The Laplace equation is; $$u_{xx}+u_{yy}=0$$the solutions satisfies the boundary conditions:$$u(x,0)=f(x), \quad u(x,1)=0, \quad u_x(0,y)=0, \quad u_x(1,y)=0$$$f(x)$ is some given integrate function. We want to solve via separation of variables so we know our solution looks like $$u(x,y)=X(x)Y(y)$$ by combining these two we get,$$\frac{X''}{X}=-\frac{Y''}{Y}=\lambda$$ where lambda is just some constant. This gives way to the following two equations:$$X''-\lambda X=0 \quad \& \quad Y''+\lambda Y =0 $$We can solve these equations and apply the boundary conditions in the normal way the first one gives us,$$X_n=A_ncos(\pi n x), \quad \lambda_n=- \pi^2n^2 $$$$Y_n=B_n \sinh(\pi n (y-1)) \quad \rm{for\;} n=0,1,2,3...$$Now we can get our normal modes which are$$u_n=C_ncos(\pi n x)sinh(\pi n(y-1))$$and our solution is$$u(x,y)=\sum_{n=0}^{\infty}E_ncos(\pi n x)sinh(\pi n (y-1))$$
so here's my problem i've been told that the solution should look like this$$u(x,y)=E_0(1-y)+\sum^{\infty}_{n=1}cos(\pi n x)sinh(\pi n (y-1))$$
And im really confused i know i must have made a mistake somewhere but for the life of me i cant tell what went wrong whats confusing me is where that $E_0(1-y)$ is coming from?
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$\begingroup$You forgot the $\lambda=0$ case. It needs to be treated seperately.
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