Question

Solve the differential equation: $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$

Answer 1

$(1+x^2)\frac{dy}{dx}+2yx=4x^2$

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

Comparing equation $\left(1\right)$ by $\frac{dy}{dx}+py=Q$

$P=\frac{2x}{1+x^2};Q=\frac{4x^2}{(1+x^{)}}$

$\therefore I.F=e^{\int \frac{2x}{1+x^2}dx}=e^{\log (1+x^2)}=1+x^2$

Multiplying equation $\left(1\right)$ by $1+x^2$,

$(1+x^2)\frac{dy}{dx}+2xy=4x^2$

$\Rightarrow \frac{d}{dx}\left[\right(1+x^2\left)y\right]=4x^2$

Integrating both sides, $w.r.t.x$

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

Dividing by $1+x^2$

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

This is required solution.