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Question Number 27801 by abdo imad last updated on 14/Jan/18

$$solvethed.e.y^{{}^{\prime }}-2xy=sinx{e}^{x^2}withinitialcondition$$$$y\left(o\right)=1.$$

Commented by abdo imad last updated on 19/Jan/18

$$he\Rightarrow y^{{}^{\prime }}-2xy=0\iff \frac{y^{{}^{\prime }}}{y}=2x\iff ln/y/=x^2+k$$$$\iff y=\lambda e^{x^2}letusethem.v.cmethod$$$$y^{{}^{\prime }}={\lambda }^{{}^{\prime }}{e}^{x^2}+2\lambda x{e}^{x^2}$$$$and\left(e\right)\iff {\lambda }^{{}^{\prime }}{e}^{x^2}+2\lambda x{e}^{x^2}-2\lambda x{e}^{x^2}=sinx{e}^{x^2}$$$$\iff {\lambda }^{{}^{\prime }}=sinx\iff \lambda =\int sinxdx+k=-cosx+kand$$$$y\left(x\right)=(-cosx+k)e^{x^2}y\left(0\right)=1\Rightarrow k-1=1\Rightarrow k=2$$$$finallyy\left(x\right)=(2-cosx)e^{x^2}.$$

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