Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the eigenvalues and eigenfunctions for $$y'' + \lambda y = 0, y(0) = 0, y'(\pi/2) = 0$$

According to my book we must check 3 cases: $\lambda < 0$, $\lambda = 0$, $\lambda > 0$.

I started with $\lambda > 0$, found the general solution, applied the boundary conditions, and found $$\lambda _n = (2n-1)^2 with~ y_n=sin((2n-1)x)$$

I moved on to $\lambda < 0$ and got $$y = c_1e^{kx}+c_2e^{-kx}$$ and didn't know how to proceed.

share|cite|improve this question
Try plugging what you know for $y$ into the differential equation. How must $k$ and $\lambda$ be related? Now look at the boundary conditions, and see what additional constraints that imposes on your unknowns. The case $\lambda=0$ should be easier to deal with. – Aaron Jul 9 '13 at 23:07
Also, a small nitpick on terminology. As it stands, you are looking for solutions to a family of differential equations. However, they will be eigenvalues/eigenfunctions of the linear operator $Ly=D^2y$ acting on the space of functions satisfying the boundary conditions. To solve the eigenvalue problem, you recast it as a boundry value problem, but the boundary value problem is not per se an eigenvalue problem. If you have not taken linear algebra or this comment doesn't quite make sense, don't worry about it. – Aaron Jul 9 '13 at 23:13
i'm sorry, I'll try to use right terminology – Frank Epps Jul 10 '13 at 6:39
Don't apologize, my intent was to inform, not to chastise. – Aaron Jul 10 '13 at 7:11
up vote 0 down vote accepted

Let $\lambda = -\mu^2$; then

$$y(x) = A e^{\mu x} + B e^{-\mu x}$$

$y(0)=0 \implies A+B=0$. Also,

$$y'\left( \frac{\pi}{2}\right) = 0 \implies 2 A \sinh{\left(\mu \frac{\pi}{2}\right)} =0 $$

The only way this happens is when $A=0$; that is, there is no nontrivial solution to this BVP for $\lambda < 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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