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Consider the DE: $y''$ + $2\lambda y'$ + $\lambda^2$$y$ = $0$ subject to boundary conditions: $y(1) + y'(1) = 0$ and $3y(2) + 2y'(2) = 0$. The problems asks to find eigenvalues and eigenfunctions of the given BVP.

My approach: Obviously, The characteristic eqn gives $r = - \lambda$. So general solution is $y(x) = c_1e^{-\lambda x} + c_2xe^{-\lambda x}$. and by applying boundary condition, I obtain the following system:

$c_1(1 - \lambda) + c_2(2 - \lambda) = 0$

$c_1(3 - 2\lambda) + 4c_2(1 - \lambda) = 0$

From here, Im having difficulties in obtaining eigenvalues. I would to ask If my approach is correct or If I am probably doing something incorrect. Is there a better way to solve this problem?

thanks

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  • $\begingroup$ You mean $y(x) = c_1 e^{-\lambda x} + c_2 xe^{-\lambda x}$. $\endgroup$ – Stefan Smith Oct 13 '12 at 0:30
  • $\begingroup$ You need to correct your system. It should contain $e^{-\lambda}$ and $e^{-2\lambda}$. $\endgroup$ – Stefan Smith Oct 13 '12 at 1:10
  • $\begingroup$ @bogus can you explain how they would be in the equation. I think they cancel out $\endgroup$ – ILoveMath Oct 13 '12 at 1:13
  • $\begingroup$ Yeah, you're right. Sorry. I'm used to problems where the exponential doesn't disappear. $\endgroup$ – Stefan Smith Oct 13 '12 at 1:17
  • $\begingroup$ @LJym89 The eigenvalues are just the zeros of the characteristic equation, and you already have those ($-\lambda$). You also found the eigenfunctions ($e^{-\lambda x}$ and $xe^{-\lambda x}$. And your general method of finding a solution for particular boundard value conditions also seems sound (didn't check if the equations are correct, though). So, once you solve for $c_1$,$c_2$, what else do you want to do? I'd say the problem is done then... $\endgroup$ – fgp Oct 13 '12 at 2:11
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Now, you have the system

$$ c_1(1 - \lambda) + c_2(2 - \lambda) = 0 \,,$$

$$ c_1(3 - 2\lambda) + 4c_2(1 - \lambda) = 0 \,. $$

Since you are solving for $c_1$ and $c_2$, then in order to get a non trivial solution for the system, you need to assume the determinant equals to zero. Doing that, you get the following values for lambda

$$ \lambda = \frac{3}{4}-\frac{i}{4}\sqrt{7}, \frac{3}{4}+\frac{i}{4}\sqrt{7}\,. $$

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  • $\begingroup$ Are you sure these are solutions for lambda? $\endgroup$ – ILoveMath Oct 13 '12 at 0:55
  • $\begingroup$ @LJym89:Try calculate the determinant. $\endgroup$ – Mhenni Benghorbal Oct 13 '12 at 0:58
  • $\begingroup$ $4(1 - \lambda)^2 - (2 - \lambda)(3 - 2 \lambda) = 0$ $\endgroup$ – ILoveMath Oct 13 '12 at 0:59
  • $\begingroup$ @LJym89: Your are right. There is a mistake. $\endgroup$ – Mhenni Benghorbal Oct 13 '12 at 0:59
  • $\begingroup$ $\lambda = 1/4 +- \sqrt{17}/4$ I think these are the only two solutions. $\endgroup$ – ILoveMath Oct 13 '12 at 1:01

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