# Solve the diﬀerential equation \frac{d^2y}{dx^2}-2\frac{dy}{dx}+4y=e^x\sin^2(\frac{x}{2})

Solve the diﬀerential equation
$\frac{{d}^{2}y}{{dx}^{2}}-2\frac{dy}{dx}+4y={e}^{x}{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)$
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The general solution will be the sum of the complementary solution and particular solution.
The complementary solution for the given equation:
$\frac{{d}^{2}y}{{dx}^{2}}-2\frac{dy}{dx}+4y={e}^{x}{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)$
${\lambda }^{2}-2\lambda +4=0$
$\lambda =\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}=\frac{-\left(-2\right)+\sqrt{{\left(2\right)}^{2}-4\left(1\right)\left(4\right)}}{2\left(1\right)}=1+\sqrt{3}i$
$\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}=\frac{-\left(-2\right)-\sqrt{{\left(2\right)}^{2}-4\left(1\right)\left(4\right)}}{2\left(1\right)}=1-\sqrt{3}i$
${y}_{p}\left(x\right)={e}^{x}\left({c}_{1}\mathrm{cos}\left(\sqrt{3}x\right)+{c}_{2}\mathrm{sin}\left(\sqrt{3}x\right)\right)$
The particular for the equation is:
$\frac{{e}^{x}{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)}{{D}^{2}-2D+4}=\frac{{e}^{x}\left(\frac{1-\mathrm{cos}x}{2}\right)}{{D}^{2}-2D+4}=\frac{1}{2}\frac{{e}^{x}\left(1-\mathrm{cos}x\right)}{{D}^{2}-2D+4}$
$=\frac{1}{2}\frac{{e}^{x}}{{D}^{2}-2D+4}-\frac{1}{2}\frac{{e}^{x}\mathrm{cos}x}{{D}^{2}-2D+4}$
$P.I=\frac{1}{f\left(D\right)}{e}^{ax}=\frac{1}{f\left(a\right)}{e}^{ax}$