ODE question - Integrating factor

In summary, the equation y' = (2y)/(t.logt) = 1/t, t > 0 can be solved using the integrating factor I = exp(\intp(x)dx), where p(t) = -2/(t.logt). By using the substitution u = log t, the indefinite integral can be found to be 0 = 1, but this result is only paradoxical because the constant of integration has been neglected.
  • #1
Pyroadept
89
0

Homework Statement


Find all solutions of the equation:

y' = (2y)/(t.logt) = 1/t, t > 0


Homework Equations


Integrating factor I = exp([tex]\int[/tex]p(x)dx)

where y' + p(x)y = q(x)


The Attempt at a Solution


Hi everyone, here's what I've done so far:

Let p(t) = -2/(t.logt)

I = exp([tex]\int[/tex]((-2)/(t.logt))dt)

Factoring out the -2, consider [tex]\int[/tex]1/(t.logt)dt

Use integration by parts:
u = 1/logt
du = -(logt)^-2.(1/t).dt

dv = (1/t)dt
v = logt

[tex]\int[/tex]u.dv = uv - [tex]\int[/tex]v.du

I end up with:

[tex]\int[/tex]1/(t.logt)dt = 1 + [tex]\int[/tex]1/(t.logt)dt

which gives me 0 = 1, which is clearly wrong.

But I can't see where I'm going wrong! I've done it five times now and I keep getting the same answer. Can anyone see where I'm going wrong or suggest another way of solving the problem?

Thanks for any help
 
Physics news on Phys.org
  • #2
Try the substitution u=log t.
 
  • #3
You didn't do anything wrong when you integrated by parts. If you were evaluating definite integrals, the first term would drop out since you'd get 1-1=0. With the indefinite integrals, the paradoxical result arises because you've neglected the constant of integration.
 

1. What is an integrating factor in an ODE?

An integrating factor is a function that is multiplied to both sides of an ordinary differential equation (ODE) to simplify the process of solving the equation. It is used to transform a non-exact ODE into an exact one, making it easier to solve.

2. How do you find the integrating factor for an ODE?

To find the integrating factor for an ODE, you first need to determine if the equation is exact or not. If the equation is not exact, you can use the formula e∫P(x)dx, where P(x) is the coefficient of the y term in the ODE, to find the integrating factor. If the equation is exact, the integrating factor is simply 1.

3. Why is an integrating factor used in solving ODEs?

An integrating factor is used in solving ODEs because it makes the equation exact, which simplifies the solution process. It also allows for the use of methods like separation of variables and integrating factors to solve ODEs that would otherwise be difficult or impossible to solve.

4. Can you use any function as an integrating factor?

No, not all functions can be used as an integrating factor. The function must have certain properties, such as being non-zero and continuous, to be a valid integrating factor for a given ODE.

5. Are there any limitations to using an integrating factor in solving ODEs?

One limitation of using an integrating factor is that it may not always be possible to find a suitable function to make the ODE exact. In these cases, alternative methods of solving the ODE may need to be used. Additionally, using an integrating factor can sometimes lead to more complex solutions, so it may not always be the most efficient method of solving an ODE.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
310
  • Calculus and Beyond Homework Help
Replies
3
Views
559
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
135
  • Calculus and Beyond Homework Help
Replies
12
Views
977
Replies
1
Views
600
  • Calculus and Beyond Homework Help
Replies
8
Views
754
  • Calculus and Beyond Homework Help
Replies
2
Views
364
  • Calculus and Beyond Homework Help
Replies
23
Views
930
  • Calculus and Beyond Homework Help
Replies
16
Views
552
Back
Top