在一个化学反应中，化合物 X 和 Y 反应生成一种产物。设 $x$ 和 $y$ 分别表示在实验开始 $t$ 分钟后，X 和 Y 的浓度，单位为摩尔每千升 (mol/kL)。 X 和 Y 的初始浓度分别由 $x_0$ 和 $y_0$ mol/kL 给出。

1. Separate the variables and integrate both sides

$\int \frac{1}{x} \, \mathrm{d}x = \int -a \, \mathrm{d}t$
$\ln x = -at + c$

2. Use the initial conditions to find the constant of integration

Given that the initial concentration of X is $x_0$, we have $x = x_0$ when $t = 0$:
$\ln x_0 = -a(0) + c$
$c = \ln x_0$

3. Substitute the constant back and make $x$ the subject

$\ln x = -at + \ln x_0$
$\ln x - \ln x_0 = -at$
$\ln\left(\frac{x}{x_0}\right) = -at$
$\frac{x}{x_0} = e^{-at}$
$x = x_0 e^{-at}$

1. Concisely obtain the solution for $x$ from the differential equation in (a)(i).

$\frac{dx}{dt} = -ax$
$\int \frac{1}{x} \, dx = \int -a \, dt$
$\ln x = -at + C$
Given $x = x_0$ when $t = 0$, we have $C = \ln x_0$.
$\ln\left(\frac{x}{x_0}\right) = -at \implies x = x_0 e^{-at}$

2. Substitute the half-life condition into the concentration equation.

For half-life $t_{0.5}$, the concentration $x$ decreases to half of its initial value, so $x = \frac{x_0}{2}$.
Substituting $x = \frac{x_0}{2}$ and $t = t_{0.5}$ into $x = x_0 e^{-at}$:
$\frac{x_0}{2} = x_0 e^{-a t_{0.5}}$

3. Solve for $t_{0.5}$ using logarithms.

Divide both sides by $x_0$:
$\frac{1}{2} = e^{-a t_{0.5}}$
Take the natural logarithm on both sides:
$\ln\left(\frac{1}{2}\right) = -a t_{0.5}$
$\ln 1 - \ln 2 = -a t_{0.5}$
$- \ln 2 = -a t_{0.5}$
$t_{0.5} = \frac{\ln 2}{a} \text{ (shown)}$

1. Express the concentration $y$ in terms of $x$ using the given initial concentrations and relationship.

Given the relationship $y_0 - y = 2(x_0 - x)$, substitute the initial concentrations $x_0 = 1$ and $y_0 = 4$:
$4 - y = 2(1 - x)$
$y = 4 - 2(1 - x)$
$y = 2 + 2x$
$y = 2(x + 1)$

2. Formulate the differential equation from the given proportionality and substitute $y$ to show the required form.

The rate of the reaction is directly proportional to $x y^2$. Since $X$ reacts to form a product, its concentration decreases over time, so its rate of change is negative. Let $k$ be a positive constant of proportionality:
$\frac{dx}{dt} = -k x y^2$
Substitute $y = 2(x+1)$ into the equation:
$\frac{dx}{dt} = -k x (2(x + 1))^2$
$\frac{dx}{dt} = -4k x (x + 1)^2$
Letting $b = 4k$, since $k > 0$, $b$ is also a positive constant. Thus:
$\frac{dx}{dt} = -bx(x + 1)^2 \quad \text{(shown)}$

1. Separate the variables and express the algebraic fraction in partial fractions.

$\frac{\mathrm{d}x}{\mathrm{d}t} = -bx(x+1)^2$\n$\int \frac{1}{x(x+1)^2} \,\mathrm{d}x = \int -b \,\mathrm{d}t$\nLet $\frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$.\n$1 = A(x+1)^2 + Bx(x+1) + Cx$\nSubstituting $x=0 \implies A=1$.\nSubstituting $x=-1 \implies C=-1$.\nComparing coefficients of $x^2 \implies A+B=0 \implies B=-1$.\n$\int \left( \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2} \right) \mathrm{d}x = \int -b \,\mathrm{d}t$

2. Integrate both sides to find the general solution.

$\ln|x| - \ln|x+1| - \frac{(x+1)^{-1}}{-1} = -bt + C$\nSince concentration $x > 0$, we can drop the modulus:\n$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = -bt + C$

3. Apply the initial condition to find $C$, and use $t=1$ to find $b$.

At the start ($t=0$), $x=1$:\n$\ln\left(\frac{1}{2}\right) + \frac{1}{2} = -b(0) + C \implies C = \frac{1}{2} - \ln 2$\nAt $t=1$, $x=0.5$:\n$\ln\left(\frac{0.5}{1.5}\right) + \frac{1}{1.5} = -b + C$\n$\ln\left(\frac{1}{3}\right) + \frac{2}{3} = -b + \frac{1}{2} - \ln 2$\n$-b = \ln\left(\frac{2}{3}\right) + \frac{1}{6} \implies b \approx 0.2388$

4. Determine the equation at $t=2$ and use a graphing calculator to solve for $x$.

At $t=2$, the right-hand side of the equation is:\n$-2b + C = 2\left(\ln\left(\frac{2}{3}\right) + \frac{1}{6}\right) + \frac{1}{2} - \ln 2$\n$-2b + C = \ln\left(\frac{4}{9}\right) + \frac{1}{3} + \frac{1}{2} - \ln 2 = \ln\left(\frac{2}{9}\right) + \frac{5}{6} \approx -0.6707$\nThus, we need to solve for $x$:\n$\ln\left(\frac{x}{x+1}\right) + \frac{1}{x+1} = \ln\left(\frac{2}{9}\right) + \frac{5}{6}$\nUsing a graphing calculator,\n$x \approx 0.314 \text{ (to 3 s.f.)}$