Find #f# ?

Given #f:[0,+oo)->RR# , differentiable with

  • #e^(f'(x))+e^(f^2(x))=f(x)+2#. #color(white)(aaa)# #AA##x##in##[0,+oo)#

  • #f(0)=0#

Find #f#

1 Answer
MattyMatty
Jun 1, 2018

#y = f(x) = 0#

Explanation:

#"Put "y = f(x)#

#=> e^(y') + e^(y^2) = y + 2#

#=> e^(y') = y + 2 - e^(y^2)#

#=> y' = dy/dx = ln(y + 2 - e^(y^2))#

#=> x + C = int dy/ln(y+2-e^(y^2))#

#"Note that "y+2-e^(y^2)" must be "> 0"."#
#"We solve "y+2-e^(y^2) = 0" first as such."#
#=> y = -0.5876088 " or " y ~~ 1.058#
#"Only between those "y" values is the ln(..) defined."#
#"We also have the problem of division by zero for "y=0.#
#"As " y(0) = 0 " was a prerequisite, we could not have a well-"#
#"defined function "y(x)" this way."#

#"By inspection we see that"#

#y = f(x) = 0#

#"is the trivial solution to the problem."#

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