The Question:
Consider the inhomogeneous ODE
$$y''(x)+y(x)=f(x) \; \; \; \; \; \; y(0)=0 \;,\;y(\pi)=1 \; \; \; \; \; \; 0<x<\pi$$
Give a condition on $f$ such that a solution exists
My Thoughts:
I get how if, for example, $f(x) = 0$ then the general solution is $y(x)=A\cos (x)+B\sin (x)$ so then no solutions exist.
By the Fredholm Alternative Theorem, the homogeneous adjoint problem
$$w''(x)+w(x)=0 \; \; \; \; \; \; w(0)=0 \;,\;w(\pi)=0 \; \; \; \; \; \; 0<x<\pi$$
has non-trivial solutions (namely $w(x)=C\sin (x)$), so that the original problem either has no solutions or non-unique solutions.
How should I proceed?
Let $$ z=y-\frac{1}{\pi}x $$ and then the equation becomes $$ z''(x)+z(x)=-\frac{1}{\pi}x+f(x), \ z(0)=z(\pi)=0. $$ Note that the corresponding homogeneous equation $$ z''(x)+z(x)=0, \ z(0)=z(\pi)=0 \tag{1} $$ has a solution $z=C\sin(x)$. Thus (1) has a solution iff $$ \int_0^\pi\left[-\frac{1}{\pi}x+f(x)\right]\sin(x)dx=0$$ or $$ \int_0^\pi f(x)\sin(x)dx=1. $$