Question

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Answer 1

$1.;frac{dy}{dx}-y=e^x\y'-y=e^x\y=uv,;y'=u'v+uv'\u'v+uv'-uv=e^x\u'v+u(v'-v)=e^x\begin{cases}v'-v=0\u'v=e^xend{cases}\v'-v=0Rightarrowfrac{dv}{dx}=vRightarrowintfrac{dv}v=int dxRightarrow ln|v|=xRightarrow v=e^x\begin{cases}v=e^x\u'e^x=e^xend{cases}\u'=1\frac{du}{dx}=1\u=int dx=x+C\\y=uvRightarrow y=(x+C)cdot e^x,;C=const$

$2.;frac{dy}{dx}+2xy=xe^{-x^2}\y'+2xy=xe^{-x^2}\y=uvRightarrow y'=u'v+uv'\u'v+uv'+2xuv=xe^{-x^2}\u'v+u(v'+2xv)=xe^{-x^2}\begin{cases}v'+2xv=0\u'v=xe^{-x^2}end{cases}\frac{dv}{dx}+2xv=0\frac{dv}{dx}=-2xv\frac{dv}v=-2xdx\intfrac{dv}v=-2int xdx\ln|v|=-x^2\v=e^{-x^2}\begin{cases}v-e^{-x^2}\u'e^{-x^2}=xe^{-x^2}end{cases}\u'e^{-x^2}=xe^{-x^2}\frac{du}{dx}=x\u=int xdx=frac{x^2}2+C\\y=uv=left(frac{x^2}2+Cright)cdot e^{-x^2},;C=const$