Why Do We Use Integrating Factors in Solving First Order Linear ODEs?

[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'.