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Show that method of variation of parameters applied to the equation $y'' + y = f(x)$ leads to a particular solution $y(x) = \int_0^x f(t) \cdot \sin (x-t) dt$. I tried like this Wronskian of $y_1$ and $y_2$ is $1$. Two solutions of corresponding homogeneous DE are $\sin(x)$ and $\cos(x)$.Then put these values into formula used in variation of parameters.Bit i did not get answer.of this form.

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  • $\begingroup$ See this for help. (-: $\endgroup$ – mrs Aug 18 '13 at 9:57
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In fact this just involves some elementary functions simplifications.

By variation of parameters, you can get the particular solution is $y_p=\sin x\int_0^xf(x)\cos x~dx-\cos x\int_0^xf(x)\sin x~dx$

Note that the particular solution can also rewrite as

$y_p=\sin x\int_0^xf(t)\cos t~dt-\cos x\int_0^xf(t)\sin t~dt$

$y_p=\int_0^xf(t)(\sin x\cos t-\cos x\sin t)~dt$

$y_p=\int_0^xf(t)\sin(x-t)~dt$

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