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The PDE question is ∂^{2}u/∂x^{2}+∂^{2}u/∂y^{2}=0

and u(x,0)=f(x), u(x,1)=0, u(0,y)=0, u(1,y)=0.

I use the method of separate the variables, with is let u(x,y)=X(x)Y(y) and get X''/X+Y''/Y=0.

Then let X''/X=-Y''/Y=-λ, i.e. X''+λX=0, and Y''-λY=0.

Since u(x,0)=f(x)=X(x)Y(0), u(x,1)=0=X(x)Y(1), u(0,y)=0=X(0)Y(y), u(1,y)=0=X(1)Y(y), we can conclude that X(0)=0, X(1)=0, Y(0)=f(x), Y(1)=0.

For X''+λX=0, we have 3 cases

1) λ=0, i.e. X''=0, i.e. X=ax+b, since X(0)=0, X(1)=0, a=b=0, which is X(x) always 0. not wavelike.

2) λ<0, Let λ=-μ^{2}, i.e. X''-μ^{2}X=0, General solution is X=a_{1}e^{μx}+a_{2}e^{-μx}. Since X(0)=0, X(1)=0, a_{1}=a_{2}=0, which is X(x) always 0. not wavelike.

3) λ>0, Let λ=μ^{2}, i.e. X''+μ^{2}X=0, General solution is X=a_{1}cos(μx)+a_{2}sin(μx). Since X(0)=0, X(1)=0, a_{2}sinμ=0.

In order to get a nontrivial solution, let μ=n∏, then X(x)=sin(n∏x).

Then we consider Y''-λY=0, i.e. Y''-n^{2}∏^{2}Y=0.

The general solution is Y=b_{1}e^{n∏y}+b_{2}e^{-n∏y}.

Then since Y(1)=0, b_{1}e^{n∏}+b_{2}e^{-n∏}=0.

Now, here is my question.

Why b_{1}+b_{2}e^{-2n∏}=0?

and How it change in the black font? Y_{h}(y)=-b_{2}e^{-2n∏}e^{n∏y}+b_{2}e^{-n∏y}

=-b_{2}e^{-n∏}e^{-n∏}e^{n∏y}+b_{2}e^{-n∏}e^{n∏}e^{-n∏y}

=-b_{2}'e^{-n∏(1-y)}+b_{2}'e^{n∏(1-y)}

=b_{2}'(e^{n∏(1-y)}-e^{-n∏(1-y)})

=2b_{2}''sin(hn∏(1-y))

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# Need help with solving PDE

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