Question

What is the general solution of the differential equation? : `xy dy/dx = x^2 - y^2`

Answer 1

`y = +-sqrt(x^2/2 - A/x^2)`

Explanation:

We have:

`xy dy/dx = x^2 - y^2` ..... [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 = (x^2 - y^2)/(xy)`

`:. dy/dx = (x^2)/(xy) - (y^2)/(xy)`

`:. dy/dx = x/y - y/x` ..... [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 = 1/v - v`
`:. x(dv)/dx = 1/v - 2v`

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

`:. v/(1-2v^2) (dv)/dx = 1/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/(1 - 2v^2) \ dv = int \ 1/x \ dx`
`:. -1/4 \ int \ (-4v)/(1 - 2v^2) \ dv = int \ 1/x \ dx`

And this is trivial to integrate to get:

`-1/4 ln | 1 - 2v^2 | = ln |x| + lnC`
`:. -1/4 ln | 1 - 2v^2 | = ln |Cx|`
`:. ln | 1 - 2v^2 | = (-4)ln |Cx|`

`:. ln | 1 - 2v^2 | = ln |1/(Cx)^4|`

And the due to the monotonicity of the logarithmic function we have:

`1 - 2v^2 = 1/(Cx)^4`

And restoring the substitution we get:

`1 - 2(y/x)^2 = 1/(Cx)^4`

Which is the General Solution of [A], and is we require an explicit solution; then:

`2(y/x)^2 = 1 - 1/(Cx)^4`
`:. y^2/x^2 = 1/2 - 1/2 1/(Cx)^4`

`:. y^2 = x^2{ 1/2 - 1/2 1/(Cx)^4 }`

`:. y^2 = x^2/2 - A/x^2`

Leading to the explicit solution:

`y = +-sqrt(x^2/2 - A/x^2)`