# Boundary conditions for my ODE give trivial answers only.

I want to solve the homogenous part of a stretched string problem where $y=y(x)$.

$$y'' + y = 0$$

with the boundary conditions such that: $y(0)=y(\pi/2)=0$

The differential equation gives rise to a solution on the form: $$y = a \cos(x) + b \sin(x)$$

But when applying the boundary conditions I end up with only trivial solution ($a=b=0$).

Have I made a mistake or does these B.C only lead to $a=b=0$?

• I do not see any question. – gerw Dec 4 '15 at 18:19
• What's the problem? – Tryss Dec 4 '15 at 18:21

## 3 Answers

As other people said, the only solution to the problem as it is written now is the trivial one. But perhaps you misread the exercise and the boundary conditions are $y(0) = y(\pi) = 0$ or $y(0) = y(2\pi)=0$? In that case you will have non-trivial solutions.

Usually this problem in physics--with such boundary conditions--has ODE $$y'' + k^2y = 0$$ The problem then turns into finding values of $k$ for which the boundary conditions are met. In this case we find infinitely many discrete $k$, corresponding to the fundamental and the harmonics above:

$$y_k(x) = A_k\sin(kx), \quad k = 2m, \text{ where } m \text{ is a positive integer }$$

If the string is constrained to move according to $$y'' + y = 0$$ alone, then the problem only admits the trivial solution. In the language of physics, $k = 1$ is not a mode of the system.

• I think in case of this equation we just get zero! :) – H. R. Dec 4 '15 at 18:55

Yes, the only solution to that boundary value problem is $y(x)=0$.