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I have a pde: $y_{xx}-y_{tt}=4$. By using the substitution $v=x-t, u=x+t$ I have boiled it down to $y(x,t)=a(x+t)+b(x-t)+x^2-t^2$ however I have initial conditions $y_t(x,0)=0$ and $y(x,0)=sin(x)$.

I just can't make them fit together! I have tried combinations of squaring $x+t$ and $x-t$ with various sines and cosines all over the place but I just can't see it. If I could have some help that would be great and also the thought process would be good too.

The solution I got with general functions may very well be wrong, I expanded out partial derivatives wrt u and v then put it all together.

Mmm
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The general solution you found is correct.

Now you should apply the approach like in http://en.wikipedia.org/wiki/D%27Alembert%27s_formula:

$y(x,0)=\sin x$ :

$a(x)+b(x)+x^2=\sin x$

$a(x)+b(x)=\sin x-x^2~......(1)$

$y_t(x,t)=a_t(x+t)+b_t(x-t)-2t=a_x(x+t)-b_x(x-t)-2t$

$y_t(x,0)=0$ :

$a_x(x)-b_x(x)=0$

$a(x)-b(x)=c~......(2)$

$\therefore a(x)=\dfrac{\sin x-x^2+c}{2}$ , $b(x)=\dfrac{\sin x-x^2-c}{2}$

$\therefore y(x,t)=\dfrac{\sin(x+t)-(x+t)^2+c}{2}+\dfrac{\sin(x-t)-(x-t)^2-c}{2}+x^2-t^2=\sin x\cos t-2t^2$

doraemonpaul
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