Question

Xy'-xy^2+y=0 решить дифференциальное уравнение

Answer 1

$xy'-xy^2+y=0|:y^2\\x\frac{y'}{y^2}+\frac{1}{y}=x\\z=\frac{1}{y};z'=-\frac{y'}{y^2}\\-xz'+z=x\\z=uv;z'=u'v+v'u\\-xu'v-xv'u+uv=x\\-xu'v+u(-xv'+v)=x\\\begin{cases}-xv'+v=0\\-u'v=1\end{cases}\\-x\frac{dv}{dx}+v=0|*\frac{-dx}{xv}\\\frac{dv}{v}=\frac{dx}{x}\\ln|v|=ln|x|\\v=x\\-\frac{du}{dx}x=1|*-\frac{dx}{x}\\du=-\frac{dx}{x}\\u=-ln|x|+C\\z=\frac{1}{y}=-xln|x|+Cx\\y=\frac{1}{x(C-ln|x|)}$