# Eigenvalue problem?

"Solve the eigenvalue problem or show that it has no solution: $y'' + 2y = x$ for $y(0) = y(\pi) = 0$"

I have managed to find a solution to the boundary value problem by finding the complementary solution by substituting $y=e^{\lambda x}$ getting $y=c_1(cos(\sqrt2 x) + c_2 sin (\sqrt2 x)$. Then finding the particular solution $y_p (x) = x/2$. Summing them and then substituting the initial values to get $y = 1/2(x - csc(\sqrt2 \pi)sin(\sqrt2 x))$.

Whilst I am confident this is the correct solution I was a bit thrown by the 'solve the eigenvalue problem.' The only time I have come across this before it included the actual term '$\lambda$' and and then solved depending if it was positive, 0 or negative. Is there a different way of solving this problem that I am missing out? Thanks.

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Is the problem statement missing something, like $\lambda$? The current solution pre-initial conditions is incorrect and the answer is also incorrect. –  Amzoti Nov 9 '13 at 22:21
My apologies, the problem should read $y′′+2y=x for y(0)=y(π)=0$. I shall edit that now. Hopefully now my answer should work, since I have double checked it and everything seems okay. However there is no $\lambda$ still, which is where I am confused. Have you ever seen a problem like this before? –  yhu Nov 9 '13 at 23:53
Are you sure it is not $2 \lambda y$? Also, they could be talking about the eigenvalues. Okay, your solution is closer, but still incorrect. There should be a $\pi$ in front of $\csc(\sqrt{2}x)$. –  Amzoti Nov 9 '13 at 23:56
Not unless there is a mistake in the question, but the next question below it is the same format so I am doubtful. And sorry I don't follow...? –  yhu Nov 9 '13 at 23:59
You solution should be $y = 1/2(x - \pi \csc(\sqrt2 \pi) \sin(\sqrt2 x))$. –  Amzoti Nov 10 '13 at 0:00