当 $N$ 单位的振金（Vibranium）与 $M$ 单位的氪石（Kryptonite）发生反应时，开始形成化学物质乌鲁（Uru），其中 $M < N$。设 $y$ 表示反应发生 $t$ 小时后生成的乌鲁单位数。 该化学反应的微分方程由下式给出： $\frac{dy}{dt} = k(N - y)(M - y),$ 其中 $0 \le y \le M$ 且 $k$ 是一个正常数。 已知 $\dfrac{M}{N} = \dfrac{3}{4}$。

1. Express the rational expression in partial fractions in terms of $N$ and $M$.

$\frac{1}{(N - y)(M - y)} = \frac{A}{N - y} + \frac{B}{M - y}$
$1 = A(M - y) + B(N - y)$
Let $y = N \implies 1 = A(M - N) \implies A = \frac{1}{M - N} = -\frac{1}{N - M}$
Let $y = M \implies 1 = B(N - M) \implies B = \frac{1}{N - M}$
$\frac{1}{(N - y)(M - y)} = \frac{-1/(N - M)}{N - y} + \frac{1/(N - M)}{M - y}$

2. Separate the variables and integrate the differential equation.

$\frac{dy}{dt} = k(N - y)(M - y)$
$\int \frac{1}{(N - y)(M - y)} \,dy = \int k \,dt$
$\frac{1}{N - M} \int \left( \frac{1}{M - y} - \frac{1}{N - y} \right) \,dy = \int k \,dt$
$\frac{1}{N - M} \left( -\ln(M - y) + \ln(N - y) \right) = kt + C$
$\frac{1}{N - M} \ln\left( \frac{N - y}{M - y} \right) = kt + C$

3. Apply the initial condition to find the constant of integration.

Since the substance Uru starts to form at $t = 0$, we have $y = 0$ when $t = 0$.
$\frac{1}{N - M} \ln\left( \frac{N}{M} \right) = k(0) + C \implies C = \frac{1}{N - M} \ln\left( \frac{N}{M} \right)$
Given $\frac{M}{N} = \frac{3}{4}$, we have $\frac{N}{M} = \frac{4}{3}$.
$C = \frac{1}{N - M} \ln\left( \frac{4}{3} \right)$

4. Express $t$ in terms of $y$, simplifying using the given ratio for $M$ and $N$.

Substitute $C$ back into the equation:
$kt = \frac{1}{N - M} \ln\left( \frac{N - y}{M - y} \right) - \frac{1}{N - M} \ln\left( \frac{4}{3} \right)$
$kt = \frac{1}{N - M} \ln\left( \frac{3(N - y)}{4(M - y)} \right)$
Since $M = \frac{3}{4}N$, we have $N - M = N - \frac{3}{4}N = \frac{1}{4}N$. Substitute this to eliminate $M$:
$kt = \frac{4}{N} \ln\left( \frac{3(N - y)}{4\left(\frac{3}{4}N\right) - 4y} \right)$
$t = \frac{4}{kN} \ln\left( \frac{3(N - y)}{3N - 4y} \right)$

1. Derive the expression for $t$ in terms of $y$ from part (i).

$\int \left( \frac{1/(N-M)}{M-y} - \frac{1/(N-M)}{N-y} \right) dy = \int k \,dt$
$\frac{1}{N-M} \ln\left(\frac{N-y}{M-y}\right) = kt + C$
Using the initial condition that $y=0$ when $t=0$ (since the substance starts to form), we find $C = \frac{1}{N-M} \ln\left(\frac{N}{M}\right)$.
Given $M = \frac{3}{4}N$, we have $N-M = \frac{1}{4}N$.
$kt = \frac{4}{N} \left( \ln\left(\frac{N-y}{\frac{3}{4}N-y}\right) - \ln\left(\frac{N}{\frac{3}{4}N}\right) \right)$
$t = \frac{4}{Nk} \ln\left(\frac{3(N-y)}{3N-4y}\right)$

2. Substitute $y = \frac{N}{4}$ into the expression for $t$ to find the required time.

$t = \frac{4}{Nk} \ln\left(\frac{3(N-\frac{N}{4})}{3N-4(\frac{N}{4})}\right)$
$t = \frac{4}{Nk} \ln\left(\frac{3(\frac{3N}{4})}{2N}\right)$
$t = \frac{4}{Nk} \ln\left(\frac{9N/4}{2N}\right)$
$t = \frac{4}{Nk} \ln\left(\frac{9}{8}\right)$

1. Obtain the expression for $t$ in terms of $y$ derived from earlier parts.

From part (i), resolving into partial fractions gives:
$\frac{1}{(N-y)(M-y)} = \frac{-1/(N-M)}{N-y} + \frac{1/(N-M)}{M-y}$
Integrating both sides:
$\int \left( \frac{-1/(N-M)}{N-y} + \frac{1/(N-M)}{M-y} \right) dy = \int k \, dt$
$\frac{1}{N-M}\ln(N-y) - \frac{1}{N-M}\ln(M-y) = kt + C$
$\frac{1}{N-M}\ln\left(\frac{N-y}{M-y}\right) = kt + C$
Using $M = \frac{3}{4}N$, we have $N - M = \frac{1}{4}N$. Initially $t=0$ and $y=0$:
$C = \frac{4}{N}\ln\left(\frac{N}{M}\right)$
Substituting $C$ back:
$\frac{4}{N}\ln\left(\frac{N-y}{M-y} \cdot \frac{M}{N}\right) = kt$
Since $\frac{M}{N} = \frac{3}{4}$, the argument of the logarithm simplifies to $\frac{3(N-y)}{4(M-y)}$. Given $M = \frac{3}{4}N$, this is $\frac{3N-3y}{3N-4y}$.
$t = \frac{4}{Nk}\ln\left(\frac{3N-3y}{3N-4y}\right)$

2. Rearrange the equation to express $y$ as a function of $t$.

$\frac{Nkt}{4} = \ln\left(\frac{3N-3y}{3N-4y}\right)$
$e^{\frac{Nkt}{4}} = \frac{3N-3y}{3N-4y}$
$e^{\frac{Nkt}{4}}(3N - 4y) = 3N - 3y$
$3Ne^{\frac{Nkt}{4}} - 4ye^{\frac{Nkt}{4}} = 3N - 3y$
$y(4e^{\frac{Nkt}{4}} - 3) = 3N(e^{\frac{Nkt}{4}} - 1)$
$y = \frac{3N(e^{\frac{Nkt}{4}} - 1)}{4e^{\frac{Nkt}{4}} - 3}$

3. Sketch the relevant part of the curve for the given context.

The relevant context requires $t \ge 0$.
Since initially $y=0$, the curve starts at the origin $(0, 0)$.
As $t \to \infty$, $e^{\frac{Nkt}{4}} \to \infty$, meaning $y$ approaches the horizontal asymptote $y = \frac{3N}{4}$.
Because $\frac{dy}{dt} = k(N-y)(M-y) > 0$ for $0 \le y < M$, the curve is strictly increasing.
(The student should draw a curve in the first quadrant, starting at $(0,0)$, concave downwards, and asymptotically approaching the horizontal dotted line $y = \frac{3N}{4}$ or $y = M$.)

4. State the behavior of $y$ for large values of $t$.

For large values of $t$, $e^{-\frac{Nkt}{4}} \to 0$.
By rewriting the equation as $y = \frac{3N(1 - e^{-\frac{Nkt}{4}})}{4 - 3e^{-\frac{Nkt}{4}}}$, we can see that as $t \to \infty$:
$y \to \frac{3N(1 - 0)}{4 - 0} = \frac{3N}{4}$
Therefore, $y$ approaches $\frac{3N}{4}$ (or $M$) units.