# First order linear differential equation

1. Jan 13, 2008

### kingwinner

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.

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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: Jan 14, 2008
2. Jan 14, 2008

### Mathdope

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. Jan 14, 2008

### kingwinner

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. Jan 14, 2008

### HallsofIvy

Staff Emeritus
You don't see a problem at t= 0?

5. Jan 14, 2008

### kingwinner

Well, at t=0, the solution is undefined, but why is there a problem for t<0?

6. Jan 14, 2008

### Kreizhn

You've used ln(t) to derive your solution, but ln(t) is not defined for $t \leq 0$. 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. Jan 14, 2008

### HallsofIvy

Staff Emeritus
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 $y(t_0)= y_0$, with $t_0> 0$, then your solution is only defined for t> 0. If $y(t_0)= y_0$, with $t_0< 0$, then your solution is only defined for t< 0.

8. Jan 14, 2008

### kingwinner

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. Jan 15, 2008

### Mathdope

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. Jan 16, 2008

### kingwinner

Thank you! It really helps!

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. Jan 17, 2008

### Mathdope

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. Jan 17, 2008

### kingwinner

How can I deal with the |s|? Is there any relationship between s and t? Does t<0 imply s<0, too?

13. Jan 17, 2008

### Mathdope

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: Jan 17, 2008
14. Jan 17, 2008

### HallsofIvy

Staff Emeritus
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. Jan 17, 2008

### kingwinner

Will the answer (gerneral solution) be the same for both the cases t>0 and t<0? What if they aren't the same?

The answer I gave is what I found

16. Jan 17, 2008

### Mathdope

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.