Question

Solve (3x+8)(y^2+4) dx +4

Answer 1

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.