## diff. equation help

1. The problem statement, all variables and given/known data
(3x-2y+1)dx+(3x-2y+3)dy=0

this a piece of my solution: (pls correct if i'm wrong)

I thought of solving it using miscellaneous substitution using (3x-2y+1) as u...
du= 3dx- 2dy ,dx=(du+2dy)/3 so,

>u[(du+2dy)/3) + udy +2dy =0

>u(du +2dy) + 3udy + 6dy =0

>udu +(5u+6)dy=0

>[u/ (5u+6)] du + dy =0
integrating it:

> (5u+6)/25 + 6/25 [ln (5u+6)] + y = 0

>(5u+6) + 6[ln(5u+6) + 25y =0

>(15x-10y+11) + 6[ln(15x-10y+11)] +25y =0

but this answer is way too different from the answer on the book the answer there is:
5(x+y+c) = 2ln[15x-10y+11] ... what could be my mistake?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
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 Recognitions: Homework Help you ans is same as the book's....simplify it! note you are basically doing the reverse of this f(x,y)=K where K is some constant then $$\frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} =\frac{df}{dt}=\frac{d(K)}{dt}=0$$ where K is 5c in your book I think
 thanks.. but uhm.. sorry I do not understand it well.. could you please explain further how can i simplify so that i could come out to the same answer as the book.. please.

## diff. equation help

Okay, you made a mistake in integration:
(u/ (5u+6))du + dy = 0
((1 - (6/5u+6))/5) du +dy = 0
On integrating:
15x-10y+11 - 6ln(15x-10y+11) +25y = k
15x+15y+11+k = 6ln(15x-10y+11)
15(x+y+c) = 6ln(15x-10y+11)
5(x+y+c) = 2ln(15x-10y+11)

Done!
 a.. ok.. wow, never noticed that.. thanks a lot!