# 1st order diff eq.

1. Oct 10, 2008

### JinM

1. The problem statement, all variables and given/known data

$\frac{dy}{dx} = \frac{3x+2y}{3x+2y+2}$

2. The attempt at a solution

I thought I'd let u = 3x+2y+2, which would give me this,

$u' = 5 - \frac{4}{u}$. This is seperable, and we thus get

$\frac{1}{5}(3x + 2y +2) - \frac{4}{25}\ln|5(3x+2y+2)-4| = x + C$

Can someone check this? I've been differentiating this beast and I keep getting different coefficients. Is what I got here correct?

Last edited: Oct 10, 2008
2. Oct 10, 2008

### gabbagabbahey

I get

$\frac{1}{5}(3x + 2y +2) + \frac{4}{25}\ln|5(3x+2y+2)-4| = x + C$

Perhaps you should show me your steps when integrating

$\int \frac{u}{5u-4} du$

3. Oct 10, 2008

### JinM

Gah, I actually missed TeXing the 5 inside the log function -- I got the same thing as you on paper. Thanks!