In a chemical reaction, a compound $X$ is formed from two compounds $Y$ and $Z$. The masses in grams of $X$, $Y$ and $Z$ present at time $t$ seconds after the start of the reaction are $x$, $15 - x$ and $25 - x$ respectively. At any time, the rate of formation of $X$ is proportional to the product of masses of $Y$ and $Z$ present at the time. It is given that $x = 0$ and $\frac{\text{d}x}{\text{d}t} = 3$ when $t = 0$. [Continued]

1. Translate the given problem statement into a differential equation using a constant of proportionality $k$.

The rate of formation of $X$ is $\frac{\text{d}x}{\text{d}t}$. It is proportional to the product of the masses of $Y$ and $Z$, which are $(15-x)$ and $(25-x)$ respectively.

$\frac{\text{d}x}{\text{d}t} = k(15 - x)(25 - x)$

2. Substitute the initial conditions to find the value of $k$.

Given $x = 0$ and $\frac{\text{d}x}{\text{d}t} = 3$ when $t = 0$:
$3 = k(15 - 0)(25 - 0)$
$3 = 375k$
$k = \frac{3}{375} = \frac{1}{125}$

1. Substitute the constant of proportionality from part (i) and separate the variables.

From part (i), the differential equation is:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{125}(15-x)(25-x)$
Separating the variables:
$\int \frac{1}{(15-x)(25-x)} \, \mathrm{d}x = \int \frac{1}{125} \, \mathrm{d}t$

2. Decompose the rational expression using partial fractions.

Let $\frac{1}{(15-x)(25-x)} \equiv \frac{A}{15-x} + \frac{B}{25-x}$
$1 \equiv A(25-x) + B(15-x)$
Substituting $x = 15 \implies 1 = 10A \implies A = \frac{1}{10}$
Substituting $x = 25 \implies 1 = -10B \implies B = -\frac{1}{10}$
Substitute back into the integral equation:
$\frac{1}{10} \int \left( \frac{1}{15-x} - \frac{1}{25-x} \right) \, \mathrm{d}x = \int \frac{1}{125} \, \mathrm{d}t$
$\int \left( \frac{1}{15-x} - \frac{1}{25-x} \right) \, \mathrm{d}x = \int \frac{2}{25} \, \mathrm{d}t$

3. Integrate both sides to find the general solution.

$-\ln|15-x| + \ln|25-x| = \frac{2}{25}t + C$
$\ln\left| \frac{25-x}{15-x} \right| = \frac{2}{25}t + C$

4. Apply the initial condition to evaluate the constant of integration.

Given $x=0$ when $t=0$:
$\ln\left| \frac{25-0}{15-0} \right| = \frac{2}{25}(0) + C \implies C = \ln\left(\frac{5}{3}\right)$
Substituting $C$ back into the equation:
$\ln\left| \frac{25-x}{15-x} \right| = \frac{2}{25}t + \ln\left(\frac{5}{3}\right)$

5. Remove the logarithm and use the physical context to determine the sign.

Since compound $X$ is formed from $Y$ (initial mass $15$ g), the mass $x$ must be less than $15$. Thus for $t \geq 0$, $15-x > 0$ and $25-x > 0$, allowing us to remove the absolute value bars.
$\frac{25-x}{15-x} = \mathrm{e}^{\frac{2}{25}t + \ln(5/3)}$
$\frac{25-x}{15-x} = \frac{5}{3}\mathrm{e}^{\frac{2}{25}t}$

6. Rearrange the equation to make $x$ the subject.

Cross-multiplying:
$3(25-x) = 5(15-x)\mathrm{e}^{\frac{2}{25}t}$
$75 - 3x = 75\mathrm{e}^{\frac{2}{25}t} - 5x\mathrm{e}^{\frac{2}{25}t}$
Group terms involving $x$:
$5x\mathrm{e}^{\frac{2}{25}t} - 3x = 75\mathrm{e}^{\frac{2}{25}t} - 75$
$x(5\mathrm{e}^{\frac{2}{25}t} - 3) = 75(\mathrm{e}^{\frac{2}{25}t} - 1)$
$x = \frac{75(\mathrm{e}^{\frac{2}{25}t} - 1)}{5\mathrm{e}^{\frac{2}{25}t} - 3}$