(1+x2)y’-xy=2x решите уравнение

Решите задачу:

$(1+x^2)y'-xy=2x\\y=uv;y'=u'v+v'u\\(1+x^2)u'v+(1+x^2)v'u-xuv=2x\\(1+x^2)u'v+u((1+x^2)v'-xv)=2x\\ \begin{cases}(1+x^2)v'-xv=0\\(1+x^2)u'v=2x\end{cases} \\\frac{(1+x^2)dv}{dx}=xv\\\frac{dv}{v}=\frac{xdx}{1+x^2}\\\int\frac{dv}{v}=\frac{1}{2}\int\frac{d(1+x^2)}{1+x^2}\\ln|v|=\frac{1}{2}ln|1+x^2|\\v=\sqrt{1+x^2}\\\frac{\sqrt{(1+x^2)^3}du}{dx}=2x\\du=\frac{2xdx}{\sqrt{(1+x^2)^3}}\\\int du=\frac{d(1+x^2)}{\sqrt{(1+x^2)^3}}\\u=-\frac{2}{\sqrt{1+x^2}}+C\\y=C\sqrt{1+x^2}-2$