Let $p, q$ be continuous functions and let $y_{1}(x)$ and $y_{2}(x)$ be, respectively, the solutions of the initial value problems

$$\begin{aligned} &y_{1}^{\prime \prime}+p(x) y_{1}^{\prime}+q(x) y_{1}=0, \quad y_{1}(0)=0, y_{1}^{\prime}(0)=1, \\ &y_{2}^{\prime \prime}+p(x) y_{2}^{\prime}+q(x) y_{2}=0, \quad y_{2}(0)=1, y_{2}^{\prime}(0)=0 . \end{aligned}$$

If $f$ is any continuous function show that the solution of

$$y^{\prime \prime}+p(x) y^{\prime}+q(x) y=f(x), \quad y(0)=0, y^{\prime}(0)=0$$

$$y(x)=\int_{0}^{x} \frac{y_{1}(s) y_{2}(x)-y_{1}(x) y_{2}(s)}{W(s)} f(s) d s,$$

where $W(x)=y_{1}(x) y_{2}^{\prime}(x)-y_{1}^{\prime}(x) y_{2}(x)$ is the Wronskian. Use this method to find $y=y(x)$ such that

$$y^{\prime \prime}+y=\sin x, \quad y(0)=0, y^{\prime}(0)=0 .$$