# ODE Integrating factors found by inspection

1. Jul 20, 2012

### equant

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

I'm reading a chapter out of Elem. DE 6th Edition by Rainville and Bedient (Ch 4 pg 61) titled integrating factors found by inspection. To explain it, the authors start with an equation, which is grouped to become:

$y dx + x dy + x^3y^2 dy = 0$

which then becomes:

$\frac{d(xy)}{(xy)^3} + \frac{dy}{y} = 0$

I am unsure of how they get from there to the next step which is this....

$-\frac{1}{2x^2y^2} + ln|y| = -ln|c|$

3. The attempt at a solution

Trying to figure out the intermediate steps I start by doing this...

$\frac{d(xy)}{x^3y^2} = - \frac{dy}{y}$

And then I'm guessing they integrate both sides. However, if that's the case, what would I be integrating the left side with respect to? I expected the equation to become this...

$\frac{1}{x^2y^2} + ln|y| = c$

Thank you.

Last edited: Jul 20, 2012
2. Jul 20, 2012

### LCKurtz

No, it doesn't. There is either a typo in the text or in your post. If your original equation was$$y dx + x dy + x^3y^2 dy = 0$$ and you divided by $x^3y^3$ it would become$$\frac {d(xy)}{(xy)^3} + \frac {dy}{y}=0$$which gives your desired answer.

3. Jul 20, 2012

### equant

The typo was in my post. I corrected it. Sorry about that. With that now corrected, I'm still not sure how they get from the one step to the next.

4. Jul 20, 2012

### LCKurtz

It is of the form $\frac {du}{u^3} = u^{-3}du$ which integrates to $\frac{u^{-2}}{-2}$. And calling the constant of integration $-\ln c$ instead of $c$ is just a convenience. They can both be anything.

5. Jul 20, 2012

### equant

ok, and in this case u = xy. I was thinking that they had to be dealt with separately, but now I see it.

Thank you very much.