<s>[INST] understanding the solution to a particular differential equation? so I ran across this problem:

$(3x^2+1)y' -2xy = 6x$

I solved it by integrating factor $ e^{\int\frac{-2x}{3x^2+1} dx}$
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$\int D[(3x^2+1)^{\frac{-1}{3}}y] = \int 6x(3x^2+1)^{\frac{-4}{3}} dx$
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$(3x^2+1)^{\frac{-1}{3}}y = -3(3x^2+1)^{\frac{-1}{3}} + C$
$y = -3 + C(3x^2+1)^{\frac{1}{3}}$
my question is interpreting the solution $y=-3 + C(3x^2+1)^{\frac{1}{3}}$. So I understand why the integrating factor method works in solving the problem. Why did the answer come out so simple after doing so much work? Could I have solved it 'faster?'
 [/INST] You could transform the equation simply as
$$
(3x^2+1)y'=2x(y+3)
\\~\\
y'=\frac{2x}{3x^2+1}(y+3)
$$
This is a separable equation and the case of the constant solution $y=-3$ appears directly as the one to exclude for performing the separation of variables. Then continue
$$
\int\frac{dy}{y+3}=\int\frac{2x\,dx}{3x^2+1}.
$$
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