Integrating Factor (Derivation)

  • #1
3,003
3
Hello all! :smile: I through a section in my text (by https://www.amazon.com/dp/0133214311/?tag=pfamazon01-20) on 1st Order linear ODEs. I am understanding the derivation of the integrating factor method pretty well; however, there are some aspects of the mathematics that I am getting hung up on. I have an "engineering" background in calculus and hence I get slowed down by the details sometimes.

In order to find the solution to an ODE of the form:

[tex]y' + p(x)y = q(x)\qquad(1)[/tex]

we first consider the homogeneous case; i.e.,
[tex]y' + p(x)y = 0\qquad(2)[/tex]

After some hootenanny, we arrive at the solution to (2) given by
[tex]y(x) = Ae^{-\int p(x)\,dx}\qquad(3)[/tex]

Now, in order to solve for the constant 'A' given the initial condition y(a) = b, it says that

"...it is convenient to re-express (3) as

[tex]y(x) = Ae^{-\int_a^x p(\xi)\,d\xi}\qquad(4)[/tex]

which is equivalent to (3) since [itex]\int p(x)\,dx \text{ and } \int_a^x p(\xi)\,d\xi[/itex] differ at most by an additive constant, say D, and the resulting [itex]e^D[/itex] can be absorbed into the arbitrary constant A.

Question 1:

I am not sure why he even brought this up? Why would changing the dummy variables cause the two integrals to differ *at all* ?


Moving on to the homogeneous case:

Basically we wish to multiply (2) by some function of x, say [itex]\mu(x)[/itex] giving

[tex]\mu(x) y' + \mu(x)p(x)y = \mu(x)q(x)\qquad(5)[/tex]

so that the left-hand-side of (5) is a derivative. After much more hootenanny, we arrive at the solution

[tex]y(x) = e^{-\int p(x)\,dx}\left(\int e^{\int p(x)\,dx}q(x)\,dx + C\right)\qquad(6)[/tex]

Now again, if we wish to solve for the constant 'C' given the initial condition y(a) = b, we are again advised to use a different expression of (6), namely

[tex]y(x) = e^{-\int_a^x p(\xi)\,d\xi}\left(\int_a^x e^{\int_a^\xi p(\zeta)\,d\zeta}q(\xi)\,d\xi + C\right)\qquad(7)[/tex]

This is just too much changing of dummy variables for my feeble mind to handle :redface: .

Question(s) 2:

There are 3 dummy variables being used here:

[tex]e^{-\int_a^x p(\xi)\,d\xi}\qquad(8)[/tex]

[tex] e^{\int_a^\xi p(\zeta)\,d\zeta}\qquad(9)[/tex]

[tex]q(\xi)\,d\xi\qquad(10).[/tex]

I am not really sure what is happening here. They are using [itex]\xi[/itex] in (8) and (10) as dummy variables and in (9) as an endpoint of the integral.

I don't expect anyone to answer all of this at once, but i would love to start a discussion about it so that I can solidify my understanding of the derivation. It's not enough anymore for me to just be able *to use* the integrating factor, but instead to *understand it.*

Thanks,
Casey
 
Last edited by a moderator:

Answers and Replies

  • #2
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Anyone have any thoughts or comments? Thanks :smile:
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
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Hello all! :smile: I through a section in my text (by https://www.amazon.com/dp/0133214311/?tag=pfamazon01-20) on 1st Order linear ODEs. I am understanding the derivation of the integrating factor method pretty well; however, there are some aspects of the mathematics that I am getting hung up on. I have an "engineering" background in calculus and hence I get slowed down by the details sometimes.

In order to find the solution to an ODE of the form:

[tex]y' + p(x)y = q(x)\qquad(1)[/tex]

we first consider the homogeneous case; i.e.,
[tex]y' + p(x)y = 0\qquad(2)[/tex]

After some hootenanny, we arrive at the solution to (2) given by
[tex]y(x) = Ae^{-\int p(x)\,dx}\qquad(3)[/tex]

Now, in order to solve for the constant 'A' given the initial condition y(a) = b, it says that



Question 1:

I am not sure why he even brought this up? Why would changing the dummy variables cause the two integrals to differ *at all* ?
Changing the dummy variables doesn't change anything and that's not what they are saying. The important change is the limits of integration.
[tex]\int p(x)dx[/tex]
without any limits of integration , is the "indefinite integral". If F(x) is an anti-derivative of p(x) then [itex]\int p(x)dx= F(x)+ C[/itex] where C is an arbitrary constant.
[tex]\int_a^x p(\xi)d\xi[/tex]
is an definite integral just with the variable x as an upper limit: with F(x) an anti-dervative as befor, it would be equal to F(x)- F(a). In other words, the "C" is now specifically -F(a).


Moving on to the homogeneous case:

Basically we wish to multiply (2) by some function of x, say [itex]\mu(x)[/itex] giving

[tex]\mu(x) y' + \mu(x)p(x)y = \mu(x)q(x)\qquad(5)[/tex]

so that the left-hand-side of (5) is a derivative. After much more hootenanny, we arrive at the solution

[tex]y(x) = e^{-\int p(x)\,dx}\left(\int e^{\int p(x)\,dx}q(x)\,dx + C\right)\qquad(6)[/tex]

Now again, if we wish to solve for the constant 'C' given the initial condition y(a) = b, we are again advised to use a different expression of (6), namely

[tex]y(x) = e^{-\int_a^x p(\xi)\,d\xi}\left(\int_a^x e^{\int_a^\xi p(\zeta)\,d\zeta}q(\xi)\,d\xi + C\right)\qquad(7)[/tex]

This is just too much changing of dummy variables for my feeble mind to handle :redface: .
One good reason for doing that is that a person seeing
[tex]e^{-\int p(x) dx}\int e^{\int p(x)dx}[/tex]
might think, mistakenly, that the two exponentials cancel. They don't of course- you cannot take a function of x "inside" the integral like that.

Question(s) 2:

There are 3 dummy variables being used here:

[tex]e^{-\int_a^x p(\xi)\,d\xi}\qquad(8)[/tex]

[tex] e^{\int_a^\xi p(\zeta)\,d\zeta}\qquad(9)[/tex]

[tex]q(\xi)\,d\xi\qquad(10).[/tex]

I am not really sure what is happening here. They are using [itex]\xi[/itex] in (8) and (10) as dummy variables and in (9) as an endpoint of the integral.
No, that is not the case. In
[tex]e^{-\int_a^x p(\xi)d\xi[/tex]
[itex]\xi[/itex] is a dummy variable- the integral will be a function of x.

In
[tex]e^{\int_a^\xi p(\zeta)d\zeta[/tex]
[itex]\zeta[/itex] is a dummy variable but [itex]\xi[/itex] is not- this integral will actually be a function of [itex]\xi[/itex] so that it can be integrated with respect to [itex]\xi[/itex] in the outer integral:
[tex]\int_a^x E(\xi)q(\xi) d\xi[/tex]
where "[itex]E(\xi)[/itex]" is the result of the previous calculation,
[tex]e^{\int_a^\xi p(\zeta)d\zeta[/tex]

In that last integral, yes, [itex]\xi[/itex] is a dummy variable- the integration is a function of x.

I don't expect anyone to answer all of this at once, but i would love to start a discussion about it so that I can solidify my understanding of the derivation. It's not enough anymore for me to just be able *to use* the integrating factor, but instead to *understand it.*

Thanks,
Casey
 
Last edited by a moderator:
  • #4
3
0
I'm learning differential equations, and although I understand the methods I have learned thus far, I often have trouble seeing what is the reasoning behind them.

Take for example, the use of the integrating factor when solving first order linear ODE's. I understand how to use it, but i'm not sure where it came from. In the resources that i'm using there's really not explanation, instead the discussion is limited to: assume there's a function by which we can multiply our differential equation to make it integrable.

Can anybody explain why we use this approach?

thanks,by (Erik006)
 
  • #5
125
2
I'm learning differential equations, and although I understand the methods I have learned thus far, I often have trouble seeing what is the reasoning behind them.

Take for example, the use of the integrating factor when solving first order linear ODE's. I understand how to use it, but i'm not sure where it came from. In the resources that i'm using there's really not explanation, instead the discussion is limited to: assume there's a function by which we can multiply our differential equation to make it integrable.

Can anybody explain why we use this approach?

thanks,by (Erik006)

I think the key is product rule. namely
(my)'=m'y+my'
Let's say you have a DE
y'+py=q
you want to make it look like product rule by multiplying both sides such that:
my'+mpy=mq
No quite product yet, but if mp=m' then
(my)'=mq
now that y can be solved by integrating both sides. m is very easy to solve given mp=m'.
 
Last edited:

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