Angelina Lai

Posted on Saturday, 15 February, 2003 - 08:26 pm:

For an electrical circuit, the current, i amperes, flowing once the switch is closed is given by
L(di/dt) + Ri = V,
where L is the inductance measured in henries, R is the resistance measured in ohms, V is the aplied voltage in volts.
Find the general solution of this differential equation.

Yatir Halevi

Posted on Saturday, 15 February, 2003 - 08:58 pm:

L,R,V are independent of t?
If so, try the following substitution:
i(t)=z(t)+V/R

Yatir

Joel Kammet

Posted on Saturday, 15 February, 2003 - 11:00 pm:

Yatir, please explain how you found that substitution.

I would have rearranged the equation to

$di / dt + (R / L) i = (V / L)$

and then use $e^{R t / L}$ as an integrating factor to get

$e^{R t / L} di / dt + (R / L) i e^{R t / L} = (V / L) e^{R t / L}$

$i e^{R t / L} = \int (V / L) e^{R t / L}$

$i e^{R t / L} = (V / R) e^{R t / L} + c$

$i = V / R + C e^{- R t / L}$

& I'm very happy to see we both get the same solution.

Joel

Yatir Halevi

Posted on Sunday, 16 February, 2003 - 06:49 am:

I once gave a very similar differential equation, here is the forum. And I remember the answer being with an aid of a substitution. Combing this with the fact that the integral of

z < quote / > / z

\ln z

the answer becomes pretty obvious (you need to get rid of the constant!). Now, you can either see what the required substitution will be or just plug in

z = y + α

and find out what

α

needs to be in order for the constant to disappear...
Joel, Wouldn't it (technically) be more correct to use e^Rt/l+D (for some constant D) as the integrating factor?

Yatir

David Loeffler

Posted on Sunday, 16 February, 2003 - 10:43 am:

May I point out that when you use a constant in the integrating factor for first-order ODE's, all you are doing is multiplying the whole equation through by a constant? There is no way whatsoever that omitting the constant could cause you to lose solutions. Omit it freely!

David

Matthew Smith

Posted on Sunday, 16 February, 2003 - 04:18 pm:

As an aside, the original equation can also be solved by calculating the complementary function and the particular integral: in this case, the former is Ce^-Rt/L and the latter V/R. This method still works when you have capacitance as well, and hence a second order equation.

Angelina Lai

Posted on Sunday, 16 February, 2003 - 09:55 pm:

Just wondering, would the constant in the integration factor matter for some other kind of DEs that are not first order?

David Loeffler

Posted on Sunday, 16 February, 2003 - 10:03 pm:

Angelina Lai

Posted on Wednesday, 19 February, 2003 - 10:40 pm:

Ok thanks!

It's just rather strange that this question appeared in DE before the chapter where it introduced integration factors. Hmm... is there another way to do it?

Colin Prue

Posted on Thursday, 20 February, 2003 - 12:35 am:

i was perplexed about this question too - i had the same thought when i came across it in DE a month or so ago...I was thinking ''why is this under separation of variables when i think it should be under integrating factor''. Answer: because the easiest way is to use separation of variables:

rearrange to give:

$L \int (V - R i)^{- 1} di = \int dt$

it looks so obvious once you see it, and it's not hard...but it caused me some head scratching at the time