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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 $a=b=0$?

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

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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.

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  • $\begingroup$ I think in case of this equation we just get zero! :) $\endgroup$ – H. R. Dec 4 '15 at 18:55
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Yes, the only solution to that boundary value problem is $y(x)=0$.

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