Question

Решить дифференциальные уравнения
y'+2xy=2xy^3

Answer 1

$y'+2xy=2xy^3|:y^3\\\frac{y'}{y^3}+\frac{2x}{y^2}=2x\\z=\frac{1}{y^2};z'=-2\frac{y'}{y^3}\rightarrow\frac{y'}{y^3}=-\frac{z'}{2}\\-\frac{z'}{2}+2xz=2x\\z'-4xz=-4x\\z=uv;z'=u'v+v'u\\u'v+v'u-4xuv=-4x\\u'v+u(v'-4xv)=-4x\\\begin{cases}v'-4xv=0\\u'v=-4x\end{cases}\\\frac{dv}{dx}-4xv=0|*\frac{dx}{v}\\\frac{dv}{v}=4xdx\\\int\frac{dv}{v}=4\int xdx\\ln|v|=2x^2\\v=e^{2x^2}\\\frac{du}{dx}e^{2x^2}=-4x|*e^{-2x^2}dx\\du=-4xe^{-2x^2}dx\\\int du=-4\int xe^{-2x^2}dx\\u=e^{-2x^2}+C\\z=1+Ce^{2x^2}$
$z=\frac{1}{y^2}=\ \textgreater \ y^2=\frac{1}{z}=\frac{1}{1+Ce^{2x^2}}\\y^2=\frac{1}{1+Ce^{2x^2}};y=0$