Question

What is the general solution of the differential equation ? `(6xy - 3y^2+2y) dx + 2(x-y)dy = 0`

Answer 1

`e^(3x) (2xy-y^2) = C`

Explanation:

`(6xy - 3y^2+2y) dx + 2(x-y)dy = 0 ..... [A]`

Suppose we have:

`M(x,y) dx = N(x,y) dy`

Then the DE is exact if `M_y-N_x=0`

`M = 6xy - 3y^2+2y => M_y = 6x-6y+2`
`N= 2(x-y) => N_x = 2`

`M_y - N_x != 0 =>` Not an exact DE

So, we seek an Integrating Factor `mu(u)` such that

`(muM)_y = (muN)_x`

So, we compute::

`(M_y-N_x)/N = (6x-6y+2 - 2)/(2(x-y)) = 3`

So the Integrating Factor is given by:

`mu(x) = e^(int \ 3 \ dx)`
`\ \ \ \ \ \ \ = e^(3x)`

So when we multiply the DE [A] by the IF we now get an exact equation:

`(6xy - 3y^2+2y)e^(3x) dx + 2(x-y)e^(3x)dy = 0`

And so if we redefine the function `M` and `N`:

`M = (6xy - 3y^2+2y)e^(3x)`
`N = 2(x-y)e^(3x)`

Then, our solution is given by:

`f_x = M` and `f_y = N` and

If we consider `f_y = N`, then:

`f = int \ 2(x-y)e^(3x) \ dy + g(x)`, where we treat `x` as constant
`\ \ = 2e^(3x) \ int x-y \ dy + g(x)`
`\ \ = 2e^(3x) (xy-1/2y^2) + g(x)`
`\ \ = e^(3x) (2xy-y^2) + g(x)`

And now we consider `f_x = M` and we differentiate the last result to get:

`f_x = (e^(3x))(2y) + (3e^(3x))(2xy-y^2) + g'(x)`
`\ \ \ = (2y + 6xy-3y^2 + g'(x))e^(3x)`

As `f_x=M` then we have:

`6xy - 3y^2+2y = 2y + 6xy-3y^2 + g'(x)`
`:. g'(x) = 0 => g(x) = K`

Leading to the GS:

`e^(3x) (2xy-y^2) = C`