2018印度理工學院入學考試高級試(JEE Advanced)的一題微分方程初始值問題

==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