Question

(1-x)(y'+y)=e^-x; y(0)=0 помогите решить пожалуйста

Answer 1

$\displaystyle (1-x)(y'+y)=e^{-x}\\y'+y=\frac{1}{e^{x}(x-1)}\\y=uv;y'=u'v+v'u\\u'v+v'u+uv=\frac{1}{e^{x}(1-x)}\\u'v+u(v'+v)=\frac{1}{e^{x}(1-x)}\\\begin{cases}v'+v=0\\u'v=\frac{1}{e^{x}(1-x)}\end{cases}\\\frac{dv}{dx}+v=0\\\frac{dv}{v}=-dx\\\int\frac{dv}{v}=-\int dx\\ln|v|=-x\\v=e^{-x}\\\frac{du}{dx}=\frac{1}{1-x}\\du=\frac{dx}{1-x}\\\int du=-\int\frac{d(1-x)}{1-x}\\u=-ln|x-1|+C\\y=e^{-x}(C-ln|1-x|)\\y(0)=0\\0=C\\y=-\frac{ln|x-1|}{e^{x}}$