Value of $\lambda$ in Boundary value problem Determine the values of $\lambda \in R$ such that the BVP
$$ \frac{d^2 u}{dx^2}+\lambda u=0$$
where $u(0)=0$ and $u(1)=0$, has a non zero solution.
I am not able to proceed in this particular question. Can someone give some hint? 
 A: $$ \frac{d^2 u}{dx^2}+\lambda u=0$$
Let $u=e^{mx}$ be the trial solution.
Hence we have that, $$\frac{du}{dx}=me^{mx}$$ and $$\frac{d^2 u}{dx^2}=m^2e^{mx}$$
So the differential equation becomes $$(m^2+\lambda)e^{mx}=0$$
$$\Rightarrow (m^2+\lambda)u=0$$
Hence we have that $$m=\pm i\sqrt{\lambda}$$
So the general solution is $$u=Ae^{i\sqrt{\lambda}x}+Be^{-i\sqrt{\lambda}x}$$
which boils down to the following using Euler's Theorem,
$$u=A(\cos \sqrt{\lambda}x+i\sin \sqrt{\lambda}x)+B(\cos \sqrt{\lambda}x-i\sin \sqrt{\lambda}x)$$ or
$$u=(A+B)\cos \sqrt{\lambda}x+(A-B)i\sin \sqrt{\lambda}x$$ or 
$$u=C\cos \sqrt{\lambda}x+D\sin \sqrt{\lambda}x$$
Now using $u(0)=0$ and $u(1)=0$, we have that 
$$0=C$$ and $$0=C\cos \sqrt{\lambda}+D\sin \sqrt{\lambda}$$
which implies both $C,D$ are $0$ or $C=0$ and $\sqrt\lambda=n\pi$ where $n \in \mathbb{N}$.
First condition is not possible, so only possible case is the second one.
Required general solution is $u=D\sin \sqrt{\lambda}x$ only if $\lambda=n^2\pi^2$.
That is, $$u=D\sin n\pi x$$   where $n \in \mathbb{N}$.
A: HINT:
$$\frac{\text{d}^2u(x)}{\text{d}x^2}+\lambda u(x)=0\Longleftrightarrow$$
$$u''(x)+\lambda u(x)=0\Longleftrightarrow$$

Assume the solution will be proportional to $e^{x\mu}$ for some constant $\mu$.
Substitute $u(x)=e^{x\mu}$ into the differential equation:

$$\frac{\text{d}^2}{\text{d}x^2}\left(e^{x\mu}\right)+\lambda e^{x\mu}=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^2}{\text{d}x^2}\left(e^{x\mu}\right)=\mu^2e^{x\mu}$:

$$\mu^2e^{x\mu}+\lambda e^{x\mu}=0\Longleftrightarrow$$
$$e^{x\mu}\left(\mu^2+\lambda\right)=0\Longleftrightarrow$$

Since $e^{x\mu}\ne0$ for any finite $\mu$, the zeros must come from the polynomial:

$$\mu^2+\lambda=0\Longleftrightarrow$$
$$\mu=\pm i\sqrt{\lambda}$$
A: The simple harmonic oscillation equation has solution
$$ u(x) = A \sin(\sqrt{\lambda} x) + B \cos(\sqrt{\lambda} $$ 
From first boundary condition an odd solution is chosen. In frequency form 
$$ \omega =  \frac{2 \pi}{T} $$
$$ A \sin {\sqrt \lambda} x =    A \sin \frac{2 \pi x }{T }   $$
The $\lambda $ here should not be confused with the time period.
You gave $ T=2 $ for one sine wave  ( double between successive roots) so that by comparing
$$ \lambda = \pi^2  $$
for first wave and in general when more waves are included for arcsin solution
$$ \lambda = (n \pi) ^2. $$
