= Solution
Differentiate the <Wronskian>:
$$
w'=uv''-u''v=u(-qv)-(-qu)v=0.
$$
It is therefore constant. <Linear dependence> plainly makes it zero. Conversely, if it is zero, the two initial-data vectors $(u(0),u'(0))$ and $(v(0),v'(0))$ are linearly dependent. Some constants $A,B$, not both zero, make $Au+Bv$ and its <derivative> zero at zero. Uniqueness for a linear second-order equation with <continuous> <coefficients> then makes $Au+Bv$ identically zero. Thus $\boxed{w=0\iff u,v\text{ are linearly dependent}}$.
For the prescribed integral put $K(x,s)=[u(x)v(s)-u(s)v(x)]/w$. Then $K(x,x)=0$, $K_x(x,x)=-1$, and $K_{xx}(x,s)=-q(x)K(x,s)$. <Differentiation> under the integral, using <continuity> of the <coefficients> and forcing, gives
$$
y'(x)=\int_0^xK_x(x,s)f(s)\,ds,\qquad y''(x)=-f(x)-q(x)y(x).
$$
This proves the required inhomogeneous equation, with the sign of the forcing as printed.
For the explicit forcing take $u=\cosh x$, $v=\sinh x$, so $w=1$ and $K(x,s)=-\sinh(x-s)$. Since $2/(e^s+e^{-s})=\operatorname{sech}s$, the integral is
$$
-\int_0^x\sinh(x-s)\operatorname{sech}s\,ds=-x\sinh x+\cosh x\log\cosh x.
$$
Indeed expand $\sinh(x-s)=\sinh x\cosh s-\cosh x\sinh s$ and integrate $1$ and $\tanh s$. Adding the entire homogeneous solution gives
$$
\boxed{y(x)=A\cosh x+B\sinh x+\cosh x\log\cosh x-x\sinh x,\quad A,B\in\mathbb R.}
$$
Finally multiply the other equation by $\sin x$ and integrate from zero to $\pi$. <Integration by parts> gives
$$
\int_0^\pi(y''+y)\sin x\,dx=[y'\sin x-y\cos x]_0^\pi=y(\pi)+y(0)=0.
$$
Thus $\boxed{\int_0^\pi f(s)\sin s\,ds=0}$ is necessary. It is also sufficient: the solution with $y(0)=0$ is $B\sin x-\int_0^x\sin(x-s)f(s)\,ds$, and the integral condition gives its value zero at $\pi$.
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