If the following list of functions defined on $x \in [0,L] \,$ $$\phi_n(x) \equiv \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)$$ where $n$ is positive integer and the functions are zero at the boundary points.
The question is to show that the following identity $$-\frac{h^2}{2m}\frac{\partial^2\phi_n(x)}{\partial x^2} = E_n\phi_n(x)$$
So here can we assume that $V$ is zero to be able to reach the desired goal ? Because in the question there is no information about $V$.
You could quite easily reformulate the Schrodinger equation for a finite potential within an infinite potential well in one dimension (Note: I do no not mean finite well height, I mean a finite potential within an infinite potential well). Here, I am using $\psi(x)$ as a trial solution.
$\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) + V(x)\psi(x) = E\psi(x)$
The spatial dependence of $V$ is then:
$V(x) = V_0$, for $x \in [0,L]$
$V(x) = \infty$, for $x < 0, x>L$
$V_0$ is a constant as the well has a constant potential at the base. Restricting the domain of the differential equation to $[0,L]$ as well as specifying the boundary conditions of $\psi(x)$ to vanish at the boundaries, the equation can then be simplified.
$\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2}\psi(x) = (E-V_0)\psi(x)$, $x \in [0,L]$
This simply transforms $E_n$ to $E-V_0$. From here you can insert your solution, the actual value of the potential inside the well becomes irrelevent as the eigenvalue energy just gives the energy with respect to the base of the well.
