I'm having trouble understanding some of the concepts related to these problems. Here's an example I'm working on:
$$y''+(\lambda+1)y=0 ; y'(0)=0,y'(1)=0$$
The characteristic equation I found was given by: $m^2 +\lambda +1=0$, which gave me $$y(x)=c_1 \cos(\sqrt{-\lambda+1}x) +c_2 \sin(\sqrt{-\lambda+1}x) $$
and
$$y'(x)= -\sqrt{-\lambda+1} c_1 \sin(\sqrt{-\lambda+1}x) + \sqrt{-\lambda+1} c_2 \cos(\sqrt{-\lambda+1}x) $$
Plugging in the boundary values I get:
$y'(0)=0$ which gives $ c_2=0$ and for $y'(1)=0$, $-\sqrt{-\lambda+1} c_1 \sin(\sqrt{-\lambda+1})=0$
But I'm not sure what to do past this point. I'm not sure what it is I'm looking for. Any insight would be great.
Best Answer
You're almost there. $$ y'' + (\lambda + 1) y = 0 \implies m^2 + (\lambda + 1) = 0 \implies m = \pm i\mu \implies \\ y = \left \{ \begin{array}{ll} A \cos \mu x + B \sin \mu x, && \text{if } \lambda \ne -1 \\ Dx + C, && \text{if } \lambda = -1 \end{array} \right . $$ where $\mu = \sqrt{\lambda + 1}$.
After applying BCs, second solution becomes trivial as $y = C$, so I assume $\lambda \ne -1$, then $$ y' = -\mu \left( -A\sin \mu x + B \cos \mu x\right) $$
Now, use BCs per your post, $$ y'(0) = 0 \implies B = 0,\quad \text{so}\ y' = -\mu A \sin \mu x \\ y'(1) = 0 \implies \mu A \sin \mu = 0 \implies \mu_n = \pi n $$ where $n \in (0, \infty)$. One can find that $\lambda_n = n^2 \pi^2 - 1$.
Trivial solution can be incorporated into this solution if we take $A_0 = C$, so $$ y = A_n \cos \mu_n x = A_n \cos n \pi x, \qquad n \in [0, +\infty). $$