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First order linear ODE-integrating factor has absolute value in it

by kingwinner
Tags: absolute, factor, linear, odeintegrating, order
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kingwinner
#1
Sep20-09, 03:37 PM
P: 1,270
1. The problem statement, all variables and given/known data
Solve the ODE y' + (3/t) y = t3.

2. Relevant equations/concepts
1st order linear ODE


3. The attempt at a solution
Integrating factor
=exp ∫(3/t)dt
=exp (3ln|t| + k)
=exp (ln|t|3) (take constant of integration k=0)
=|t|3

If the integrating factor were |t|2 = t2, I wouldn't have any problem with it, the absolute value is gone, luckily.
But now here in this case, multiplying both sides of the ODE by |t|3, the absolute value is giving me trouble. How can I proceed? Can I just forget about the absolute value and mutliply the ODE simply by t3? (I've seen a lot of people doing this, but I don't think it's correct...)

I am never able to understand how to deal with problems like this. What is the correct way to handle these problems where there is an absolute value sign in the integrating factor?

Any help in this matter is greatly appreciated!
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rock.freak667
#2
Sep20-09, 04:15 PM
HW Helper
P: 6,202
In those types of questions you don't need to always have to use |t|, you can use it as just t or in your case t3
kingwinner
#3
Sep20-09, 04:24 PM
P: 1,270
Quote Quote by rock.freak667 View Post
In those types of questions you don't need to always have to use |t|, you can use it as just t or in your case t3
But that automatically means that you're assuming t>0, and in the problem there is no restriction on t (t can be negative, right?).

rock.freak667
#4
Sep20-09, 04:36 PM
HW Helper
P: 6,202
First order linear ODE-integrating factor has absolute value in it

Quote Quote by kingwinner View Post
But that automatically means that you're assuming t>0, and in the problem there is no restriction on t (t can be negative, right?).
I believe if you take both cases for t>0 |t|=t and for t<0 |t|= -t, your general solution is the same.
kingwinner
#5
Sep20-09, 05:55 PM
P: 1,270
Quote Quote by rock.freak667 View Post
I believe if you take both cases for t>0 |t|=t and for t<0 |t|= -t, your general solution is the same.
um...why? Can we always ignore the absolute value that appears in the integrating factor?
rock.freak667
#6
Sep20-09, 06:18 PM
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P: 6,202
Quote Quote by kingwinner View Post
um...why? Can we always ignore the absolute value that appears in the integrating factor?
because when you multiply the ODE by |t3| we get:

y|t3|=∫|t3|t3 dt

Check what happens for t>0 (|t| = t) and t<0 (|t|=-t)
kingwinner
#7
Sep20-09, 06:24 PM
P: 1,270
Quote Quote by rock.freak667 View Post
because when you multiply the ODE by |t3| we get:

y|t3|=∫|t3|t3 dt

Check what happens for t>0 (|t| = t) and t<0 (|t|=-t)
So in this example, if t>0, we multiply the whole ODE equation by t^3
And if t<0, we multiply the whole ODE equation by -t^3, which gives the exact same equation as the above case since we can cancel out the negatives from both sides.
Therefore, in either case, the general solution must be the same, right?
kingwinner
#8
Sep21-09, 04:31 PM
P: 1,270
I've read an example on the web in which the integrating factor is |t|^2 = t^2 and they commented: "We were able to drop the absolute value bars here because we were squaring the t, but often they can’t be dropped so be careful with them and don’t drop them unless you know that you can. Often the absolute value bars must remain."

But I don't know why the absolute value bars must remain in most cases?? Can someone please give me an example in which the absolute value bars must reamin in the integrating factor and would get different answers for the different cases??

Help...I am confused...


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