Eigenvalues of Differential Equation with Boundary Condition Here is a problem from my homework assignment that I am struggling with:
Consider the differential equation $\frac{d^2\phi}{dx^2}+\lambda\phi=0 $.
Determine the eigenvalues $\lambda$ if $\phi$ satisfies the following boundary conditions:
$\phi(a)=0$
$\phi(b)=0$
I have been able to successfully complete this problem with 6 other sets of boundary conditions, but this one is giving me trouble.  The question also states we only need to consider $\lambda>1$.  Here is what I have done so far:
Because $\lambda>1$, we get our general solution of:
$\phi = C_1cos(\sqrt\lambda x)+C_2sin(\sqrt\lambda x)$.
We can plug in our first boundary condition and get:
$\phi(a) = C_1cos(\sqrt\lambda a)+C_2sin(\sqrt\lambda a)=0$
Usually I would solve for $C_1$ or $C_2$ to plug into the second equation, so here I solve for $C_1$:
$C_1=\frac{-C_2 sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}$
Now I substitute $C_1$ in and plug in the $b$ value to get:
$\phi(a) = \frac{-C_2 sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}cos(\sqrt\lambda b)+C_2sin(\sqrt\lambda b)=0$
Now I will factor our $C_2$ to get:
$\phi(a) = C_2[\frac{sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}cos(\sqrt\lambda b)+sin(\sqrt\lambda b)]=0$
Here we assume that $C_2$ cannot equal zero, or we would also get $C_1$ equal to zero, which would give us the trivial solution.  Thus we conclude that:
$\frac{sin(\sqrt\lambda a)}{cos\sqrt\lambda a)}cos(\sqrt\lambda b)+sin(\sqrt\lambda b)=0$
And that is as far as I can get....
The back of the book says the answer is:
$\lambda = (\frac{n \pi}{b-a})^2$
Any help would be greatly appreciated!
Thanks.
 A: Up to a scalar multiple, the eigenfunction $\phi$ satisfying $\phi(a)=0$, $\phi'(a)\ne 0$ is
$$
               \sin(\sqrt{\lambda}(x-a)).
$$
This is an actual eigenfunction iff $\sin(\sqrt{\lambda}(b-a))=0$. Equivalently,
$$
                       \sqrt{\lambda}(b-a)=\pm\pi,\pm 2\pi,\pm 3\pi,\cdots,\\
                            \lambda = \frac{n^{2}\pi^{2}}{(b-a)^{2}},\;\;\; n=1,2,3,\cdots.
$$
$\lambda = 0$ cannot be considered using this method for $\lambda=0$ because $\sin(\sqrt{\lambda}(x-a))\equiv 0$ in that case. The correct solution for $\lambda=0$ is
$C(x-a)$, but this function does cannot vanish at $b$. So $\lambda=0$ is not an eigenvalue.
A: It is also possible to solve this with your original method. From
$$-\frac{\sin(\sqrt{λ}a)}{\cos(\sqrt{λ}a)}\cos(\sqrt{λ}b)+\sin(\sqrt{λ}b)=0$$
(beware, you forgot a minus there) you obtain (by division with $\cos(\sqrt{λ}b)$)
$$-\tan(\sqrt{λ}a)+\tan(\sqrt{λ}b)=\frac{\sin(\sqrt{λ}(b-a))}{\cos(\sqrt{λ}a)\cos(\sqrt{λ}b)}= 0\ .$$
And from $\sin(\sqrt{λ}(b-a))=0$ you get desired solution.
This works under conditions
$$\lambda_n\neq\frac{\pi(2n+1)}{2a}\ ,\quad \lambda_n\neq\frac{\pi(2n+1)}{2b}\ ,$$
which you must take into account when dealing with special intervals e.g. $b=-a$. It results in different eigenfunctions for even and odd $n$.
A: Write $$\lambda = \mu^{2},\ \ \ \mu > 0$$
Then as you found the general solution would be $\phi = c_{1}\ cos(\mu x) + c_{2} \ sin(\mu x)$. 
As for you final answer you may write in terms of $tan\  (\mu x) =\frac{sin\  (\mu x)}{cos \ (\mu x)} $.
