First order linear differential equation

  • Thread starter kingwinner
  • Start date
  • #1
1,270
0
1) Solve y' + (1/t) y = t^3.

Integrating factor
=exp ∫(1/t)dt
=exp (ln|t| + k)
=exp (ln|t|) (take constant of integration k=0)
=|t|
...
and then I've found that the gerenal solution is:
y = 1/|t| + [c + ∫(from 0 to t) |s| s^3 ds]
Is this the correct final answer and is there any way to simply this? Can I get rid of all the absolute values?




2) Solve the initial value problem ty'+2y=4t^2, y(1)=2.

Integrating factor=t^2
...
General solution is y = t^2 + c/t^2
Put y(1)=2 => c=1
So the solution to the initial value problem is y=t^2 + 1/t^2, t>0.
Note that the function y=t^2 + 1/t^2, t<0 is NOT part of the solution of this initial value problem.

==============================
I have no idea (red part) why you have to put the restriction t>0, and why is the part for t<0 definitely NOT part of the solution? What's the problem here? This example is driving me crazy...


I am a beginner of this subject, and I hope that someone would be nice enough to explain these. Thanks!
 
Last edited:

Answers and Replies

  • #2
139
0
the initial value problem ty'+2y=4t^2, y(1)=2.

Integrating factor=t^2
...
General solution is y = t^2 + c/t^2
Put y(1)=2 => c=1
So the solution to the initial value problem is y=t^2 + 1/t^2, t>0.
Note that the function y=t^2 + 1/t^2, t<0 is NOT part of the solution of this initial value problem.

==============================
I have no idea (red part) why you have to put the restriction t>0, and why is the part for t<0 definitely NOT part of the solution? What's the problem here? This example is driving me crazy...


I am a beginner of this subject, and I hope that someone would be nice enough to explain these. Thanks!
As to this one, your integrating factor is correct but your simplification of it's original form is what hides the problem from you regarding the t > 0 restriction. Look at the exponential form of the integrating factor and notice that it has an ln |t| in it. This has a discontinuity at t = 0, and since your initial value is assigned as t = 1, we can only use the interval containing t = 1 (which is precisely the interval t > 0).

In other words, that is the largest allowable interval upon which you can define a continuous integrating factor which contains the initial value.
 
  • #3
1,270
0
As to this one, your integrating factor is correct but your simplification of it's original form is what hides the problem from you regarding the t > 0 restriction. Look at the exponential form of the integrating factor and notice that it has an ln |t| in it. This has a discontinuity at t = 0, and since your initial value is assigned as t = 1, we can only use the interval containing t = 1 (which is precisely the interval t > 0).

In other words, that is the largest allowable interval upon which you can define a continuous integrating factor which contains the initial value.
2) Sorry, I still don't fully understand what you mean...
y=t^2 + 1/t^2, t<0 and y=t^2 + 1/t^2, t>0 are just the left and right branches of a single function y=t^2 + 1/t^2, so shouldn't y=t^2 + 1/t^2 be the full and complete answer? I don't get what's the problem with the t<0 part at all...
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
You don't see a problem at t= 0?
 
  • #5
1,270
0
You don't see a problem at t= 0?

Well, at t=0, the solution is undefined, but why is there a problem for t<0?
 
  • #6
743
1
You've used ln(t) to derive your solution, but ln(t) is not defined for [itex] t \leq 0 [/itex]. The fact that you know ln(t) is continuous in it's domain and that there is a singularity at t=0 is a bit of a tip-off.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
41,847
966
There isn't a problem with t< 0 but you cannot extend a solution across t= 0. Typically, for a first order differential equation, you also have an "initial condition" which here must be given at a non-zero value of t. If you initial condition is [itex]y(t_0)= y_0[/itex], with [itex]t_0> 0[/itex], then your solution is only defined for t> 0. If [itex]y(t_0)= y_0[/itex], with [itex]t_0< 0[/itex], then your solution is only defined for t< 0.
 
  • #8
1,270
0
There isn't a problem with t< 0 but you cannot extend a solution across t= 0. Typically, for a first order differential equation, you also have an "initial condition" which here must be given at a non-zero value of t. If you initial condition is [itex]y(t_0)= y_0[/itex], with [itex]t_0> 0[/itex], then your solution is only defined for t> 0. If [itex]y(t_0)= y_0[/itex], with [itex]t_0< 0[/itex], then your solution is only defined for t< 0.
2) Are there any intuitive/theoretical reasons for choosing only one branch, but not the other?

As far as I know, in this case, the left and right branches are defined by the SAME function, then why should we even consider rejecting part of it??
 
  • #9
139
0
There are theoretical reasons that derive from the existence and uniqueness theorem regarding solutions to ODEs. Intuitive? Not that I can see. The theorem states that in a first order ODE of the form in this example that the derivative (y') must be a continuous function of t and y on some interval containing the initial value (t = 1 in this case), i.e. y' = f(t,y) where f(t,y) is continuous in some interval containing t = 1. If you put your equation in that form (i.e. solve it for y') you have a problem at t = 0. Thus any solution is valid on an interval a < t < b only if (1) t = 1 is in the interval and (2) f(t,y) is continuous on the interval. t > 0 is that interval.
 
  • #10
1,270
0
Thank you! It really helps!

How about Q1?
I am stuck at the point y = 1/|t| + [c + ∫(from 0 to t) |s| s^3 ds], how can I proceed from here?
The absolute values look unpleasant........
 
  • #11
139
0
1) Solve y' + (1/t) y = t^3.

Integrating factor
=exp ∫(1/t)dt
=exp (ln|t| + k)
=exp (ln|t|) (take constant of integration k=0)
=|t|
...
and then I've found that the gerenal solution is:
y = 1/|t| + [c + ∫(from 0 to t) |s| s^3 ds]
Is this the correct final answer and is there any way to simply this? Can I get rid of all the absolute values?
OK. First, since you have a similar problem at t = 0, a general solution will have to be on either t > 0 or t < 0. Notice what happens in each case with your integrating factor |t|. If t > 0, |t| = t. If t < 0 |t| = -t. Do each general solution separately, and then see if they can be "recombined" into the one that you gave above.
 
  • #12
1,270
0
OK. First, since you have a similar problem at t = 0, a general solution will have to be on either t > 0 or t < 0. Notice what happens in each case with your integrating factor |t|. If t > 0, |t| = t. If t < 0 |t| = -t. Do each general solution separately, and then see if they can be "recombined" into the one that you gave above.
How can I deal with the |s|? Is there any relationship between s and t? Does t<0 imply s<0, too?
 
  • #13
139
0
s goes from 0 to t in your integral. If t > 0 |s| = s. If t < 0 you can reverse the order of the integral so that s goes from t to 0 and replace |s| with -s (notice the effect of both of these will produce something nice).

By the way, is the answer you gave what you found or is it the known solution?
 
Last edited:
  • #14
HallsofIvy
Science Advisor
Homework Helper
41,847
966
2) Are there any intuitive/theoretical reasons for choosing only one branch, but not the other?

As far as I know, in this case, the left and right branches are defined by the SAME function, then why should we even consider rejecting part of it??
In the problem you gave, there is a very practical reason, neither "intuitive" not "theoretical"! You were asked to solve the equation with the initial condition y(1)=2. The domain of your solution must include x= 1 so you must choose the "branch" that includes x= 1, the function defined on the positive numbers. You cannot then extend your solution solution, in a unique way, to the negative numbers because you cannot pass 0.

You could define a function having the correct solution (with y(1)= 2) for the positive numbers and the given formula with any value for the undetermined constant for the negative numbers and have an infinite number of solutions to the equation satisfying y(1)= 2.

By the way, a "function" is NOT a "formula". The same formula is used on the left and right branches, but not the same function because they have different domains.
 
  • #15
1,270
0
s goes from 0 to t in your integral. If t > 0 |s| = s. If t < 0 you can reverse the order of the integral so that s goes from t to 0 and replace |s| with -s (notice the effect of both of these will produce something nice).
Will the answer (gerneral solution) be the same for both the cases t>0 and t<0? What if they aren't the same?

By the way, is the answer you gave what you found or is it the known solution?

The answer I gave is what I found
 
  • #16
139
0
Will the answer (gerneral solution) be the same for both the cases t>0 and t<0? What if they aren't the same?
They aren't necessarily the same. If they weren't the same you would define it in branches i.e. for t > 0 f(t) = ..., for t < 0 f(t) = ... etc.
 

Related Threads on First order linear differential equation

Replies
1
Views
2K
Replies
7
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
9
Views
3K
Replies
3
Views
1K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
1
Views
2K
Top