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## Homework Statement

Compute the eigenvalues/functions of the given regular S-L problem

f''(x)+λf(x)=0

0<x<π

f(0)=0

f'(π)=0

**2. The attempt at a solution**

First off, why is π not included in the given boundary if it tells you f'(x) at π?

Now for my attempt:

assuming λ=0

f''(x)+0*f(x)=0

f''(x)=0

f'(x)=a, and f'(π)=0 so a=0

f(x)=ax+b, but a=0 so f(x)=b, and f(0)=0 so b also =0

therefore

f(x)=ax+b=0+0=0

so this doesn't really help

assuming λ>0

f''(x)+λf(x)=0

char eqn: r^2+λ=0

r=±√(λ)i

f(x)=c1*cos(√(λ)x)+c2*sin(√(λ)x)

f(0)=0 so plug in 0 ->f(0)=c1*cos(0)+c2*sin(0)c1=c1=0, therefore c1=0

now we are left with

f(x)=c2*sin(√(λ)x)

f'(x)=c2*√(λ)*sin(√(λ)x)

f'(π)=c2*√(λ)*sin(√(λ)π)=0

we can conclude than c2=0 or λ=0 but that wouldn't get us anywhere

instead, we assume

sin(√(λ)π)=0

√(λ)π=n*π (where n=1,2,...)

√(λ)=n

λ=n^2

so for λ>0 we have an eigenvalue of λ=n^2 where n=1,2,... and an eigenfunction of f(x)=c2*sin(nx) (plug n^2 into λ)

the last case is λ<0

f''(x)+λf(x)=0

char eqn: r^2+λ=0

r=±√(λ)

f(x)=c1*e^(√(λ)x)+c2*x*e^(√(λ)x)

f(0)=c1*e^0+c2*0*e^0=c1=0, therefore c1=0

we are left with

f(x)=c2*x*e^(√(λ)x)

f'(x)=c2*x*√(λ)*e^(√(λ)x)+c2*e^(√(λ)x)

f'(π)=c2*π*√(λ)*e^(√(λ)π)+c2*e^(√(λ)π)=0

we can factor it to make it a little more manageable

f'(π)=c2*e^(√(λ)π)*(√(λ)π+1)

once again, we can see that c2 and λ=0, but are trivial

assume that the last term =0

√(λ)π+1=0

√(λ)π=-1

√(λ)=-1/π

λ=1/π^2

(here, can I specify that only the negative root is used and not the positive one?)

so this gives us an eigenvalue of λ=1/π^2 and an eigenfunction of f(x)=c2*x*e^(±x/π)

The answer in the back of the book is

λ=(2n-1)^2 /4

f(x)=sin((2n-1)x/2) n=1,2.....

My answer looks nothing like that so what am I misunderstanding?