Find ff ?

Find ff for which we have :

  • f'(1)=f(1)=1
  • f(x)>0
  • x^3f''(x)-xf'(x)+2f(x)=0 x>0

1 Answer
Jim S
Jan 10, 2018

f(x)=e^((x-1)/x) x>0

Explanation:

We have

x^3f''(x)-xf'(x)+2f(x)=0

x>0 , x^3>0

so

f''(x)-1/x^2f'(x)+(2f(x))/x^3=0 <=>

f''(x)=1/x^2*f'(x)-2/x^3*f(x) <=>

f''(x)=1/x^2*f'(x)+(1/x^2)'f(x) <=>

(f'(x))'=(1/x^2*f(x))'

f'(x)=1/x^2*f(x)+c_1

For x=1

f'(1)=f(1)+c_1 <=> c_1=0

Therefore, f'(x)=f(x)/x^2

f'(x)-1/x^2*f(x)=0

f'(x)+(1/x)'f(x)=0

e^(1/x)f'(x)+e^(1/x)(1/x)'f(x)=0

(e^(1/x)f(x))'=0

e^(1/x)f(x)=c_2

For x=1 -> e^1f(1)=c_2 <=>c_2=e

As a result ,

e^(1/x)f(x)=e

f(x)=e*e^(-1/x)

f(x)=e^((x-1)/x) , x>0

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