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I understand that the below is a non-homogeneous second order O.D.E.

$y'' + 4y = sin^2(x)$

and that I can re-arrange the R.H.S to get

$y'' + 4y = \frac{1}{2}(1 - cos2x)$

However, may I know how to proceed from here on?

Thank you.

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    First, solve $y''+4y=0$ – 2017-02-20
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    what have you got for the solution of the homogeneous solution? – 2017-02-20
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    @JJacquelin I got $y = Acos2x + Bsin2x$ for that equation. – 2017-02-20
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    @Dr.SonnhardGraubner Hi, as above my solution is $y = Acos2x + Bsin2x$. – 2017-02-20
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    for $$y_P$$ make the ansatz $$y_p=A+B\cos(2x)+Cx\sin(2x)$$ – 2017-02-20
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    @Dr.SonnhardGraubner I understand that $y(x) = y_{h}(x) + y_{p}(x)$ and that $y_{p}(x)$ is a solution of $y'' + 4y = sin^2(x)$ without arbitrary constant. However, I am lost at what to do with the ansatz. Could you explain to me how you derived at the ansatz and what's my next course of action so that I can add $y_{h}$ and $y_{p}$ eventually? – 2017-02-20
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    you must this plug in your equation to determine the constants $$A,B,C$$ – 2017-02-20
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    @stephchia: I think it will be of great help for you to read/review http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx. – 2017-02-20

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Here is the substantial way of doing it $$\frac{d^2}{dx^2}y+4y=\frac{d}{dx}\Big(\frac{d}{dx}y+2iy\Big)-2i\Big(\frac{d}{dx}y+2iy\Big)$$ Let $$\xi(x)=\frac{d}{dx}y+2iy$$ Then $$\frac{d}{dx}\xi-2i\xi=\sin^{2}(x)$$ Using the integrating factor thing you have $$\xi(x)=\frac{d}{dx}y+2iy=e^{2ix}\Big\{\int_{0}^{x}\sin^{2}(z)e^{-2iz}dz+\xi_{0}\Big\}$$ Using the integrating factor the second time you get $$y(x)=e^{-2ix}\Big\{\int_{0}^{x}e^{4iw}\Big\{\int_{0}^{w}\sin^{2}(z)e^{-2iz}dz+\xi_{0}\Big\}dw+x_{0}\Big\}$$ Then do the integrals)