Wronskian Second Solution/Differential Equations

  • Thread starter Thread starter tracedinair
  • Start date Start date
  • Tags Tags
    Wronskian
Click For Summary
SUMMARY

The discussion centers on solving for Φ2 in the differential equation given by Φ2 = Φ1 * ∫ e^(-∫a(x)dx) / (Φ1)^2 dx, where Φ1 = cos(ln(x)) and a = 1/x. The integral simplifies to Φ2 = cos(ln(x)) * ∫ e^(-ln(x)) / cos^(2)(ln(x)) dx. A critical error was identified regarding the exponential expansion, where e^(-ln(x)) should be interpreted as 1/x, not -x. The correct approach involves using a simple u-substitution to resolve the integral.

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with integration techniques, specifically integration by parts
  • Knowledge of exponential functions and logarithms
  • Experience with u-substitution in calculus
NEXT STEPS
  • Practice solving differential equations involving integrals
  • Study integration by parts in depth
  • Review u-substitution techniques for integrals
  • Explore applications of the Wronskian in solving differential equations
USEFUL FOR

Students and educators in mathematics, particularly those focused on calculus and differential equations, as well as anyone seeking to enhance their problem-solving skills in this area.

tracedinair
Messages
47
Reaction score
0

Homework Statement



Given that Φ2 = Φ1 * ∫ e^(-∫a(x)dx)) / (Φ1)^2 dx

and Φ1 = cos(ln(x)), a = 1/x, solve for Φ2.

Homework Equations



The Attempt at a Solution



Φ2 = cos(ln(x)) * ∫ e^(-∫1/x dx)) / cos^(2)(ln(x)) dx

= cos(ln(x)) * ∫ e^(-ln(x)) / cos^(2)(ln(x)) dx

= cos(ln(x)) * - ∫ x / cos^(2)(ln(x)) dx

My problem begins here with trying to solve for that integral. I don't have the slightest idea where to begin, except maybe integration by parts.
 
Physics news on Phys.org
I believe you have expanded your brackets in the exponential wrong e^{-ln(x)} does not equal -x but rather e^{ln(x^{-1})} which is of course \frac{1}{x} in which case a simple u substitution will work
 
djeitnstine said:
I believe you have expanded your brackets in the exponential wrong e^{-ln(x)} does not equal -x but rather e^{ln(x^{-1})} which is of course \frac{1}{x} in which case a simple u substitution will work

Thanks for catching that mistake.
 

Similar threads

Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 26 ·
Replies
26
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K