Find the equation of a curve passing through origin and satisfying the differential equation $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = 4{x^2}$.

Given that, $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy = 4{x^2}$

$\Rightarrow$ $\frac{{dy}}{{dx}} + \frac{{2x}}{{1 + {x^2}}} \cdot y = \frac{{4{x^2}}}{{1 + {x^2}}}$

which is a linear differential equation.

On comparing it with
$\frac{{dy}}{{dx}} + Py = Q,$

we get
$P = \frac{{2x}}{{1 + {x^2}}},Q = \frac{{4{x^2}}}{{1 + {x^2}}}$

$\therefore$ ${\rm{IF}} = {e^{\int P dx}} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}} dx}}$

Put $1 + {x^2} = t \Rightarrow 2xdx = dt$

${\rm{IF}} = 1 + {x^2} = {e^{\int {\frac{{dt}}{t}} }} = {e^{\log t}} = {e^{\log \left( {1 + {x^2}} \right)}}$

The general solution is
$y \cdot \left( {1 + {x^2}} \right) = \int {\frac{{4{x^2}}}{{1 + {x^2}}}} \left( {1 + {x^2}} \right)dx + C$

$\Rightarrow$ $y \cdot \left( {1 + {x^2}} \right) = \int 4 {x^2}dx + C$

$\Rightarrow$ $y \cdot \left( {1 + {x^2}} \right) = 4\frac{{{x^3}}}{3} + C$
………(i)

Since, the curve passes through origin, then substituting
$x = 0$ and $y = 0$ in Eq. (i),

we get
$C = 0$
The required equation of curve is
$y\left( {1 + {x^2}} \right) = \frac{{4{x^3}}}{3}$

$\Rightarrow$ $y = \frac{{4{x^3}}}{{3\left( {1 + {x^2}} \right)}}$