Solving 1st Order Diff EQ: Check My Work for dy/dx = (3x+2y)/(3x+2y+2)

[JinM]

Homework Statement

\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?

 

[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

 

[JinM]

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