IE19

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Convert to an integral equation: y''+y=\cos(x),y(0)=0,y'(0)=0\,

y''=-y+\cos(x)\,

\int _{0}^{x}y''(\xi )d\xi =\int _{0}^{x}-y(\xi )+\cos(\xi )d\xi \,

y'(x)-y'(0)=\int _{0}^{x}-y(\xi )+\cos(\xi )d\xi \,

y'(x)=-\int _{0}^{x}y(\xi )d\xi +\left[\sin(x)-\sin(0)\right]\,

\int _{0}^{x}y'(\xi )d\xi =-\int _{0}^{x}\!\!\!\int _{0}^{s}y(\xi )d\xi ds+\int _{0}^{x}\sin(\xi )d\xi \,

y(x)-y(0)=-\int _{0}^{x}y(\xi )(x-\xi )d\xi +\left[-\cos(x)+\cos(0)\right]\,

The solution is

y(x)=-\int _{0}^{x}y(\xi )(x-\xi )d\xi +(1-\cos(x))\,

To check the solution, start differentiating.

y'(x)=-{\frac  {d}{dx}}\int _{0}^{x}y(\xi )(x-\xi )d\xi +\sin(x)\,

y'(x)=-\int _{0}^{x}{\frac  {\partial }{\partial x}}y(\xi )(x-\xi )d\xi +\sin(x)\,

y'(x)=-\int _{0}^{x}y(\xi )d\xi +\sin(x)\,

y''(x)=-{\frac  {d}{dx}}\int _{0}^{x}y(\xi )d\xi +\cos(x)\,

y''(x)=-y(x)+\cos(x)\,

y''(x)+y(x)=\cos(x)\,


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