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Question Number 115367 by Bird last updated on 25/Sep/20

$$solve{xy}^{{}^{″}}-(x^2+1)y^{{}^{\prime }}=x^{2}sin\left(2x\right)$$

Answered by Olaf last updated on 26/Sep/20

$$$$$${xy}^{″}-(x^2+1)y^{\prime }=x^2\sin 2x$$$$\mathrm{Let}u=\frac{y^{\prime }}{x}$$$$y^{″}={xu}^{\prime }+u$$$$x({xu}^{\prime }+u)-(x^2+1)xu=x^2\sin 2x$$$${xu}^{\prime }+u-(x^2+1)u=x\sin 2x$$$${xu}^{\prime }-x^{2}u=x\sin 2x$$$$u^{\prime }-xu=\sin 2x$$$$e^{-\frac{x^2}{2}}{u}^{\prime }-{xe}^{-\frac{x^2}{2}}u=e^{-\frac{x^2}{2}}\frac{e^{ix}-e^{-ix}}{2i}$$$$\Rightarrow \mathrm{Integration}\mathrm{by}\mathrm{parts}.$$$$\mathrm{The}\mathrm{function}u\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{combination}$$$$\mathrm{of}\mathrm{functions}\mathrm{erf}.$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{a}\mathrm{simple}\mathrm{expression}!$$

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