ODE question - Integrating factor

Click For Summary
SUMMARY

The discussion focuses on solving the ordinary differential equation (ODE) given by y' = (2y)/(t.logt) = 1/t for t > 0. The user attempts to find the integrating factor using the formula I = exp(∫p(x)dx) with p(t) = -2/(t.logt). The integration by parts approach leads to confusion due to neglecting the constant of integration, resulting in an incorrect conclusion of 0 = 1. A suggested solution is to use the substitution u = log t to clarify the integration process.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with integrating factors and their application
  • Knowledge of integration techniques, specifically integration by parts
  • Basic concepts of logarithmic functions and their properties
NEXT STEPS
  • Study the method of integrating factors in ODEs
  • Learn about integration by parts with a focus on indefinite integrals
  • Research the substitution method in integration, particularly u-substitution
  • Explore common pitfalls in solving ODEs and how to avoid them
USEFUL FOR

Students studying differential equations, educators teaching ODE concepts, and anyone looking to improve their integration techniques in calculus.

Pyroadept
Messages
82
Reaction score
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
Try the substitution u=log t.
 
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.
 

Similar threads

Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
12
Views
2K
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K