Question

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`

Answer 1

`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."`