在学习理论中，一个人在时间 $t$ 分钟内记忆的材料量 $y$ 单位满足微分方程 $\frac{\mathrm{d}y}{\mathrm{d}t} = k(10 - y),$ 其中 $k$ 是一个正常数。 最初，记忆的材料量为零单位。

1. Separate variables and integrate both sides of the differential equation.

$\frac{\mathrm{d}y}{\mathrm{d}t} = k(10 - y)$
$\int \frac{1}{10 - y} \,\mathrm{d}y = \int k \,\mathrm{d}t$
$-\ln|10 - y| = kt + C$

2. Apply the initial condition $t=0, y=0$ to find the constant of integration.

$-\ln|10 - 0| = k(0) + C$
$C = -\ln 10$

3. Rearrange the equation to express $y$ as the subject in terms of $t$ and $k$.

$-\ln|10 - y| = kt - \ln 10$
$\ln|10 - y| = \ln 10 - kt$
$|10 - y| = e^{\ln 10 - kt}$
$|10 - y| = 10e^{-kt}$
Since $y = 0$ initially, $10 - y > 0$, allowing us to drop the modulus:
$10 - y = 10e^{-kt}$
$y = 10(1 - e^{-kt})$

4. Use the second given condition $t=2, y=5$ to find the value of the constant $k$.

$5 = 10(1 - e^{-2k})$
$\frac{1}{2} = 1 - e^{-2k}$
$e^{-2k} = \frac{1}{2}$
$-2k = \ln\left(\frac{1}{2}\right) = -\ln 2$
$k = \frac{1}{2}\ln 2$

5. Substitute $k$ back into the equation and express $y$ entirely in terms of $t$.

$y = 10\left(1 - e^{-\left(\frac{1}{2}\ln 2\right)t}\right)$
Alternatively, simplifying the exponential expression:
$y = 10\left(1 - e^{\ln(2^{-t/2})}\right)$
$y = 10\left(1 - 2^{-\frac{t}{2}}\right)$

1. Identify the key features of the graph from the solution to the differential equation.

From part (i), the solution is of the form $y = 10(1 - e^{-kt})$ where $k > 0$.
For the graph of $y$ against $t$ for $t \ge 0$:

• When $t = 0$, $y = 10(1 - e^0) = 0$.
• As $t \to \infty$, $e^{-kt} \to 0$, which means $y \to 10$. Thus, $y = 10$ is a horizontal asymptote.
• Since $\frac{\mathrm{d}y}{\mathrm{d}t} = k(10 - y) > 0$ for $0 \le y < 10$, the curve is strictly increasing.

2. Sketch the graph showing the origin and asymptote.

• Draw a set of axes with $t$ on the horizontal axis and $y$ on the vertical axis (for $t \ge 0$).
• Mark the starting point at the origin $(0, 0)$.
• Draw a dashed horizontal line at $y = 10$.
• Draw a smooth curve starting from $(0,0)$, curving upwards but flattening out to approach the line $y = 10$ from below without touching it.

3. Explain the eventual behaviour of the amount of material memorised.

Mathematically, as $t \to \infty$, $y \to 10$.
In the context of the model, this means the amount of material memorised will eventually approach a maximum limit of 10 units.

1. Express the rate of forgetting mathematically

The rate at which material is forgotten is proportional to the amount memorised, $y$. Thus, the rate of forgetting is $my$, where $m$ is a positive constant of proportionality.

2. Formulate the net differential equation

The net rate of change of $y$ is the learning rate minus the forgetting rate.
$\frac{\mathrm{d}y}{\mathrm{d}t} = k(10 - y) - my$