Solve `(x-1)dy+ydx=x(x-1)y^(1/3)dxdot`

Explanation

Dividing by `dxy^(1//3)(x-1)`, the given equation reduces to `y^(-1//3)(dy)/(dx)+1/(x-1)y^(2//3)=x`
put `y^(2//3)`=z, so that, `2/3y^(-1//3)(dy)/(dx)=(dz)/(dx)`
Then given equation reduces to
`(dz)/(dx) + 2/(3(x-1))z = 2/3x` (linear form)
Hence, solution is given by
`z(x-1)^(2/3)=2/3intx(x-1)^(2//3) dx+c`
Putting `(x-1)=t^(3)` in the R.H.S, we get
`intx(x-1)^(2//3)dx = intt^(3)+1(t^(2)3t^(2))dt`
`=3int(t^(7)+t^(4))dt`
`=3[(1//8)t^(8)+(1//5)t^(5)]`
`=(3//8)(x-1)^(8//3)+(3//5)(x-1)^(5//3)`
Hence, the solution is `y^(2//3) = 1/4(x-1)^(2)+2/5(x-1)+c(x-1)^(-2//3)`.