# Finding Eigenvalues for $y''+\lambda y=0$ with boundary conditions.

Given the equation $y''+\lambda y=0$ and boundary conditions $y(1)=0$ and $y(0)+y'(0)=0$.

Let $r=\pm\sqrt{-\lambda}.$

If $\lambda >0$ we have $y=C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x).$ Applying our boundary conditions I get to this point: $-\sqrt{\lambda}=\tan(\sqrt{\lambda}).$ At this point I'm not sure what to do .

Similarly for the case of $\lambda <0$. This means that $r=\pm\sqrt{\lambda}$. I believe this means I have to apply the conditions to $y=C_1e^{\sqrt{\lambda}x}+C_2e^{-\sqrt{\lambda}x}$.

I want to make sure my approach for the second case is correct and I'm not sure how to finish up the first case.

• I think the derivative was messed up. $y'(0)=C_1+C_2\sqrt{\lambda}$. Also, $y(1)=0$. – emka Nov 13 '14 at 15:48
• I have those equations. I'm not sure how to handle the cases that arise from $\lambda$ being positive and negative. – emka Nov 13 '14 at 15:59
• I see that I made a sign error with. I'm still yielding $\sqrt{\lambda}=tan(\sqrt{\lambda})$. – emka Nov 13 '14 at 16:16