Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck on how to approach this question:

enter image description here

enter image description here


enter image description here

Any help is appreciated. Leo

share|cite|improve this question
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


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?

share|cite|improve this answer
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.