I have the following:
$ y_{tt} = 25y_{xx},~0<x<\pi ,~t>0,;~y(0,t)=y(\pi,t)=0,~y(x,0)=y_{t}(x,0)=sin^2(x) \\25X''(x)+\lambda X(x)=0~~~~(1) \\T(t)''+\lambda T(t)=0~~~~(2) \\ X= sin(nx)~~~~(3) \\T=C_{1}cos(5nt)+C_{2}sin(5nt)~~~~(4) $ So: I obtained (3) and (4) by solving (1) and (2), respectively. I am not sure how to process (4) according to the initial conditions. Note: $ C_{1} and C_{2}$ are arbitrary constants. What I have been doing is, after plugging in $t=0$ getting $T=C_{2}sin(5nt)$ by assuming that since $u(x,t)=XT(0)$ that $T(0)$ must be equal to zero, and then solving the resulting homogenous equation. The next thing I think to do is $T'=C_{2}5nt*cos(5nt)$ but as far as figuring out how to make this equal $sin^2(x)$ as the conditions require I am lost. I am familiar with the use of Fourier sine series to determine what $C_{2}$ would be but I don't get the correct answer.
I do: $sin^2(x)=sin(nx) \\\frac{2}{\pi}\int_{0,\pi}sin^2(x)sin(nx)dx= C_{2} \\\frac{1}{\pi}\int_{0,\pi}sin(nx)-cos(2x)sin(nx)dx \\\frac{1}{\pi}[[\frac{1}{n}-cos(n\pi)+1]-\int_{0,\pi}[sin(2x+nx)+sin(2x-nx)dx]$
After this I don't know.
The solution to the homogeneous wave equation with fixed ends is
$$ \begin{align}\begin{cases} u_{tt} - c^{2} u_{xx} =0 & t > 0 , 0 < x < L \\ u(0,t) = u(L, t) = 0 & t> 0 \\ u(x,0) = f(x) , u_{t}(x,0) = g(x) & 0 < x < L \end{cases} \end{align} $$
$$ u(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \sin(\frac{n \pi x}{L})\cos(\frac{ n c\pi t}{L} )+ B_{n} \sin(\frac{n \pi x}{L}) \sin(\frac{n c \pi t}{L})\bigg) $$
$$ u(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \cos(\frac{ n c\pi t}{L} )+ B_{n} \sin(\frac{n c \pi t}{L})\bigg) \sin(\frac{n \pi x}{L}) $$
noting that $c^{2} = 25 $ and $L = \pi$ we have
$$ u(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \cos(\frac{ 5 n \pi t}{\pi} )+ B_{n} \sin(\frac{5n \pi t}{\pi})\bigg) \sin(\frac{n \pi x}{\pi}) $$
$$ u(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n} \cos(5 n t )+ B_{n} \sin(5nt)\bigg) \sin(n x) $$
The coefficients can be determined when
$$ A_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin^{2}(x) \sin(nx) dx = \frac{2}{\pi} \frac{2 \cos( \pi n) -2}{n^{3} -4n} = \frac{4}{\pi} \frac{\cos(n \pi) -1}{n(n^{2}-4)}$$
So for odd $n$ , $A_{n} = 0$ and $n =0, 2, -2$ don't exist.
Similarly
$$ 5 n B_{n} = \frac{2}{\pi} \int_{0}^{\pi} \sin^{2}(x) \sin(nx) dx = \frac{2}{5 \pi n} \frac{2 \cos( \pi n) -2}{n^{3} -4n} = \frac{4}{ 5 \pi n} \frac{\cos(n \pi) -1}{n(n^{2}-4)} $$