Solving the ODE $(x^2 y-1)y'-(xy^2-1)=0$ I need to solve $(x^2 y-1)y'-(x y^2-1)=0$.
Not sure how to approach this ODE, would love some help regarding solving it or any useful resources.
 A: Too long for a comment
Using @projectilemotion's answer, we can easily solve the cubic equation
$$y^3-3x y^2+\frac{3 (2 C+1) x^2}{2 C}y-\frac {(2 C+1) x^3+2 }{2C}=0$$
For $C=1$, the solution is
$$y=x \left(1-\sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\sqrt{2}\frac{
   \left(x^3-1\right)}{x^3}\right)\right)\right)$$
For $C=2$,
$$y=x \left(1-\sinh \left(\frac{1}{3} \sinh ^{-1}\left(2-\frac{2}{x^3}\right)\right)\right)$$
So, trying the substitution $y=x \left(1-e^{z}\right)$, the differential equation becomes
$$\left(x^4 \left(e^{z}-1\right)+x\right) z'+1=0$$ Switch variables
$$\frac{\left(e^z-1\right) x^4+x}{x'}+1=0$$ Let $x=\frac 1{\sqrt[3]t}$
$$t'-3 t-3 e^z+3=0\quad \implies t=c_1 \,e^{3 z}+\frac{1}{2} \left(2-3 e^z\right)$$
Now, back to $x$ and $y$ for the implicit solution.
A: Substitute $y(x)=xv(x)$. Then $y'(x)=v(x)+xv'(x)$ and hence we obtain the ODE
$$v'(x)=\frac{v(x)-1}{x^4 v(x)-x}$$
An obvious solution to this differential equation is $v(x)=1$, i.e. $y(x)=x$. To find other solutions, note that the form of the ODE suggests that we write $v'(x)=\frac{dv}{dx}=\frac{1}{dx/dv}=\frac{1}{x'(v)}$ and treat $v$ as the independent variable. This gives a Bernoulli ODE
$$x'(v)+\frac{1}{v-1}x(v)=\frac{v}{v-1}(x(v))^4.$$
The canonical substitution $u(v)=\frac{1}{x(v)^3}$ reduces this to the linear ODE
$$\frac{du}{dv}-\frac{3}{v-1}u(v)=-\frac{3v}{v-1}.$$
Solving this using your favorite method (e.g. integrating factors) gives
$$u(v)=\frac{1}{2}(3v-1)+C(v-1)^3$$
for some arbitrary constant $C$. Plugging everything back in and simplifying gives the general solution (in implicit form)
$$\frac{3}{2}yx^2-\frac{x^3}{2}+C(y-x)^3=1. \tag{$\ast$}$$
Note that in theory you could make these solutions explicit by solving some cubic equation, but they do not have simple expressions.
For visualisation, one can plot these solutions using e.g. Desmos. Note that I have verified that these solutions coincide with Mathematica's numerical solutions.
Remark: In a certain sense the trivial solution $y(x)=x$ is a limiting case ($C\to \infty$) of the general solution found in $(*)$. One can see this intuitively by setting $C$ to be very large in the Desmos plot.
