# Find f ?

## Given $f : \left[0 , + \infty\right) \to \mathbb{R}$ , differentiable with ${e}^{f ' \left(x\right)} + {e}^{{f}^{2} \left(x\right)} = f \left(x\right) + 2$. $\textcolor{w h i t e}{a a a}$ $\forall$$x$$\in$$\left[0 , + \infty\right)$ $f \left(0\right) = 0$ Find $f$

Jun 1, 2018

$y = f \left(x\right) = 0$

#### Explanation:

$\text{Put } y = f \left(x\right)$

$\implies {e}^{y '} + {e}^{{y}^{2}} = y + 2$

$\implies {e}^{y '} = y + 2 - {e}^{{y}^{2}}$

$\implies y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(y + 2 - {e}^{{y}^{2}}\right)$

$\implies x + C = \int \frac{\mathrm{dy}}{\ln} \left(y + 2 - {e}^{{y}^{2}}\right)$

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

$\text{By inspection we see that}$

$y = f \left(x\right) = 0$

$\text{is the trivial solution to the problem.}$