$x''+\lambda x=0$ for $x\left(0\right)=x\left(L\right)=0\:$ for $L>0$ - it turns out that L must be dependent on t, but that shouldn't be the case For $\lambda =0$ and $\lambda <0$ the solution is the trivial solution $x\left(t\right)=0$
So we have to calculate for  $\lambda >0$
The general solution here is
$x\left(t\right)=C_1cos\left(\sqrt{\lambda }t\right)+C_2sin\left(\sqrt{\lambda \:}t\right)$
Because $0=C_1\cdot cos\left(0\right)+C_2\cdot sin\left(0\right)=C_1$ we know that
$x\left(t\right)=C_2sin\left(\sqrt{\lambda }t\right)$
$\sqrt{\lambda }t=n\pi$
$\sqrt{\lambda }=\frac{n\pi }{t}$
But does there a solution for lambda exist which is not dependent on t?
 A: $\lambda >0$ the general solution is $x(t)=C_{1}cos(\sqrt{\lambda}t)+C_{2}sin(\sqrt{\lambda}t)$ then the condition $x(0)=0$ gives $C_{1}=0$ as you've computed.
Then the condition $x(L)=0$ gives $C_{2}sin(\sqrt{\lambda}L)=0$ and since we are looking for a non-trivial solution, we have $\sqrt{\lambda}L=n\pi$ so this gives $\lambda=(\frac{n\pi}{L})^{2}$ for some integer n which is independent of t.
A: By saying $x(L) = 0$
you mean $\sqrt{\lambda}*L = n \pi$
A: There only exist solutions for specific values of $\lambda$, and those $\lambda$ are the eigenvalues of the underlying problem. So the goal of finding non-trivial solutions of the following equations is to (a) find the $\lambda$ for which there exists a non-trivial solution, and (b) find the actual non-trival solutions. In order to normalize the problem, first find the solutions of
$$
          x''+\lambda x = 0,\;\;\; x(0)=0,\; x'(0)=1.
$$
Every non-zero solution of $x''+\lambda x=0$ subject to $x(0)=0$ must be a non-zero scalar multiple of the above solution. $x'(0)=1$ can be imposed because $x$ an be scaled to achieve that, unless $x(0)=x'(0)=0$, which is only the case for $x\equiv 0$, which is case we're not interested in. The solution of the above equation is
$$ 
             x(t)= \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}
$$
This is valid even when $\lambda=0$, if you take the limit as $\lambda\rightarrow 0$, which gives $x(t)=t$. Finally in order to solve the full equation where $x(L)=0$, you must solve the following equation in $\lambda$:
$$
               \frac{\sin(\sqrt{\lambda}L)}{\sqrt{\lambda}} = 0.
$$
$\lambda=0$ is not a solution because $x(t)=t$ is not a solution of $x(L)=0$. So $\lambda=0$ does not work. However, $\sqrt{\lambda}L=n\pi$ for $n=1,2,3,\cdots$ or $\lambda=n^2\pi^2/L^2$ for $n=1,2,3,\cdots$ are valid values of $\lambda$ for which the original system has have a solution, and that solution is
$$
              x_n(t) = \frac{\sin(n\pi t)}{n\pi}
$$
These are valid solutions for $n=1,2,3,\cdots$, and there are no others.
