Differential Equations | Part IA, 2006

Let p,qp, q be continuous functions and let y1(x)y_{1}(x) and y2(x)y_{2}(x) be, respectively, the solutions of the initial value problems

y1+p(x)y1+q(x)y1=0,y1(0)=0,y1(0)=1,y2+p(x)y2+q(x)y2=0,y2(0)=1,y2(0)=0.\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 ff is any continuous function show that the solution of

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

y(x)=0xy1(s)y2(x)y1(x)y2(s)W(s)f(s)ds,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)=y1(x)y2(x)y1(x)y2(x)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)y=y(x) such that

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

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