutions $X(x)T(t)$, then you end up with $$ X''(x)=\lambda X(x),\;\; T''(t)=\lambda T(t) \\ X(-\pi)=X(\pi)=0. $$ One of the problems with such problems is that there a scaling degree a freedom, but you can eliminate that scale factor by choose $X$ so that $X'(-\pi)=1$. That also makes the the eigenfunctions have a power series expansion in $\lambda$, which is a good thing. So, start by solving $$ X''(x)=\lambda X,\;\;\; X'(-\pi)=1,\; X(-\pi)=0 \\ \implies X(x) = \frac{1}{\sqrt{\lambda}}\sin(\sqrt{\lambda}(x+\pi)). $$ The eigenvalues are then determined by the following equation in $\lambda$: $$ 0=X(\pi)=\frac{1}{\sqrt{\lambda}}\sin(2\pi\sqrt{\lambda})=0. $$ So $\lambda=0$ is not an eigenvalue because the above equation remains valid in the limiting case as $\lambda\rightarrow 0$ and gives $2\pi$, which is non-zero; this follows because of how the eigenfunctions have a power series expansion in $\lambda$ when normalized. In fact, you can get the special case solution for $\lambda=0$ as a limit of that power series in $\lambda$ by using L'Hospital's rule: $$ X(x) = x+\pi. $$ Proper normalization always gets rid of the pesky special cases; they become limiting cases in $\lambda$. The above is a solution of the equation $X''=0X$ with $X(-\pi)=0$ and $X'(-\pi)=1$.
The zeros of $\sin$ in the complex plane are where $$ 2\pi\sqrt{\lambda} = n\pi \\ \lambda = \frac{n^2}{4},\;\;\; n=1,2,3,\cdots. $$ The corresponding eigenfunctions are $$ X_n = \sin(\frac{n}{2}(x+\pi)),\;\;\; n=1,2,3,\cdots \\ T_n = A_n\cos(\frac{n}{2}t)+B_n\sin(\frac{n}{2}t). $$ The full solution is $$ u(t,x)=\sum_{n=0}^{\infty}T_n(t)X_n(x). $$ The constants $A_n$, $B_n$ are determined by the initial conditions