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Question Number 37363 by math khazana by abdo last updated on 12/Jun/18

$$solve{y}^{{}^{\prime }}+{xe}^{-x^2}y=e^{-x}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jun/18

$$\frac{dy}{dx}+{xe}^{-x^2}y=e^{-x}$$$$intregatingfactor$$$$e^{\int {xe}^{-x^2}dx}$$$$t=x^{2}dt=2xdx$$$$e^{\int e^{-t}\times \frac{dt}{2}}$$$$e^{\frac{e^{-t}}{-2}}=e^{\frac{e^{-x^2}}{-2}}$$$$e^{\frac{e^{-x^2}}{-2}}\frac{dy}{dx}+{xe}^{-x^2}\times e^{\frac{e^{-x^2}}{-2}}\times y=e^{-x}\times e^{\frac{e^{-x^2}}{-2}}$$$$\frac{d}{dx}\left({ye}^{\frac{e^{-x^2}}{-2}}\right)=e^{-x}\times e^{\frac{e^{-x^2}}{-2}}$$$$\int d\left({ye}^{\frac{e^{-x^2}}{-2}}\right)=\int e^{-x}\times e^{\frac{e^{-x^2}}{-2}}dx$$$$contd$$$$k=x+\frac{e^{-x^2}}{2}dk=1+\frac{1}{2}\times e^{-x^2}\times -2x$$$$$$

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