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.


Application of your boundary condition should give you \begin{equation} 0 = C_{1} \cos(\sqrt{\lambda})+C_{2}\sin(\sqrt{\lambda}) \qquad (y(1)=0) \end{equation} and \begin{equation} 0 = C_{1} + C_{2}\sqrt{\lambda} \qquad (y(0) + y'(0)=0) \end{equation} Take it from here.

EDIT: Can't add comment so I'll re-edit me partial solution.

Your roots are of the form \begin{equation} \pm i \sqrt{\lambda} \end{equation} Remembering that \begin{eqnarray} e^{z} &=& e^{a+ib} \\ &=& e^{a}(\cos(b)+i\sin(b)) \end{eqnarray} You can then look at substituing in these forms and plug in your boundary conditions.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.