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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:

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

we first consider the

[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

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

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 .

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

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 .

**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

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