Question

y''+ky=0 y(L)=0 y'(0)=0

Answer 1

You've got initial conditions and all set up. You just need to apply integration.

Answer 2

@sara_setareh you probably won't be around to check this, but...

Three methods come to mind on how to approach this.

The first: You're given a homogeneous DE with constant coefficients, so you have the characteristic polynomial
\[r^2+k=0\]
which has roots \(r=\pm ki\). So, you get the general solution
\[y=C_1\cos kx+C_2\sin kx\]
Differentiate:
\[y'=-kC_1\sin kx+kC_2\cos kx\]Plug in initial conditions:
\[0=C_1\cos kL+C_2\sin kL\\
0=-kC_1\sin 0+kC_2\cos 0~~\iff~~C_2=0~~\Rightarrow~~C_1=?\text{ (lot of cases to consider)}\]
(assuming \(k\not=0\))

Second method: Laplace Transform. (You probably won't return to check on this, so I won't both describing the process.)

Third method: Treat as a separable equation (which is what @BonjourSalut is referring to)