# How to solve Sturm-Liouville problem $y'' + (1 + \lambda)y = 0$?

Find all the eigenvalues and eigenfunctions of Sturm-Lioville problem: $$y'' + (1 + \lambda)y = 0$$ $$y(0) = y \left(\frac{\pi}{2}\right) = 0$$ Can someone please tell me how to solve this? Because most of the time, I mostly solve this type of Sturm-Liouville problem $y'' + \lambda y = 0$. And, my book is base on theory more than example. Thank you in advance.

• The only difference between the problem you usually solve and this one is the $1+\lambda$? OK, so let $1+\lambda = \mu$, solve the problem with $\mu$, and then take $\lambda = \mu - 1$. – Robert Israel Apr 27 '15 at 5:40

The general solution to $y''+(1+\lambda)y=0$ is $y=A\sin(kx)+B\cos(kx)$

where $k^2 = 1 + \lambda$

the condition $y(0) = 0$ gives $B=0$

then $y(\frac{\pi}{2})=0$ requires that k be an even integer $k=2n$

So for every positive integer $n$

$y_n = A \sin (2nx)$

$\lambda_n = 4n^2 -1$

• can you please be more specific. Because, I prefer how to solve rather than an answer – Alexander Apr 27 '15 at 5:20

It's exactly the same as $$y'' + \lambda y = 0$$

except that $\lambda$ is replaced by $\lambda + 1$

So the solution is essentially the same, just treat $\lambda +1$ as $\lambda$

For $\lambda + 1 < 0$, $y = e^{\sqrt{\lambda+1}\,x} + e^{-\sqrt{\lambda+1}\,x}$

For $\lambda+1 = 0$, $x^2 + x$

For $\lambda_1 > 0$, $y = \cos\sqrt{\lambda+1}\,x + \sin\sqrt{\lambda+1}\,x$

And so on