Boundary Value ODE: Eigenvalues & Functions

[WendysRules]

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.

 

[LCKurtz]

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 $\lambda = \frac{n^2\pi^2}{4L^2}$

$\cos\sqrt{\lambda}L = 0$ is what you want. But the cosine is not zero for all integer multiples of $\frac \pi 2$, just for odd multiples. Try an even multiple for example.

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.