Question

Уравнение Бернулли
$y'+y=x \sqrt{y}$

Answer 1

$y'+y=x \sqrt{y}|:\sqrt y\\\frac{y'}{\sqrt y}+\sqrt{y}=x\\z=\sqrt y;z'=\frac{y'}{2\sqrt y}\rightarrow \frac{y'}{\sqrt y}=2z'\\2z'+z=x\\z=uv;z'=u'v+v'u\\2u'v+2v'u+uv=x\\2u'v+u(2v'+v)=x\\\begin{cases}2v'+v=0\\2u'v=x\end{cases}\\\frac{2dv}{dx}+v=0|*\frac{dx}{2v}\\\frac{dv}{v}=-\frac{dx}{2}\\\int\frac{dv}{v}=-\frac{1}{2}\int dx\\ln|v|=-\frac{x}{2}\\v=e^{-\frac{x}{2}}\\2\frac{du}{dx}e^{-\frac{x}{2}}=x|*\frac{e^\frac{x}{2}dx}{2}\\du=\frac{xe^\frac{x}{2}}{2}dx\\\int du=\frac{1}{2}\int xe^\frac{x}{2}dx$
$\int xe^\frac{x}{2}dx=2xe^\frac{x}{2}-2\int e^\frac{x}{2}dx=2xe^\frac{x}{2}-4e^\frac{x}{2}+C\\u=x=\ \textgreater \ du=dx\\dv=e^\frac{x}{2}dx=\ \textgreater \ v=2e^\frac{x}{2}\\u=xe^\frac{x}{2}-2e^\frac{x}{2}+C\\z=\sqrt y=x-2+Ce^{-\frac{x}{2}}\\y=(x-2+Ce^{-\frac{x}{2}})^2$