Answer
$y=C_1 \arctan x + C_2$
Work Step by Step
We are given:
$(1+x^2)y''=-2xy'$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}$
Subsitute to the equation:
$(1+x^2)\frac{du}{dx}=-2xu$
$-\frac{du}{u}=\frac{2x}{(1+x^2)}dx$
Integrating both sides:
$-\ln(u)=\ln(1+x^2)+\ln(C)$
$u^{-1}=(2+x^2)C$
$u=\frac{1}{C(1+x^2)}$
$\frac{dy}{dx}=\frac{C}{1+x^2}$
$dy=\frac{C}{1+x^2}dx$
Integrating the last equation,
$y=C_1 \arctan x + C_2$
The final solution is:
$y=C_1 \arctan x + C_2$
