<s>[INST] Check my general solution to the differential equation? Given differential equation:
$$y' = \frac{(2xy^{3}+4x)}{(x^{2}y^{2}+y^{2})}$$
This is the general solution that I got for the above differential equation:
$$\frac{1}{3} \ln{\lvert y^3+2\rvert}=\ln{\lvert x^2+1\rvert}+\ln C$$
Please check if it is correct?

And for the solution satisfying y(1)=0, this is what I got:
$$(y^3+2)^{\frac{1}{3}}=(x^2+1)+C$$
 [/INST] Rearranging the differential equation gives
$$ \frac{y^2 y'}{y^3+2} = \frac{2x}{1+x^2} $$
Integrating this does give you the general solution you have,
$$ \frac{1}{3}\log{(2+y^3)} = \log{(1+x^2)}+A, $$
$A$ a constant. Exponentiating both sides gives
$$ y^3+2 = e^{3A}(1+x^2)^3 = B(1+x^2)^3, $$
where I have relabelled the constant as $B$.
To find the solution that satisfies y(1)=0, you have to put $y=0$ and $x=1$ to find the constant $A$ (your solution should not have an undetermined constant if you have one first-order equation and one boundary condition!):
$$ 2 = B \cdot 2^3, $$
so we find that $B=1/4$, or
$$ y^3+2 = \frac{1}{4}(1+x^2)^3. $$
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