= Solution
The <Wronskian> satisfies <Abel identity>. Differentiating and substituting the homogeneous equations gives
$$
W'=y_1y_2''-y_1''y_2=-pW,\qquad W(0)=-1,
$$
so $W(x)=-\exp[-\int_0^xp(r)\,dr]\ne0$. The two homogeneous solutions are therefore independent.
Define the <variation of parameters> kernel
$$
K(x,s)=\frac{y_1(s)y_2(x)-y_1(x)y_2(s)}{W(s)}.
$$
For fixed $s$, it satisfies the homogeneous differential equation in $x$. Direct evaluation gives $K(s,s)=0$ and $K_x(s,s)=1$. For $y(x)=\int_0^xK(x,s)f(s)\,ds$, differentiation under the integral therefore yields
$$
y'=\int_0^xK_x(x,s)f(s)\,ds,\qquad
y''=f(x)+\int_0^xK_{xx}(x,s)f(s)\,ds.
$$
Combining these identities proves $y''+py'+qy=f$; both initial values vanish. Continuity of the coefficients gives uniqueness of this <initial value problem>, establishing the proposed formula rather than merely recognizing it as a standard method.
For the oscillator, $y_1=\sin x$, $y_2=\cos x$ and $W=-1$, so the kernel is $\sin(x-s)$. Hence
$$
y(x)=\int_0^x\sin(x-s)\sin s\,ds
=\sin x\int_0^x\cos s\sin s\,ds-\cos x\int_0^x\sin^2s\,ds.
$$
Using $\int_0^x\cos s\sin s\,ds=\sin^2x/2$ and $\int_0^x\sin^2s\,ds=x/2-\sin(2x)/4$ gives
$$
\boxed{y(x)=\frac12(\sin x-x\cos x).}
$$
The factor $x\cos x$ reflects <resonance> with the homogeneous oscillator frequency.
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