<s>[INST] Initial value problem with separation of variables 
I have this initial value problem. Also I am a beginner on the topic.
As far as I know, I should separating the variables and integrate and with using initial condition I should have get rid off from the constant(C).
But the problem is, the equation kinda looks scary and messy.
Is there a way to verify that If the equation is separable or not ?
If not, how do I proceed ?
 [/INST] $$x^3y'-3x^2y=\dfrac {2x^7(4x^3-3y)}{3x^5+3x^3+2y}$$
Substitute $w=\dfrac y {x^3}$:
$$\left(\dfrac y {x^3}\right)'=\dfrac {2x(4-3y/x^3)}{3x^2+3+2y/x^3}$$
$$w'=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
Then take $v=x^2+1$
$$\dfrac {dw}{dx}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$\dfrac {dw}{dx^2+1}\dfrac {dx^2+1}{dx}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$2x\dfrac {dw}{dx^2+1}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$\dfrac {dw}{dv}=\dfrac {4-3w}{3v+2w}$$
$$({3v+2w}) {dw}-(4-3w)dv=0$$
$$2w {dw}+3dwv-4dv=0$$
Integrate and unsubstitute $v,w$:
$$w^2+3wv-4v=C$$
$$\dfrac {y^2} {x^6}+\dfrac {3y(x^2+1)} {x^3}-4(x^2+1)=C$$
$$y(1)=1 \implies C=-1$$
$$y^2 +{3y(x^2+1)} {x^3}-4x^8-3x^6=0$$
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