I am stuck on how to approach this question:
Solution:
Any help is appreciated. Leo
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
Hints: The characteristic equation is: $m^2 +\lambda = 0$ At $\lambda = 0$, we have: $m^2 = 0$, so the solution is: $y = x_0 + x_1 t$. At $\lambda < 0$, we have: $m^2 - \lambda = 0$, thus we have an eigenvalue of $\pm \lambda$, so the solution is: $y = x_0 e^{\sqrt{-\lambda} t} + x_1e^{-\sqrt{-\lambda} t}$, where $-\lambda$ and $\sqrt{-\lambda}$ are positive. At $\lambda > 0$, we have $m^2 + \lambda = 0$, thus we have a complex eigenvalue of $\pm i\lambda$, so the solution is: $y = x_0 \sin (\sqrt{\lambda}~t) + x_1\cos (\sqrt{\lambda}~t)$. Can you use the BC's to find $x_0$ and $x_1$, find all three cases together to determine the value of $\lambda$ and finish this off? |
|||||
|
|
I suggest you follow the advice in the question and analyze the three cases $\lambda > 0$, $\lambda = 0$ and $\lambda < 0$. To get you started, assume $\lambda = 0$. Then $\phi'' = 0$ and the solutions are linear, i.e. $\phi = Ax + B$. This can't satisfy the boundary conditions (except for $A=B=0$, but that's not very interesting so we disregard it) so $\lambda = 0$ is not an eigenvalue. Now check the other two cases. |
|||
|
|
