# ODE Integrating factors found by inspection

## Homework Statement

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....

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

## 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:

LCKurtz
Science Advisor
Homework Helper
Gold Member

## Homework Statement

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:

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

which then becomes:

$\frac{d(xy)}{x^3y^2} + \frac{dy}{y} = 0$

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

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

## 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.

No, it doesn't. There is either a typo in the text or in your post.

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.

LCKurtz
Science Advisor
Homework Helper
Gold Member
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.

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.

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.