Answer
$y=2 \sqrt {x^2+1}$
Work Step by Step
We are given that $y(x^2+1) \dfrac{dy}{dx}=xy^2$
We will separate the variables to obtain:
$\dfrac{ \ dy}{y}=\dfrac{1}{2}(\dfrac{2x}{x^2+1})\ dx$
Integrate to obtain:
$\int \dfrac{ \ dy}{y}=\dfrac{1}{2}\int(\dfrac{2x}{x^2+1})\ dx$
This implies that $\ln|y|=\dfrac{1}{2} \ln (x^2+1)+C$
After applying the initial conditions, $y=2$ when $x=0$, we get $C=\ln (2)$
Therefore, we have: $\ln|y|=\dfrac{1}{2} \ln (x^2+1)+2 \implies y=e^{\ln 2} e^{\ln (x^2+1)^{1/2}}$
or, $y=2 \sqrt {x^2+1}$
