$f(x)= e^{f'(x)}$. Solve the DE for $f(x)$

Question

f differentiable function in R. $f(x)= e^{f'(x)}$ $f(0)=1$

I have proved that $f(x)=1$ for every $x\lt0$. im stuck for $x\gt0 $

Answer 1

Take $f(x)=1$ and then apply Picard–Lindelöf theorem

Edit.

Well, for short, this is the theorem that guarantees local existence and unicity of solutions of Cauchy problems for ODE (i.e. one of the most important theorems in theory of ODE) under some hypothesis on the equation and initial data. You can look the correct formulation on wiki, but here what we do here: in the neighborhood of initial data the logarithm function is well-defined and is a differmorphism, so we can safely apply it to both sides of the equation:

$$f'(x)=\ln(f(x)), f(0)=1.$$ We check the conditions of our theorem: the application $(x,f)\to \ln(f)$ is continuous in $x$ (in fact, it doesn't depend on it) and is Lipschitz in $f$ in some neighborhood of $f(0)=1 $(it's sufficient to check that it's $\mathcal C^1$ in that neighborhood). Thus, we can guarantee the existence and inicity of solution $f(x)$ of our ODE on $x\in (-\varepsilon,\varepsilon)$ for some $\varepsilon>0$. On the other hand, the function $f(x)=1\,\forall x$ fits our ODE; thus we conclude that this function is indeed the only solution to our ODE.

Answer 2

In the domain $H:=\{(x,y)\ |\ y>0\}\subset{\mathbb R^2}$ your differential equation is equivalent to

$$y'=\log y=:\psi(x,y)\ .\tag{1}$$

As $\psi$ is a continuously differentiable function in $H$ it is locally Lipschitz with respect to $y$ in $H$. It follows that for any $(x_0,y_0)\in H$ the initial value problem $$y'=\log y,\qquad y(x_0)=y_0\tag{2}$$ has a unique solution $x\mapsto y=\phi(x)$ whose graph will leave any given compact subset $K\subset H$. Given the initial point $(x_0,y_0):=(0,1)$ it is obvious that $y=\phi(x)\equiv1$ solves the IVP $(2)$ for $-\infty<x<\infty$, so there is not more to it.

The right side $\psi(x,y)$ in $(1)$ being independent of $x$ it follows that in order to obtain a picture of all solutions of $(1)$ it is enough to consider initial points $(x_0,y_0):=(0,b)$ for $b>0$. Separation of variables will lead to solutions that can be expressed in terms of the logarithmic integral function ${\rm Li}$.