Question

(1+x^2)dy/dx+2xy-4x^2=0 bu using Bernoulli's Equation

Answer 1

Not of Bernoulli type, but still linear.

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

There's no need to find an integrating factor, since the left hand side already represents a derivative:

$\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=(1+x^2)\dfrac{\mathrm dy}{\mathrm dx}+2xy$

So, you have

$\dfrac{\mathrm d}{\mathrm dx}[(1+x^2)y]=4x^2$

and integrating both sides with respect to $x$ yields

$(1+x^2)y=\displaystyle\int4x^2\,\mathrm dx$
$(1+x^2)y=\dfrac43x^3+C$
$y=\dfrac{4x^3}{3(1+x^2)}+\dfrac C{1+x^2}$