Question

Solve the differential equation :

(2xy^2-x^3)dy+(y^3-2yx^2)dx=0

Answer 1

Multiply both sides by
`x^(-3)` to get a homogeneous equation.

`(2xy^2 - x^3) \, dy + (y^3 - 2yx^2) \, dx = 0 \\\\ \implies \left(2(y^2)/(x^2) - 1\right) \, dy + \left((y^3)/(x^3) - 2\frac yx\right) \, dx = 0`

Substitute
`y=vx` and
`dy = x\,dv + v\,dx`. This makes the equation separable.

`(2v^2 - 1) (x\,dv + v\,dx) + (v^3 - 2v) \, dx = 0`

`x (2v^2 - 1) \,dv + (2v^3 - v) \, dx + (v^3 - 2v) \, dx = 0`

`x (2v^2 - 1) \,dv + (3v^3 - 3v) \, dx = 0`

Separate the variables.

`x (2v^2 - 1) \, dv = (3v - 3v^3) \, dx`

`(2v^2-1)/(3v-3v^3) \, dv = \frac{dx}x`

`\frac13 (1 - 2v^2)/(v(v-1)(v+1)) \, dv = \frac{dx}x`

Expand the left side into partial fractions.

`(1-2v^2)/(v(v-1)(v+1)) = \frac av + \frac b{v-1} + \frac c{v+1} \\\\ (1-2v^2)/(v(v-1)(v+1)) = (a(v^2-1) + bv(v+1) + cv(v-1))/(v(v-1)(v+1)) \\\\ 1-2v^2 = (a+b+c) v^2 + (b-c) v - a`

Solve for the coefficients
`a,b,c`.

`\begin{cases} a + b + c = -2 \\ b-c = 0 \\ -a = 1 \end{cases} \implies a=-1, b=c=-\frac12`

Thus our equation becomes

`\left(-\frac1{3v} - \frac1{6(v-1)} - \frac1{6(v+1)}\right) \, dv = \frac{dx}x`

Integrate both sides.

`\displaystyle \int \left(-\frac1{3v} - \frac1{6(v-1)} - \frac1{6(v+1)}\right) \, dv = \int \frac{dx}x`

`\displaystyle -\frac13 \ln|v| - \frac16 \ln|v-1| - \frac16 \ln|v+1| = \ln|x| + C`

Solve for
`v` (as much as you can, anyway).

`\displaystyle -\frac16 \left(2\ln|v| + \ln|v-1| + \ln|v+1|\right) = \ln|x| + C`

`\displaystyle \ln\left|\frac1{\sqrt[6]{v^2(v^2-1)}}\right| = \ln|x| + C`

`\displaystyle \frac1{\sqrt[6]{v^4-v^2}} = Cx`

`\sqrt[6]{v^4-v^2} = \frac Cx`

`\left(\sqrt[6]{v^4-v^2}\right)^6 = \left(\frac Cx\right)^6`

`v^4-v^2 = \frac C{x^6}`

Put the solution back in terms of
`y`.

`\left(\frac yx\right)^4-\left(\frac yx\right)^2 = \frac C{x^6}`

`y^4 - x^2y^2 = \frac C{x^2}`

`\boxed{x^2y^4 - x^4y^2 = C}`

which is about as simple as we can hope to get this.