# PDE separation of variables problem

$$u_{xx}+4u_{yy}=0, \quad u(0,y)=u(1,y)=0, \quad u=u(x,y)$$ This is what I've tried; $$u(x,y)=X(x)Y(y), \quad PDE: \frac{X''(x)}{X(x)}=-4\frac{Y''(y)}{Y(y)}=\lambda$$

Since I only have boundary conditions for $$X$$, I will only run through cases for this;

$$\lambda=0 \Longrightarrow X''(x)=0 \Longrightarrow X(x)=c_1 x+c_2$$ $$X(0)=0 \Longrightarrow c_2=0, \quad X(1)=0 \Longrightarrow c_1=0$$

We have shown that $$\lambda=0$$ gives a trivial solution.

$$\lambda=\omega^2 \Longrightarrow X''(x)-\omega^2X(x)=0 \Longrightarrow X(x)=c_1e^{\omega x}+c_2e^{-\omega x}$$ $$X(0)=0 \Longrightarrow c_1=-c_2, \quad X(x)= c_1(e^{\omega x}-e^{-\omega x})$$ $$X(1)=0 \Longrightarrow c_1(e^{\omega} - e^{-\omega})=0 \Longrightarrow c_1=0$$

We have shown that $$\lambda>0$$ gives a trivial solution.

$$\lambda=-\omega^2 \Longrightarrow X''(x)+\omega^2X(x)=0 \Longrightarrow X(x)=c_1cos\omega x+c_2sin\omega x$$ $$X(0)=0 \Longrightarrow c_1=0$$ $$X(1)=0 \Longrightarrow c_2sin\omega = 0 \Longrightarrow \omega=n\pi, n=0,1,2,...$$

We have shown that $$\lambda>0 \Longrightarrow u(x,y)=csin(n\pi x)Y(y)$$, now we just need to find $$Y(y)$$.

$$Y''(y)-4\omega^2 Y(y)=0 \Longrightarrow Y(y)=c_1 e^{2\omega y} + c_2 e^{-2\omega y}$$

Finally, I would like to write my solution as

$$u(x,y)=X(x)Y(y)=c_1 sin(n\pi x)(c_2e^{2\omega y} + c_3e^{-2\omega y})$$

When I look at the solution, it says;

$$u^{\pm}_n = sin(n\pi x)e^{\pm \frac{n\pi y}{2}}$$

I can't seem to understand where I have gone wrong, the exponent doesn't seem to match the answer, and neither does the fact that I have 2 terms but they only have 1.

The equation for $$Y$$ is $$Y''=-\lambda Y /4$$. You have taken it as $$Y''=-4\lambda Y$$. When they gave the solution they just mentioned two linearly independent solutions. The general solution is the set of all linear combinations of the two solutions and that is what you have obtained.