Question

Дифференциальное уравнение

Answer 1

$y'-frac{2x}{1+x^2}y=1+x^2\y=uv;y'=u'v+v'u\u'v+v'u-frac{2x}{1+x^2}uv=1+x^2\u'v+u(v'-frac{2x}{1+x^2}v)=1+x^2\begin{cases}v'-frac{2x}{1+x^2}v=0\u'v=1+x^2end{cases}\frac{dv}{dx}-frac{2x}{1+x^2}v=0\frac{dv}{v}=frac{2x}{1+x^2}dx\intfrac{dv}{v}=intfrac{d(1+x^2)}{1+x^2}\ln|v|=ln|1+x^2|\v=1+x^2\u'(1+x^2)=1+x^2\u'=1\du=dx\u=x+C\y=uv=(1+x^2)(x+C)$