# Boundary value problem

Consider the DE: $y''$ + $2\lambda y'$ + $\lambda^2$$y = 0 subject to boundary conditions: y(1) + y'(1) = 0 and 3y(2) + 2y'(2) = 0. The problems asks to find eigenvalues and eigenfunctions of the given BVP. My approach: Obviously, The characteristic eqn gives r = - \lambda. So general solution is y(x) = c_1e^{-\lambda x} + c_2xe^{-\lambda x}. and by applying boundary condition, I obtain the following system: c_1(1 - \lambda) + c_2(2 - \lambda) = 0 c_1(3 - 2\lambda) + 4c_2(1 - \lambda) = 0 From here, Im having difficulties in obtaining eigenvalues. I would to ask If my approach is correct or If I am probably doing something incorrect. Is there a better way to solve this problem? thanks - You mean y(x) = c_1 e^{-\lambda x} + c_2 xe^{-\lambda x}. – Stefan Smith Oct 13 '12 at 0:30 corrected...... – Mann Oct 13 '12 at 0:30 You need to correct your system. It should contain e^{-\lambda} and e^{-2\lambda}. – Stefan Smith Oct 13 '12 at 1:10 @bogus can you explain how they would be in the equation. I think they cancel out – Mann Oct 13 '12 at 1:13 Yeah, you're right. Sorry. I'm used to problems where the exponential doesn't disappear. – Stefan Smith Oct 13 '12 at 1:17 show 1 more comment ## 1 Answer Now, you have the system$$ 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}\,.$$- Are you sure these are solutions for lambda? – Mann Oct 13 '12 at 0:55 @LJym89:Try calculate the determinant. – Mhenni Benghorbal Oct 13 '12 at 0:58$4(1 - \lambda)^2 - (2 - \lambda)(3 - 2 \lambda) = 0$– Mann Oct 13 '12 at 0:59 @LJym89: Your are right. There is a mistake. – Mhenni Benghorbal Oct 13 '12 at 0:59$\lambda = 1/4 +- \sqrt{17}/4\$ I think these are the only two solutions. –  Mann Oct 13 '12 at 1:01
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