The question is Solve the one dimensional wave equation

$\displaystyle \frac{1}{c^2}$

on $\displaystyle 0<=x<=3$, subject to the conditions

$\displaystyle u(x,0) = \frac{1}{2}x$, $\displaystyle u_t(x,0)=x(3-x), u(0,t)=u(3,t)=0$

well i started it, but im not sure if this is right coz i don't really understand wave or heat equations SIGH...

$\displaystyle

u_tt(x,t) = X''(x)T(t) $ and $\displaystyle u_tt(x,t) = X(x)T''(t)

$

$\displaystyle \frac{1}{c^2} $

Case 1let $\displaystyle v = w^2 > 0 $

$\displaystyle

\frac{X''(x)}{X(x)} = w^2 $

$\displaystyle

X''(x) = w^2 X(x) = 0 $

Using Auxilliary Equations:

$\displaystyle

m^2 - w^2 = 0$

So $\displaystyle m = +- w $

And $\displaystyle X(x) = C_1 e^{wx} + C_2 e^{-wx}$

As $\displaystyle u(0,t) = u(3,t) = 0, X(0) = V(3) = 0$

Therefore, $\displaystyle C_1 + C_2 = 0$ and $\displaystyle C_1 = - C_2$

Therefore, no solutions...

Case 2$\displaystyle v = w^2=0$

$\displaystyle X''(x) - 0X(x) = 0$

Auxilliary Equations $\displaystyle m^2 - 0 = 0$, $\displaystyle m = +- 0 $

Therefore, $\displaystyle X(x) = C_3 e^0 + C_4 e^{-0} $

As $\displaystyle X(0) = X(2) = 0, X(0) = C_3, X(3) = 2C_3 + C_4 $

So $\displaystyle 2C_3 + C_4 = 0 $

Therefore, no solution as $\displaystyle 2C_3 = -C_4$

Case 3$\displaystyle v = -w^2$ as $\displaystyle v<0$

$\displaystyle

\frac{X''(x)}{X(x)} = -w^2 $

$\displaystyle X''(x) + w^2X(x) = 0 $

Auxilliary Equation: $\displaystyle m^2 + w^2 = 0, m = +-iw $\

So $\displaystyle C_5coswx +C_6sinwx = 0$

$\displaystyle C_5 = 0, $ So $\displaystyle

C_5cos3w + C_6sin3w = 0$

$\displaystyle sin3w = 0$

Therefore, $\displaystyle 3w = \frac{n\Pi}{3}$ and $\displaystyle V = \frac{n^2\Pi^2}{9}$

Thus, $\displaystyle X(x) = B sin $ $\displaystyle \frac{xn\Pi}{3}$

Now Im lost, am i doing it right so far, if so, then how do i go about completing this, coz i have no idea what to do now.. Thank you