Question

What is the general solution of the differential equation? `(3x^2+2y)dx+2xdy=0`

Answer 1

`y = -1/2x^2 + A/x`

Explanation:

We can write the equation `(3x^2+2y)dx+2xdy=0` as:

`\ \ \ \ \ \ \ 2xdy = -(3x^2+2y)dx`
`:. 2xdy/dx = -3x^2 - 2y`
`:. dy/dx + y/x = -3/2x`

This is a First Order DE of the form:

`y'(x) + P(x)y = Q(x)`

Which we know how to solve using an Integrating Factor given by:

`IF = e^(int P(x) \ dx)`

And so our Integrating Factor is:

`IF = e^(int 1/x \ dx)`
`\ \ \ \ = e^(ln|x|)`
`\ \ \ \ = x`

If we multiply by this Integrating Factor we will (by its very design) have the perfect differential of a product:

`\ \ \ \ \ dy/dx + y/x = -3/2x`

`:. xdy/dx +y = -3/2x^2`

`:. \ \ d/dx(xy) = -3/2x^2`

Which is now a separable DE, and we can separate the variables to get:

`" "xy = int \ -3/2x^2 \ dx`
`:. xy = -1/2x^3 + A`
`:. y = -1/2x^2 + A/x`

Answer 2

`y = C/x - x^2/2`

Explanation:

An alternative approach

The equation is exact. ie there exists a potential function `f(x,y) = C` such that `df = f_x dx + f_y dy = 0`

to verify this we check the mixed partials

`f_x = 3x^2 + 2y implies f_(xy) = 2`

`f_y = 2x implies f_(yx) = 2 qquad star`

`f_(xy) = f_(yx)` !!!

For reasons that become apparent, we can now choose to integrate either term ... we will integrate `f_x` wrt x

`f_x = 3x^2 + 2y implies f = x^3 + 2xy + alpha(y)`

....where `alpha(y)` is the function in y only that may have been lopped off by a differentiation wrt x.

We then look at the other partial that follows by now differentiating this f wrt y:

`f = x^3 + 2xy + alpha(y) implies f_y = 2x + alpha'(y)`

And from `star`

`alpha'(y) = 0 implies alpha(y) = C`

`implies f = x^3 + 2xy + C = tilde C`

Final bit of algebra:

`x(x^2 + 2y) = C`

`2y = C/x - x^2`

`y = C/x - x^2/2`