[Cambridge] In a chemical reaction a compound $X$ is formed from a compound $Y$. The masses in grams of $X$ and $Y$ present at time $t$ seconds after the start of the reaction are $x$ and $y$ respectively. The sum of the two masses is equal to 100 grams throughout the reaction. At any time, the rate of formation of $X$ is proportional to the mass of $Y$ at that time. When $t = 0$, $x = 5$ and $\frac{\text{d}x}{\text{d}t} = 1.9$.

1. Translate the problem statement into a differential equation with an unknown constant of proportionality.

We are given that the sum of the masses is $100$, so $x + y = 100 \implies y = 100 - x$.

The rate of formation of $X$ is proportional to the mass of $Y$ ($y$), which gives:
$\frac{\text{d}x}{\text{d}t} = ky$

Substituting $y = 100 - x$ into the equation yields:
$\frac{\text{d}x}{\text{d}t} = k(100 - x)$

2. Substitute the initial conditions to find the constant $k$ and deduce the final differential equation.

Substitute the initial conditions ($t = 0$, $x = 5$, $\frac{\text{d}x}{\text{d}t} = 1.9$) into the equation:
$1.9 = k(100 - 5)$
$1.9 = 95k$
$k = \frac{1.9}{95} = 0.02$

Therefore, we have shown that:
$\frac{\text{d}x}{\text{d}t} = 0.02(100 - x)$

1. Separate the variables $x$ and $t$.

$\int \frac{1}{100 - x} \, \text{d}x = \int 0.02 \, \text{d}t$

2. Integrate both sides with respect to their variables.

$-\ln(100 - x) = 0.02t + C$

3. Use the initial conditions to find the constant of integration $C$.

Substitute $t = 0$ and $x = 5$:
$-\ln(100 - 5) = 0.02(0) + C$
$C = -\ln 95$

4. Rearrange the equation to express $x$ explicitly in terms of $t$.

Substitute $C$ back into the equation:
$-\ln(100 - x) = 0.02t - \ln 95$
$\ln(100 - x) = \ln 95 - 0.02t$
$100 - x = e^{\ln 95 - 0.02t}$
$100 - x = e^{\ln 95} \cdot e^{-0.02t}$
$100 - x = 95e^{-0.02t}$
$x = 100 - 95e^{-0.02t}$

1. Determine the limit of $x$ as $t \to \infty$ using the solution to the differential equation.

From part (ii), solving $\frac{\text{d}x}{\text{d}t} = 0.02(100 - x)$ gives $-\ln(100-x) = 0.02t + c$. Applying the initial conditions $t=0, x=5$ yields $x = 100 - 95\text{e}^{-0.02t}$.

As $t$ becomes very large ($t \to \infty$), the exponential term $\text{e}^{-0.02t} \to 0$.
Therefore, $x \to 100 - 95(0) = 100$.