Originally Posted by

**Ackbeet** I've pored over your post # 2 now, and I think I can finally discern the logic you're employing there. Basically, it comes down to this: $\displaystyle b = 0$, or you get $\displaystyle a$ having to be both an odd-integer multiple of $\displaystyle \pi/2$, and a multiple of $\displaystyle \pi$, which can't be. Therefore, $\displaystyle b=0$. I think your overall logic works, provided that the form of your solution hasn't already assumed that $\displaystyle \lambda>0.$. You have not shown that lambda can't be zero. In order to do that, you have to re-solve the DE with that assumption in mind (the solutions you get for that case are not obtainable with any selection of the integration constants for the lambda not zero case), and show that there are no eigenvectors.

I would probably solve the problem this way: break it up into three cases, according to the dichotomy law: $\displaystyle \lambda<0,\;\lambda=0,\;\lambda>0.$ For $\displaystyle \lambda<0,$ let $\displaystyle \lambda=-\alpha^{2}.$ For $\displaystyle \lambda=0,$ do the obvious. And for $\displaystyle \lambda>0,$ let $\displaystyle \lambda=\alpha^{2}.$ In each case, you get a different form of the solution, with which you work to see if it's an allowed case. For $\displaystyle \lambda<0$, you get exponentials. For $\displaystyle \lambda=0,$ you get straight lines. For $\displaystyle \lambda>0,$ you get sinusoids. Remember that, by definition, eigenvectors cannot be identically zero. That fact, in this case, rules out the $\displaystyle \lambda<0$ and $\displaystyle \lambda=0$ cases.

Make sense?