<s>[INST] How to solve the differential equation $(2x^3y)\:\text{dy}+(1-y^2)(x^2y^2+y^2-1)\:\text{dx}=0$? 
Solve
  $$(2x^3y)\:\text{dy}+(1-y^2)(x^2y^2+y^2-1)\:\text{dx}=0$$  

I tried the substitution $y^2=t$ ; $2y\:\text{dy}=\text{dt}$ to get
$$(x^3)\:\text{dt}+(1-t)[(x^2+1)t-1]\:\text{dx}=0$$  
However, I don't know how to proceed further.  
 [/INST] since
$$2x^3yy'+(1-y^2)(x^2y^2+y^2-1)=0$$
then we have
$$2x^3yy'=(y^2-1)^2+(y^2-1)x^2y^2=(y^2-1)^2+(y^2-1)x^2(y^2-1)+x^2(y^2-1)$$
so
$$x^3(y^2-1)'=(y^2-1)^2+(y^2-1)^2x^2+x^2(y^2-1)$$
let
$y^2-1=u$,then we have
$$\dfrac{du}{dx}=\left(\dfrac{1}{x}+\dfrac{1}{x^3}\right)u^2+\dfrac{u}{x}$$
this is Bernoulli equation
$$-\dfrac{d(1/u)}{dx}=\dfrac{1}{x}\cdot\dfrac{1}{u}+\dfrac{1}{x^3}+\dfrac{1}{x}$$
then It is easy solve it
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