Simple question - Derivation for obtaining Intgrating Factor

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In summary, the person is seeking clarification on the derivation for obtaining the integrating factor and is confused about the use of ln|\mu(t)| in the next step. They are then given an explanation using the chain rule and shown how the derivative of ln|\mu(t)| is equal to 1/\mu(t).
  • #1
jellicorse
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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 [tex]\mu(t)[/tex] such that the LHS of the DE is recognizable as the derivative of some function.


Need to choose [tex]\mu(t)[/tex] to satisfy:

[tex]\frac{d\mu(t)}{dt}=2[/tex] for this particular example.

[tex]\frac{d\mu(t)/dt}{\mu(t)}=2[/tex]


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

[tex]\frac{d}{dt}ln|\mu(t)|=2[/tex]

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

Can anyone tell me how this works?
 
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  • #2
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##,
$$\frac{d}{dt}\ln(-\mu(t))=(\ln\circ(-\mu))'(t)=\ln'((-\mu)(t))(-\mu)'(t)=\frac{1}{-\mu(t)}(-\mu'(t))=\frac{1}{\mu(t)}\mu'(t).$$
 
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  • #3
Thanks for explaining that Frederik... I think I can see it now.
 

Related to Simple question - Derivation for obtaining Intgrating Factor

1. What is the purpose of using an integrating factor?

An integrating factor is used to simplify the process of solving a differential equation by making it easier to integrate. It essentially allows us to convert a non-exact equation into an exact one, making it easier to find a solution.

2. How do you determine the integrating factor for a given differential equation?

The integrating factor for a given differential equation can be determined by multiplying both sides of the equation by an appropriate function that will make the left-hand side an exact differential. This function is known as the integrating factor.

3. Can any function be used as an integrating factor?

No, not all functions can be used as integrating factors. The function must satisfy certain conditions, such as being continuous and non-zero, in order to be a valid integrating factor for a given differential equation.

4. What is the general form of the integrating factor for a first-order differential equation?

The general form of the integrating factor for a first-order differential equation is e∫P(x)dx, where P(x) is the coefficient of the y' term in the differential equation. This form can be modified for specific types of differential equations, such as linear or homogeneous equations.

5. Can an integrating factor be used for higher-order differential equations?

Yes, an integrating factor can also be used for higher-order differential equations. However, the process may be more complex as it involves finding multiple integrating factors for each derivative term in the equation.

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