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

User MattClarke
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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.


User Abdul Rashid
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