It's been four years since I last did a problem of this particular nature. :)
Try this: divide both sides of the given equation by (x^2-5x+6), obtaining
(3x+8)(y^2+4) + 4y(x^2+5x+6) dy
-------------------- dx + ------------------------- dy = 0
x^2+5x + 6 x^2+5x + 6
This gives us:
(3x+8)(y^2+4)
-------------------- dx + 4ydy = 0
(x+2)(x+3)
Now divide all 3 terms by y^2+4:
3x+8
--------------- dx + 4ydy = 0 This completes the separation of variables.
(x+2)(x+3)
Now we'll need to integrate each of the 3 terms.
3x+8
--------------- dx + 4ydy = 0 can be re-written in the form
(x+2)(x+3)
3x +8 A B
--------------- = --------- + ---------- We will need to determine the values
(x+2)(x+3) x+2 x+3 of A and B (partial fraction expansion).
I trust you know how to do this; if not, let me know and I'll help you with that also.
I obtained B=2 and A = 1. You might want to verify that 1/(x+2) + 2/(x+3) =
3x+8
-------------- .
(x+2)(x+3)
Integrating the left side of this d. e. results in ln(x+2) + ln(x+3).
Integrating 4ydy results in 4*y^2/2, or 2y^2.
Integrating 0 results in a constant, C.
Thus,
ln(x+2) + ln(x+3) + 2y^2 = C is the solution of this d. e.
units²