# Simple question - Derivation for obtaining Intgrating Factor

1. Mar 8, 2014

### jellicorse

I was wondering if somebody could clear up some confusion I have regarding this.

I've been going over the derivation for obtaining the integrating factor again in my book and there is one step I don't understand.

There's no point going through the whole thing from scratch, but I've got to the point where we need to multiply the whole DE by a function $$\mu(t)$$ such that the LHS of the DE is recognizable as the derivative of some function.

Need to choose $$\mu(t)$$ to satisfy:

$$\frac{d\mu(t)}{dt}=2$$ for this particular example.

$$\frac{d\mu(t)/dt}{\mu(t)}=2$$

But I don't see how the next step follows from the previous one:

$$\frac{d}{dt}ln|\mu(t)|=2$$

In particular, I don't see where the $$ln|\mu(t)|$$ has come from.

Can anyone tell me how this works?

2. Mar 8, 2014

### Fredrik

Staff Emeritus
You need to use the chain rule and that $\frac{d}{dx}\ln x=\frac 1 x$.

For all t such that $\mu(t)>0$,
$$\frac{d}{dt}\ln \mu(t)=(\ln\circ\mu)'(t)=\ln'(\mu(t))\mu'(t)=\frac{1}{\mu(t)}\mu'(t).$$ For all t such that $\mu(t)<0$,
$$\frac{d}{dt}\ln(-\mu(t))=(\ln\circ(-\mu))'(t)=\ln'((-\mu)(t))(-\mu)'(t)=\frac{1}{-\mu(t)}(-\mu'(t))=\frac{1}{\mu(t)}\mu'(t).$$

3. Mar 8, 2014

### jellicorse

Thanks for explaining that Frederik... I think I can see it now.