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I am stuck on how to approach this question:

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Solution:

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Any help is appreciated. Leo

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What is $L$? is there a part of the question missing? –  Amzoti Mar 4 '13 at 4:55
    
Well, what have you tried? What are your thoughts? –  Antonio Vargas Mar 4 '13 at 4:56
    
Well, I know that for only certain values of lambda I have non-trivial solutions; the formula in the book is: lambda=(n*pi/L)^2...This has to somehow be part of the solution. PS: My book is very poor at explaining and has absolutely no examples so I am stuck at square 1. –  RealityDysfunction Mar 4 '13 at 4:58
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2 Answers

Hints:

The characteristic equation is: $m^2 +\lambda = 0$

At $\lambda = 0$, we have: $m^2 = 0$, so the solution is: $y = x_0 + x_1 t$.

At $\lambda < 0$, we have: $m^2 - \lambda = 0$, thus we have an eigenvalue of $\pm \lambda$, so the solution is: $y = x_0 e^{\sqrt{-\lambda} t} + x_1e^{-\sqrt{-\lambda} t}$, where $-\lambda$ and $\sqrt{-\lambda}$ are positive.

At $\lambda > 0$, we have $m^2 + \lambda = 0$, thus we have a complex eigenvalue of $\pm i\lambda$, so the solution is: $y = x_0 \sin (\sqrt{\lambda}~t) + x_1\cos (\sqrt{\lambda}~t)$.

Can you use the BC's to find $x_0$ and $x_1$, find all three cases together to determine the value of $\lambda$ and finish this off?

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Well... the last case has imaginary roots. (I added solution in the back of the book to the question). –  RealityDysfunction Mar 4 '13 at 5:17
    
Great hints, Amzoti! –  amWhy Apr 26 '13 at 0:09
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I suggest you follow the advice in the question and analyze the three cases $\lambda > 0$, $\lambda = 0$ and $\lambda < 0$.

To get you started, assume $\lambda = 0$. Then $\phi'' = 0$ and the solutions are linear, i.e. $\phi = Ax + B$. This can't satisfy the boundary conditions (except for $A=B=0$, but that's not very interesting so we disregard it) so $\lambda = 0$ is not an eigenvalue.

Now check the other two cases.

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