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Решите задачу:

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

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