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Double differential equation (Posted on 2008-06-26) Difficulty: 3 of 5
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)

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Solution Natural logarithm Comment 2 of 2 |


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
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