- #1
WendysRules
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Homework Statement
Find the eigenvalues and eigenfunctions of the following boundary-value problem.
## y''+\lambda y = 0 ; y(0) = 0, y'(L) = 0 ##
Homework Equations
The Attempt at a Solution
So, we have to test when lambda is equal to, less than and greater than 0.
Let ## \lambda = 0## thus, the ODE becomes ## y'' = 0 ## which implies solutions of the form ## y(t) = C_1t+C_2 ## which would make the derivative ## y'(t) = C_1 ##. When we apply our boundary conditions we see that ## y(0) = C_2 = 0 ## and ##y'(L) = C_1 = 0 ## which are trivial solutions.
Let ## \lambda < 0 ## thus, the ODE becomes ## y''- \lambda y = 0 ## which implies solutions of the form ## y(t) = C_1e^{\sqrt{\lambda}t}+C_2e^{-\sqrt{\lambda}t} ## which would make the derivative ## y'(t) = \sqrt{\lambda}C_1e^{\sqrt{\lambda}t}-\sqrt{\lambda}C_2e^{-\sqrt{\lambda}t} ##. When we apply our boundary conditions we see that ## y(0) = C_1+C_2 = 0 ## therefore ## C_2=-C_1 ## and ## y'(L) = \sqrt{\lambda}C_1e^{\sqrt{\lambda}L}-\sqrt{\lambda}C_2e^{-\sqrt{\lambda}L} = \sqrt{\lambda}C_1e^{\sqrt{\lambda}L}+\sqrt{\lambda}C_1e^{-\sqrt{\lambda}L} = C_1[e^{\sqrt{\lambda}L}+e^{-\sqrt{\lambda}L} = 0 ## Since the exponentials can't ever be zero, it implies that ## C_1 = 0 ## this, these solutions are trivial.
Let ## \lambda > 0 ## thus, the ODE becomes ## y''+\lambda y = 0 ## which implies solutions of the form ## y(t) = C_1\cos{\sqrt{\lambda}t}+ C_2\sin{\sqrt{\lambda}t}. ##Which would make the derivative ## y'(t) = -\sqrt{\lambda}C_1\sin{\sqrt{\lambda}t} + \sqrt{\lambda}C_2\cos{\sqrt{\lambda}t} ##. When I apply my boundary conditions, we see that ## y(0) = C_1 = 0 ## and ## y'(L) = C_2\cos\sqrt{\lambda}L = 0 ## Now we solve for ## \lambda ## to see that ## \sqrt{\lambda} L = \frac{n\pi}{2} ## which makes ## \lambda = \frac{n^2\pi^2}{4L^2} ##
Thus, our solution for this boundary problem would be ## y(t) = C\sin\frac{n\pi}{2L}t ## for ## \lambda = \frac{n^2\pi^2}{4L^2}## but the solution in the book is ## y(t) = C\sin\frac{(2n+1)\pi}{2L}t ## for ## \lambda = \frac{(2n+1)^2\pi^2}{4L^2}## which I don't understand why we are limiting our n to be odd?
Any help would be appreciated.