I have to solve$ U_{xx}+U_{yy}=0$ with $u(0,y)=u(a,y)=u(x,b)=0,u(x,0)=f(x)$.
$0<=x<=a$,$0<=y<=b$ .
I let $u(x,t)=X(x)Y(y)$.
Then $X''(x)Y(y)+X(x)Y''(y)=0$.
Then I took the minus sign to the $X''$ side and wrote as
$-X''(x)-\lambda X(x)=0 $
$Y''(y)-\lambda Y(y)=0$.
Is there a difference as to where I include the minus sign.
Should I take it to $Y''$ part.
In this article page 4
they have done it with $Y''$.
To get the non trivial solution when $\lambda>0$ letting $\lambda=\omega^2$
I came up with $\omega={n\pi\over a}$.
$X(x)=Bsin({n\pi x\over a})$.
Now I am struggling with Y(y) to bring it to $sinh$ as is in that article.
Since $Y''(y)-\lambda Y(y)=0$
$Y(y)=C e^{\omega y}+D e^{ -\omega y}$.
Since $u(x,b)=0$, I get $Y(b)=0$.
$Y(b)=C e^{\omega b}+D e^{ -\omega b}=0$. gives $C=-De^{-2\omega b}$.
Thus $Y(y)=De^{\omega y}(-e^{-2\omega b}+e^{-2\omega y})$.
Have I done correctly so far.What can I do to bring it to the form Y(y)=$sinh(\omega(y-b))$ as is the answer in that article