I know how to use the Rayleigh-Ritz method when given a sturm-Liouville problem. But I am not sure where to start when asked questions like the one above. I know i need to plug the trial functions in to the integral, and then determine c, but i feel like i am supposed to be doing something else first. Any help is much appreciated.
The Galerkin-Ritz method consists in using an approximation for $y$ as
$$ y\approx \phi_n(x) = \phi_0(x)+\sum_{k=1}^n c_k\omega_k(x) $$
with $\phi_0(x)$ obeying the boundary conditions and $\omega_k(x)$ linearly independent functions such that $\omega_k(0) = \omega_k(1)= 0$. As the BC are $y(0) = 0, y(1) = 1$ we will choose
$$ \phi_0(x) = \frac{(1-0)}{(1-0)}(x-0)+0 = x $$
and
$$ \omega_k(x) = (x-0)(x-1)x^k $$
then
$$ J[\phi_n]=\int_0^1\left((\phi_n'(x))^2+x\phi_n^2(x)+2x^2\phi_n(x)\right)dx=\Phi[c] $$
Here $\Phi[c]$ is a quadratic in $c$ and it's minimum is associated with $c_k$'s such that
$$ \frac{\partial\Phi}{\partial c_k} =0\Rightarrow \int_0^1\left(\phi_n''(x)-x\phi_n(x)-x^2\right)\omega_k(x) dx = \phi_n'(x)\omega_k(x)|_0^1 $$
now choosing $n=3$ we have
$$ \left\{ \begin{array}{rcl} \frac{39 c_1}{280}+\frac{131 c_2}{1260}+\frac{199 c_3}{2520}&=&\frac{1}{15} \\ \frac{131 c_1}{1260}+\frac{223 c_2}{2520}+\frac{509c_3}{6930}&=&\frac{1}{21} \\ \frac{199 c_1}{2520}+\frac{509 c_2}{6930}+\frac{901 c_3}{13860}& = & \frac{1}{28} \\ \end{array} \right. $$
and after solving
$$ c_1= \frac{25206}{25511},\ \ c_2= -\frac{33836}{25511},\ \ c_3= \frac{21626}{25511} $$
Follows the plot showing in blue the solution for
$$ y''-xy-x^2=0,\ \ \{y(0) = 0,\ y(1)=1\} $$
and in red $\phi_3(x)$
