Question

Can someone explain to me how to use the Integrating Factor?

I have\[u_t=ku_{xx}+Q(x,t)\]on \(-\infty 12 years ago

Answer 1

so I take the FT and get\[\bar{U}_t=-kw^2\bar{U}+\bar{Q}\]and now my solutions say that the "usual" integrating factor can be used to obtain a particular solution. What is this "integrating factor"?

Answer 2

Does this look familiar from ordinary DE?
\( y' + p(x) y = q(x) \) Integrating Factor: \( \mu = e^{\int p(x) \ dx} \) which has property
\(\mu' = p(x) e^{\int p(x) dx} = p(x) \mu(x)\)

\( \mu y' + \mu p y = \mu q \)
\( \mu y' + \mu' y = \mu q \)
\( \left( \mu y \right) ' = \mu q \)

Answer 3

Essentially the integrating factor is created with that specific property in mind so that after multiplying both sides by it, the side with y and its first derivative can be pulled together by product rule and then the integral of both sides will remove the combined result.

Answer 4

hmm. i'm not really getting anywhere. my ODE textbook has an explanation of this but i'm getting lost trying to follow the steps.

Answer 5

I have \[\frac{dU}{dt}+kwU=Q(w)\]so I got an integrating factor\[\mu=e^{kw^2/2}\]

Answer 6

then my book tells me to multiply both sides of the equation by this factor.

Answer 7

\[e^{kw^2/2}\frac{dU}{dt}+kwe^{kw^2/2}U=Qe^{kw^2/2}\]

Answer 8

You want to look at this equation more like an ordinary differential equation in t.
So our integrating factor would also be integrated with respect to t.
\( \bar{U} _t = -k \omega ^2\bar{U} + \bar{Q} \)
\( \bar{U}_t + k \omega ^2 \bar{U} = \bar {Q} \)
We have dU/dt, so we want to multiply everything by a function mainly of t.
\( \large \mu (t) = e^{ \int k \omega ^2 dt } \)
Note that this looks a bit more like what the end result's exponent has.

Answer 9

And the property that I mentioned earlier is that:
\( \mu ' (t) = \dfrac{d}{dt} e^{ k \omega ^2 t } = k \omega ^2 e^{k \omega^2 t} = k \omega^2 \color{green}{\mu}\)

Answer 10

So multiplying both sides by \( \mu \) yields:
\( \mu \bar{U}_t + \color{red}{k \omega ^2 \mu} \bar{U} = \bar{Q} \)
This is where the usefulness of integrating factor comes into play.

Answer 11

the right side is \( \bar{Q} \mu \) ***

Answer 12

ah, thanks. I think I get it now.

Answer 13

so I end up with like\[\frac{d}{dt}(e^{kw^2t})=Q(w,t)e^{kw^2t}\]

Answer 14

oops
\[\frac{d}{dt}(e^{kw^2t}U)=Q(w,t)e^{kw^2t}\]

Answer 15

yup, that looks good.

Answer 16

\[e^{kw^2t}U=\int Q(w,t)e^{kw^2t}dt\]
\[U=e^{-kw^2t}\int Q(w,t)e^{kw^2t}dt\]

Answer 17

and then the tau is just a dummy variable because the book used a definite integral?

Answer 18

Yes. In fact, the definite integral should give you the "particular" solution rather than an indefinite integral with any constants of integration.

Answer 19

right. ok thanks a lot for your help. hopefully my ODE skills don't screw me like this on my final exam!

Answer 20

mm, glad to help!
and imo integrating factors are the trickiest thing i can remember for methods in ODE. lol