 Solution

Differentiate the Wronskian:It is therefore constant. Linear dependence plainly makes it zero. Conversely, if it is zero, the two initial-data vectors $(u(0),u^{\prime }(0))$ and $(v(0),v^{\prime }(0))$ 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 $\boxed{w=0⟺u,v\text{ are linearly dependent}}$.

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, givesThis 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 $2/(e^s+e^{-s})=\mathrm{sech}s$, the integral isIndeed expand $\sinh (x-s)=\sinh x\cosh s-\cosh x\sinh s$ and integrate $1$ and $\tanh s$. Adding the entire homogeneous solution givesFinally multiply the other equation by $\sin x$ and integrate from zero to $\pi$. Integration by parts givesThus $\boxed{{\int }_0^{\pi }f(s)\sin sds=0}$ 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 $\pi$.