All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Double differential equation (Posted on 2008-06-26)
Find all injective functions f, defined for all x>0, satisfying f(f'(x))=-f(x)

 No Solution Yet Submitted by Jonathan Lindgren Rating: 2.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Natural logarithm Comment 2 of 2 |

Hi!

1. x--> f'(x) so f(f'(f'(x))) = -f(f'(x)) = -(-f(x))=f(x)

Because f is injective f'(f'(x)) = x

2. From the initial relation if we aplly ( )' operator we find

f'(f'(x))*f''(x) = -f'(x) because f(g(x))' = f'(g(x))*g'(x)

After replace f'(f'(x)) with x we find that

x*f''(x) = -f'(x) or simply x*f''+f'=0 or

(x*f)' = 0 so

x*f' = C (constant), f'=C/x , f=c1*ln(x) + c2, and after some calculus c1=1 and c2=0

So f(x)=ln(x)

 Posted by Chesca Ciprian on 2008-06-27 13:57:21
Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information