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

kingwinner

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!

rock.freak667

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

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

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

um...why? Can we always ignore the absolute value that appears in the integrating factor?

rock.freak667

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

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

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...