How to solve $y''(t)+λy(t)=0 $

Question

How to solve this Sturm-Liouville problem: $$y''(t)+λy(t)=0 $$ $$y(0)=y'(\pi) ,y'(0)=y(\pi)$$

Would really appreciate a solution or a significant hint because I couldn't find anything

Answer 1

Whats the first thing that comes to mind? maybe make a guess? define $$\mathbb{R} \ni t \mapsto y(t):=\sum_{n=0}^{\infty}a_nt^n$$ and see what happens when you plug that in, and start comparing coefficients

Answer 2

I'm assuming that $\lambda$ here is a real constant. In that case, break down the problem into three cases, depending on whether $\lambda$ is positive, negative, or zero. In the last case, you will get a linear function which is identically zero due to the boundary conditions. In the first and second cases, the general form of solutions is given by $c_1 \cos(\sqrt{\lambda}t)+c_2 \sin(\sqrt{\lambda}t)$ and $c_1 e^{\sqrt{-\lambda}t}+c_1 e^{-\sqrt{-\lambda}t}$, respectively. Applying the boundary conditions gives $2 \times 2$ linear systems for the coefficients $c_1, c_2$, which may be solved explicitly (and uniquely) conditional on the Jacobian being non-singular.

The boundary conditions also seem to give you a hint that you're probably looking for trigonometric solutions. But also, try and see how these behave when applied to exponentially growing and decaying solutions and whether that gives a triviality, as is often the case in such posed boundary value problems.

Answer 3

As the equation is linear with constant coefficients, we solve the characteristic polynomial

$$r^2+\lambda=0,$$ which gives the general solution

$$y(t)=c_+e^{\sqrt{-\lambda}t}+c_-e^{-\sqrt{-\lambda}t}.$$

We also have

$$y'(t)=\sqrt{-\lambda}\,c_+e^{\sqrt{-\lambda}t}-\sqrt{-\lambda}\,c_-e^{-\sqrt{-\lambda}t}.$$

Now plugging the given conditions,

$$c_++c_-=\sqrt{-\lambda}\,c_+e^{\sqrt{-\lambda}\pi}-\sqrt{-\lambda}\,c_-e^{-\sqrt{-\lambda}\pi},$$

$$\sqrt{-\lambda}\,c_+-\sqrt{-\lambda}\,c_-=c_+e^{\sqrt{-\lambda}\pi}+c_-e^{-\sqrt{-\lambda}\pi}.$$

This homogeneous linear system has non-trivial solutions when

$$\begin{vmatrix}1-\sqrt{-\lambda}\,e^{\sqrt{-\lambda}\pi}&1+\sqrt{-\lambda}\,e^{-\sqrt{-\lambda}\pi}\\\sqrt{-\lambda}-e^{\sqrt{-\lambda}\pi}&-\sqrt{-\lambda}-e^{-\sqrt{-\lambda}\pi}\end{vmatrix}=0$$

or

$$(\lambda+1)(e^{\sqrt{-\lambda}2\pi}-1)=0.$$