y''+λy = 0; y'(-π)=0, y'(π)=0 ... y''+λy = 0; y(-2)=0; y'(2)=0?

Question

I'm stuck on these two problems.

Someone showed me how to do one of these the other day when I asked, and I thought I would be able to do the rest after getting an answer, but I don't see what to do when the initial conditons don't have a zero in the parentheses.

btw - the answers the book gives are:

for the first one:

Eigenvalue λ[subscript 0]=0 with eigenfunction y(x)[subscript 0]≡1, and positive eigenvalues {n^2/4} for n = 1,2,3 ...

The nth eigenfunction y[subscript n](x) is sin(nx/2) if n is odd, cos(nx/2) if n is even.

for the second one:

Only positive eigenvalues {n^2π^2/64} for n = 1,2,3...

The nth eigenfunction y[subscript n](x) is cos(nπx/8)+sin(nπx/8) if n is odd, cos(nπx/8)-sin(nπx/8) if n is even.

Any help would be greatly appreciated.

Answer 1

Again if lamda is positive =w^2
the general solution is
y=A cos(w x-phi)
y´= -wAsin m(w x-phi)
y(-2)= 0 so -2w-phi= (2n+1)pi/2
y´(2)=-wA sin(2w-phi)=0 so at 2 the solution has a max or min
so y(2) =A or -A
2w-phi =k*pi
2w-kpi = -2w -(2n+1)pi/2
4w= kpi-npi -pi/2 =pi*(k-n-1/2) so 4w/pi +1/2 must be integer
which in general is not true.
If lambda is negative = -w^2
y= A e^wx +B e^-wx
y´= wA e^wx -wB e^wx

0= Ae^-2w +B e^2w
0=wA e^2w -wB e^-2w

Ie^-2w e^2w I
Iw e^2w -w e^-2wI = w(-e^-4w -e^4w) not zero so you can calculater A and B
we^2w -we^-2w I

Answer 2

wwwwwwwwwoooooooooooooowwwwwwwwww. I have no idea i am so sorry i would help you if i could. :(