Consider the Boundry Value ODE:
$(d^2\ T)/(dy^2\ )=\sin(y^2\ )$
Derive the equation for approximating the interior points using the central difference method.
I'm kinda stuck on this one. I was able to derive the equation down to this form:
$T_{i-1}[\frac{1}{∆y^2}]+T_i [\frac{-2}{∆y^2}]+T_{i+1} [\frac{1}{∆y^2} ]=sin(y_i)^2$
With the forcing function on the right, but I feel like I am missing something or doing something wrong. i.e. Am i supposed to do something with the sin at the end?
As you said, you get the linear equations $$ \dfrac{T_{i-1}-2T_i + T_{i+1}}{\Delta y^2} = \sin (y_i^2), \quad i=1, \cdots, n $$
Now you just need to solve this (tridiagonal) linear system, getting the values of $T_0$ and $T_{n+1}$ from the boundary conditions. The $\sin (y_i^2)$ terms are know and correspond to the rhs of the linear system.