Integrating Factor (Derivation)

[Saladsamurai]

Hello all! 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:

$$y' + p(x)y = q(x)\qquad(1)$$

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

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

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

$$y(x) = Ae^{-\int_a^x p(\xi)\,d\xi}\qquad(4)$$

which is equivalent to (3) since $\int p(x)\,dx \text{ and } \int_a^x p(\xi)\,d\xi$ differ at most by an additive constant, say D, and the resulting $e^D$ 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 $\mu(x)$ giving

$$\mu(x) y' + \mu(x)p(x)y = \mu(x)q(x)\qquad(5)$$

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

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

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

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

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

Question(s) 2:

There are 3 dummy variables being used here:

$$e^{-\int_a^x p(\xi)\,d\xi}\qquad(8)$$

$$e^{\int_a^\xi p(\zeta)\,d\zeta}\qquad(9)$$

$$q(\xi)\,d\xi\qquad(10).$$

I am not really sure what is happening here. They are using $\xi$ 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

 

[Saladsamurai]

Anyone have any thoughts or comments? Thanks

 

[HallsofIvy]

Hello all! 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:

$$y' + p(x)y = q(x)\qquad(1)$$

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

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

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.
$$\int p(x)dx$$
without any limits of integration , is the "indefinite integral". If F(x) is an anti-derivative of p(x) then $\int p(x)dx= F(x)+ C$ where C is an arbitrary constant.
$$\int_a^x p(\xi)d\xi$$
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 $\mu(x)$ giving

$$\mu(x) y' + \mu(x)p(x)y = \mu(x)q(x)\qquad(5)$$

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

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

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

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

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

One good reason for doing that is that a person seeing
$$e^{-\int p(x) dx}\int e^{\int p(x)dx}$$
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:

$$e^{-\int_a^x p(\xi)\,d\xi}\qquad(8)$$

$$e^{\int_a^\xi p(\zeta)\,d\zeta}\qquad(9)$$

$$q(\xi)\,d\xi\qquad(10).$$

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

No, that is not the case. In
$$e^{-\int_a^x p(\xi)d\xi$$
$\xi$ is a dummy variable- the integral will be a function of x.

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

In that last integral, yes, $\xi$ 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

 

[Sherard]

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)

 

[klondike]

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'.