## Problem 4

Determine the differentiable function $f:(0, \infty) \rightarrow \mathbf{R}$ knowing that $f^{\prime}(x)=f(x)+\frac{f(x)}{x}+e^{x}$, for any $x>0$ and $f(1)=e$.

Gazeta Matematica

Selected by: Astalus Niculina, Ilie Stefan, Matefi Istvan

## INSPECTORATUL SCOLAR JUDETEAN MURES

S.S.M.R - BRANCH MURES

Mathematics Olympiad

Local phase
## Problem 4

Determine the differentiable function $f:(0, \infty) \rightarrow \mathbf{R}$ knowing that $f^{\prime}(x)=f(x)+\frac{f(x)}{x}+e^{x}$, for any $x>0$ and $f(1)=e$.

Gazeta Matematică

## Solution

Let the function $g:(0, \infty) \rightarrow \mathbf{R}, g(x)=\frac{f(x)}{x \cdot e^{x}}-\ln x . \quad \ldots \ldots \ldots \ldots . .3$ points

We have $g^{\prime}(x)=\frac{x \cdot e^{x} f^{\prime}(x)-\left(e^{x}+x \cdot e^{x}\right) f(x)}{\left(x \cdot e^{x}\right)^{2}}-\frac{1}{x}=\frac{f^{\prime}(x)-\left(1+\frac{1}{x}\right) f(x)}{x \cdot e^{x}}-\frac{1}{x}=\frac{1}{x}-\frac{1}{x}=0$. .2 points

Thus, $g$ is constant. Since $g(1)=\frac{f(1)}{e}=1$, it follows that $g(x)=1$ for any $x>0$, hence

$$
f(x)=(1+\ln x) x e^{x}, x \in(0, \infty) \ldots \ldots \ldots \ldots . . .2 \text { points }
$$