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(\mathrm dy)/(\mathrm dx)=(3y^4-5x^2y^2-6x^4)/(4xy^3)`

Multiplying both the numerator and denominator by
`\frac1{x^4}`, we get


`(\mathrm dy)/(\mathrm dx)=(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
`\frac yx`, the ODE is homogeneous. This means substituting
`y=xv` will be an effective approach. Indeed, we have
`(\mathrm dy)/(\mathrm dx)=x(\mathrm dv)/(\mathrm dx)+v`, and the ODE can be rewritten as the separable equation


`x(\mathrm dv)/(\mathrm dx)+v=(3v^4-5v^2-6)/(4v^3)`

`x(\mathrm dv)/(\mathrm dx)=-(v^4+5v^2+6)/(4v^3)`

`(4v^3)/(v^4+5v^2+6)\,\mathrm dv=-\frac{\mathrm dx}x`

`\displaystyle\int(4v^3)/(v^4+5v^2+6)\,\mathrm dv=-\int\frac{\mathrm dx}x`

`\displaystyle\int\left((12v)/(v^2+3)-(8v)/(v^2+2)\right)\,\mathrm dv=-\ln|x|+C`

`6\ln(v^2+3)-4\ln(v^2+2)=-\ln|x|+C`

`\ln((v^2+3)^6)/((v^2+2)^4)=-\ln x+C`

`((v^2+3)^6)/((v^2+2)^4)=\frac Cx`


`(\left((y^2)/(x^2)+3\right)^6)/(\left((y^2)/(x^2)+2\right)^4)=\frac Cx`

`((y^2+3x^2)^6)/(x^4(y^2+2x^2)^4)=\frac Cx`

`((y^2+3x^2)^6)/((y^2+2x^2)^4)=Cx^3`

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