Find #f# ?

Find #f# 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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