The points $(x,y)$ on a curve satisfy the equations $x = \tfrac{1}{2}t^2 - 2t$ and $\dfrac{dy}{dx} = \dfrac{1}{2t - 1}$ where $t$ is a parameter, $t > 2$. It is given that $y = 3 - \tfrac{3}{4}\ln 7$ when $t = 4$.

1. Find the derivative of $x$ with respect to $t$ and apply the chain rule to obtain $\frac{dy}{dt}$.

Given $x = \frac{1}{2}t^2 - 2t$, differentiating with respect to $t$ gives:
$\frac{dx}{dt} = t - 2$
Using the chain rule, we find $\frac{dy}{dt}$:
$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
$\frac{dy}{dt} = \left( \frac{1}{2t - 1} \right) (t - 2) = \frac{t - 2}{2t - 1}$

2. Rewrite the expression for $\frac{dy}{dt}$ as a proper fraction suitable for integration.

To integrate $\frac{t - 2}{2t - 1}$, we can perform algebraic long division or split the fraction since the numerator and denominator have the same degree:
$\frac{t - 2}{2t - 1} = \frac{\frac{1}{2}(2t - 1) + \frac{1}{2} - 2}{2t - 1}$
$\frac{t - 2}{2t - 1} = \frac{\frac{1}{2}(2t - 1) - \frac{3}{2}}{2t - 1} = \frac{1}{2} - \frac{3}{2(2t - 1)}$

3. Integrate $\frac{dy}{dt}$ with respect to $t$ to find the general equation for $y$.

Now, integrate to find $y$ in terms of $t$:
$y = \int \left( \frac{1}{2} - \frac{3}{2(2t - 1)} \right) dt$
$y = \frac{1}{2}t - \left( \frac{3}{2} \right) \left( \frac{1}{2} \right) \ln(2t - 1) + C$
$y = \frac{1}{2}t - \frac{3}{4}\ln(2t - 1) + C$
(Note: Since $t > 2$, it follows that $2t - 1 > 0$, so the modulus sign in the natural logarithm is not needed.)

4. Use the given initial condition to determine the constant of integration and state the final expression for $y$.

Substitute the given condition that $y = 3 - \frac{3}{4}\ln 7$ when $t = 4$:
$3 - \frac{3}{4}\ln 7 = \frac{1}{2}(4) - \frac{3}{4}\ln(2(4) - 1) + C$
$3 - \frac{3}{4}\ln 7 = 2 - \frac{3}{4}\ln 7 + C$
$3 = 2 + C \implies C = 1$
Therefore, the equation for $y$ is:
$y = \frac{1}{2}t - \frac{3}{4}\ln(2t - 1) + 1$

1. Find $\dfrac{dx}{dt}$ from the given equation for $x$.

$x = \frac{1}{2}t^2 - 2t \implies \frac{dx}{dt} = t - 2$

2. Differentiate $\dfrac{dy}{dx}$ with respect to $t$.

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left((2t - 1)^{-1}\right) = -1(2t - 1)^{-2}(2) = \frac{-2}{(2t - 1)^2}$

3. Use the formula for the second derivative of parametric equations to obtain the final expression.

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{-2}{(2t - 1)^2}}{t - 2} = \frac{-2}{(2t - 1)^2(t - 2)}$

1. Find the value of the parameter $t$ when $x = 0$.

For a Maclaurin's series, we need to evaluate the function and its derivatives at $x = 0$.
Substitute $x = 0$ into the parametric equation for $x$:
$\frac{1}{2}t^2 - 2t = 0$
$t(t - 4) = 0$
Since it is given that $t > 2$, we have $t = 4$.

2. Evaluate $y$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ at $x = 0$ (which corresponds to $t = 4$).

Substitute $t = 4$ into the given expressions for $y$ and its derivatives:
$y(0) = 3 - \frac{3}{4}\ln 7$ (given in the question specifically for $t = 4$)

$y'(0) = \left. \frac{dy}{dx} \right|_{t=4} = \frac{1}{2(4) - 1} = \frac{1}{7}$

$y''(0) = \left. \frac{d^2y}{dx^2} \right|_{t=4} = \frac{-2}{(2(4) - 1)^2(4 - 2)} = \frac{-2}{(7)^2(2)} = -\frac{1}{49}$

3. Construct the Maclaurin's series for $y$ up to the term in $x^2$.

Using the standard Maclaurin's series expansion formula $y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$

$y = \left(3 - \frac{3}{4}\ln 7\right) + \left(\frac{1}{7}\right)x + \frac{-\frac{1}{49}}{2}x^2$
$y = 3 - \frac{3}{4}\ln 7 + \frac{1}{7}x - \frac{1}{98}x^2$