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## Homework Statement

u

_{xx}+u

_{yy}=0

u(x,0)=u(x,pie)=0

u(0,y)=0

u

_{x}(5,y)=3siny-5sin4y

## Homework Equations

## The Attempt at a Solution

Using separable method I get

Y"-kY= 0 and X"+kX=0

For Case 1 and Case 2 where k>0 and k=0 there are no eigenvalues

So Case 3 k<0 gives

Y=ccos(sqrk x y) + dsin(sqrk x y)

y=0, then c=0

y=pie, then dsin(pie x sqrk)=0

k= -n

^{2}, Y

_{n}=sin(ny)

X"-n

^{2}X=0, X=ccosh(nx)+dsinh(nx)

u(x,y)= summation (from n=1 to n=infinity) sin(ny)[c

_{n}cosh(nx)+d

_{n}sinh(nx)]

u(0,y)= summation (from n=1 to n=infinity) c

_{n}sin(ny)

Using Fourier's Series to expand 0, I got 0 so c

_{n}is 0

u

_{x}(x,y)= summation (from n=1 to n=infinity) sin(ny)[d

_{n}ncosh(nx)]

u

_{x}(x,0)= summation (from n=1 to n=infinity) sin(ny)[d

_{n}n]

Using Fourier's Series to expand 3siny-5sin4y, I also got 0 so d

_{n}is 0 as well

Therefore u(x,y)=0

However the answer given in the book is

u(x,y)=(3/cosh5)(siny)(sinhx) - (5/cosh20)(sin4y)(sinh4x)

I realized that the book didn't used Fourier's Series but compared the coefficients of the two equations instead. But my question is what happened to the n because if we compare the coefficients aren't we supposed to get

[3/ncosh(5n)] n=1 so we get [3/cosh(5)]

[-5/nsinh(5n)] n=4 so we get [-5/4sinh(20)] so u(x,y) should be

[3/cosh(5)](siny)(sinhx) - [5/4sinh(20)](sin4y)(sinh4x)

Did I do it correctly or is it just a printing error?