Solve differential equations $(x^2y^2-1)dy + 2xy^3dx = 0$ Help me solve this:
$$(x^2y^2-1)dy + 2xy^3dx = 0$$
We now what:
$$\frac{\partial M}{\partial y}=2x^2y$$  and $$\frac{\partial N}{\partial x}=2y^3$$
I try to make it exact but get this:
$\frac{x^2-y^2}{xy^2}$
Help me!
may be substituchion $z = \frac{x^2-y^2}{xy^2}$ or we can be it more simple
 A: The ODE
$$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$$
is exact if
$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
Here we have
$$M(x,y)=2xy^3\implies\frac{\partial M}{\partial y}=6xy^2\\N(x,y)=x^2y^2-1\implies\frac{\partial N}{\partial x}=2xy^2$$
so the equation is not exact. Introduce an integrating factor $\mu(x,y)$ such that the ODE
$$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$$
is exact, i.e.
$$\frac{\partial(\mu M)}{\partial y}=\frac{\partial(\mu N)}{\partial x}\implies\frac{\partial\mu}{\partial y}M+\mu\frac{\partial M}{\partial y}=\frac{\partial\mu}{\partial x}N+\mu\frac{\partial N}{\partial x}\\
\implies\frac1\mu\left(\frac{\partial\mu}{\partial y}(2xy^3)-\frac{\partial\mu}{\partial x}(x^2y^2-1)\right)=-4xy^2$$
Notice that if $\mu(x,y)=\mu(y)$ (independent of $x$), then $\frac{\partial\mu}{\partial x}=0$ and
$$\frac1\mu\frac{\partial\mu}{\partial y}=-\frac{4xy^2}{2xy^3}=-\frac2y\implies\ln|\mu|=-2\ln|y|\implies\mu=\frac1{y^2}$$
Can you take it from here?
A: Another approach:
$$(x^2y^2-1)dy + 2xy^3dx = 0$$
$$x^2y^2dy + 2xy^3dx = dy$$
$$x^2y^2dy + y^3d(x^2) = dy$$
$$y^2(x^2dy + yd(x^2)) = dy$$
$$d(yx^2) = \frac {dy}{y^2} \implies yx^2+\frac 1 y =C$$
