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[More information]Q:

Find the general solution of $y'+2xy=4x^2+6y^2$

Find the general solution of $y'+2xy=4x^2+6y^2$
I tried for the homogenous case:
$y'+2xy=0$,$y=ce^x$
For the inhomogenous case:
$y'+2xy=4x^2+6y^2$
Solving:
$y'=4x^2+6y^2-2xy$
$y'+2xy=4x^2+6y^2$
$y'-2y=4x^2+6y^2$
$y'-2y=4x^2$
$y'+4xy-2y=4x^2$
$y'-4xy+2y=4x^2$
$y'-4xy+2y=4x^2+6y^2$
$y'+4xy-2y=4x^2+6y^2$
$y'+4xy=4x^2+6y^2$
The book tells me I should end up with something like
$(y+2x)e^{ -4x}=(4x^2+6y^2)e^{ -4x}$
I'm unsure on what to do from here to end up with the correct answer of $(-2x+c)e^{4x}$

A:

We have
$$y'+2xy=4x^2+6y^2.$$
Since
$$(y'+2xy)e^{ -4x}=(4x^2+6y^2)e^{ -4x},$$
$$\int (y'+2xy)e^{ -4x}\,dx= 70238732e0 sachincr
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