<s>[INST] 1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram 
$$y=\frac{1}{\sqrt{x^2+1}}\cdot C$$
But when I try to calculate $y'$, I get a strange equation. I think I had to be wrong somewhere. I will be grateful for your help.
 [/INST] You must have asked Wolfram Alpha to solve the homogeneous equation, i.e. with $0$ instead of $x$ on the right side.  According to the standard method for solving first-order equations, your integrating factor is
$$ \eqalign{\mu(x) &= \exp \left(\int \frac{x\; dx}{1+x^2}\right) \cr
&= \exp\left(\frac{1}{2} \log(1+x^2)\right)\cr
&= \sqrt{1+x^2}}$$
and then the general solution is
$$ \eqalign{y &= \frac{1}{\sqrt{1+x^2}} \left(\int x \sqrt{1+x^2} \; dx + C \right) \cr
&= \frac{1}{\sqrt{1+x^2}} \left( \frac{(1+x^2)^{3/2}}{3} + C \right)\cr
&= \frac{1+x^2}{3} + \frac{C}{\sqrt{1+x^2}}}$$
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