==問題==
(譯文)
設可微函數$f : \mathbb{R} \rightarrow \mathbb{R}$滿足$f(0) = 1$且對於任意實數$x, y$有
$$f(x+y) = f(x)f'(y) + f'(x)f(y).$$
試求出$\ln f(4)$之值。
(原文)
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function with $f(0) = 1$ and satisfying the equation
$$f(x+y) = f(x)f'(y) + f'(x)f(y) \,\, \text{for all}\,\, x, y \in \mathbb{R}.$$
Then the value of $\ln f(4)$ is .
==解答==
($1^\circ$)
$$\begin{align*} &f(0) = f(0+0) = f(0)f'(0) + f'(0)f(0) \\ \Rightarrow &f(0) = 2f(0)f'(0) \\ \Rightarrow &1 = 2\cdot 1 \cdot f'(0) \\ \Rightarrow &f'(0) = \frac{1}{2}. \end{align*}$$
($2^\circ$)
$$\begin{align*} &f(x) = f(0+x)=f(0)f'(x)+f'(0)f(x) = 1\cdot y' + \frac{1}{2} y \\ \Rightarrow &y = y' + \frac{1}{2} y \\ \Rightarrow &y' = \frac{1}{2}y \\ \Rightarrow & y = e^{\frac{1}{2}x} + C. \end{align*}$$
$x = 0$代入，得
$$1 = e^{\frac{1}{2}\cdot 0} + C.$$
故$C = 0$，即
$$y = e^{\frac{1}{2}x}.$$
所以
$$\ln f(4) = \ln e^{\frac{1}{2} \cdot 4} = \ln e^2 = 2.$$
==評論==
輕而易舉
