Confirming Wronskian Determinants with Basic Functions

Click For Summary
SUMMARY

The discussion confirms the results of Wronskian determinants for specific functions. The Wronskian of the functions e^x, e^-x, and sinh(x) is established as 0, due to sinh(x) being a linear combination of e^x and e^-x. Additionally, the Wronskian of cos(ln(x)) and sin(ln(x)) is correctly calculated as 1/x, derived from the identity cos^2(ln(x)) + sin^2(ln(x)) = 1.

PREREQUISITES
  • Understanding of Wronskian determinants
  • Knowledge of hyperbolic functions
  • Familiarity with trigonometric identities
  • Basic calculus concepts
NEXT STEPS
  • Study the properties of Wronskian determinants in linear algebra
  • Explore the relationship between hyperbolic and exponential functions
  • Learn about trigonometric identities and their applications
  • Investigate advanced applications of Wronskian in differential equations
USEFUL FOR

Students in calculus or linear algebra, educators teaching Wronskian determinants, and anyone interested in the applications of differential equations.

sapiental
Messages
110
Reaction score
0

Homework Statement



Hi, could someone please confirm my results. I just put my answers because the procedure is so long. let me know if you get the same results.

1) Wronskian(e^x, e^-x, sinh(x)) = 0

2) Wronskian(cos(ln(x)), sin(ln(x)) = 1/x * [cos^2(ln(x)) + sin^2(ln(x))] = 1/x

thanks in advance:smile:
 
Physics news on Phys.org
The one is super easy. sinh(x) is a linear combination of e^x and e^(-x). So 0. You don't even have to compute anything. And you are right on the second one as well.
 

Similar threads

Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
4
Views
1K
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K