Question

Solve the equation $(3 y^{4}-5 x^{2} y^{2} -6 x^{4} ) dx-(4x y^{3}) dy=0$ leaving your answer in implicit form.

Answer 1

$(3y^4-5x^2y^2-6x^4)\,\mathrm dx-4xy^3\,\mathrm dy=0\iff\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3y^4-5x^2y^2-6x^4}{4xy^3}$

Multiplying both the numerator and denominator by $\dfrac1{x^4}$, we get

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{4\left(\frac yx\right)^4-5\left(\frac yx\right)^2-6}{4\left(\frac yx\right)^3}$

Since the derivative can be expressed as a function of $\dfrac yx$, the ODE is homogeneous. This means substituting $y=xv$ will be an effective approach. Indeed, we have $\dfrac{\mathrm dy}{\mathrm dx}=x\dfrac{\mathrm dv}{\mathrm dx}+v$, and the ODE can be rewritten as the separable equation

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{3v^4-5v^2-6}{4v^3}$
$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v^4+5v^2+6}{4v^3}$
$\dfrac{4v^3}{v^4+5v^2+6}\,\mathrm dv=-\dfrac{\mathrm dx}x$
$\displaystyle\int\dfrac{4v^3}{v^4+5v^2+6}\,\mathrm dv=-\int\dfrac{\mathrm dx}x$
$\displaystyle\int\left(\frac{12v}{v^2+3}-\dfrac{8v}{v^2+2}\right)\,\mathrm dv=-\ln|x|+C$
$6\ln(v^2+3)-4\ln(v^2+2)=-\ln|x|+C$
$\ln\dfrac{(v^2+3)^6}{(v^2+2)^4}=-\ln x+C$
$\dfrac{(v^2+3)^6}{(v^2+2)^4}=\dfrac Cx$

$\dfrac{\left(\frac{y^2}{x^2}+3\right)^6}{\left(\frac{y^2}{x^2}+2\right)^4}=\dfrac Cx$
$\dfrac{(y^2+3x^2)^6}{x^4(y^2+2x^2)^4}=\dfrac Cx$
$\dfrac{(y^2+3x^2)^6}{(y^2+2x^2)^4}=Cx^3$

You're welcome to unpack this further, but I would stop here.