Consider the following boundary value problem: $$-(p(x)u')'+q(x)u=g(x) \hspace{35pt} (1)\\ 0<x<1, u(0)=u_l, u(1) = u_r\hspace{35pt}$$
where $p\in C^1[0,1], p(x) \ge p_0 >0; q\in C[0,1], q(x)\ge 0; g\in C[0,1]$
I want to show that if $u(x)$ is a solution of (1) then $$y(x):=u(x)-u_l(1-x) -u_rx $$
is a solution to the system (2) below.
$$-(p(x)y')'+q(x)y=f(x) \hspace{35pt} (2)\\ 0<x<1, y(0)=0, y(1) = 0\hspace{35pt}$$
So I proceed as follows:
$$y(0) = u(0)-u_l = 0\\ y(1) = u(1) - u_r = 0,$$
thus the boundary conditions are satisfied.
Now I substitute $y(x)$ into (1) and obtain
$$-(p(x)u')'+q(x)u=f(x)+p'(x)(u_l+u_r)+q(x)(u_l(1-x)-u(x)+u_rx)\hspace{35pt} (3)$$
Now how does one show that the RHS of (3) is $g(x)$? I'm quite puzzled. I'm likely supposed to somehow use the properties given for $q$ and $p$, but I don't see how these can be used. Would appreciate some hint(s).
Thank you for your attention.