Simple question - Derivation for obtaining Intgrating Factor

[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?

 

[Fredrik]

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

 

[jellicorse]

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