Question

What is the general solution of the differential equation `dy/dx=(y^3 - yx^2) / (x^3 + y^2x)`?

(portions of this question have been edited or deleted!)

Answer 1

`y^2/(2x^2) + ln |xy| + C = 0`

Explanation:

Presumably you require the solution, of the Differential equation rather than the equation, which is of course in the question.

We have:

`dy/dx=(y^3 - yx^2) / (x^3 + y^2x)` ..... [A]

Note, that we cannot isolate terms in `x` and `y` alone so let us attempt some manipulation in an attempt to simplify the equation:

`dy/dx = (y(y^2 - x^2)) / (x(y^2 + x^2))`
`\ \ \ \ \ = (y/x) (y^2 - x^2) / (y^2 + x^2) xx (1/x^2)/(1/x^2)`
`\ \ \ \ \ = (y/x) ((y/x)^2 - 1) / ((y/x)^2+1)` ..... [B]

So Let us try a substitution, Let:

`v = y/x => y=vx => dy/dx = v + x(dv)/dx`

And substituting into the above DE [B], to eliminate `y`:

`v + x(dv)/dx = (v) (v^2 - 1) / (v^2+1)`

`:. x(dv)/dx = v{ (v^2 - 1) / (v^2+1) - 1 }`

`:. x(dv)/dx = v{ ( (v^2-1) - (v^2+1) ) / (v^2+1) }`

`:. x(dv)/dx = ( -2v ) / (v^2+1)`

`:. (v^2+1)/v (dv)/dx = -2/x`

Which has indeed helped as it has reduced the initial equation [A] to a separable equation, so we can now "seperate the variables" to get:

`int \ (v^2+1)/v \ dv = int \ -2/x \ dx`

`:. int \ v+1/v \ dv = int \ -2/x \ dx`

And this is trivial to integrate to get:

`1/2v^2 + ln |v| = -2ln|x| - C`

And restoring the substitution we get:

`1/2(y/x)^2 + ln |y/x| = -2ln|x| - C`
`:. y^2/(2x^2) + ln |y| -ln|x| = -2ln|x| - C`
`:. y^2/(2x^2) + ln |y| +ln|x| + C = 0`
`:. y^2/(2x^2) + ln |xy| + C = 0` QED