Question

Дифференциальное уравнение

Answer 1

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