It is given that $y^3 + y^2 + 2y = x^2 - 3x$, find the Maclaurin series for $y$ up to and including the term in $x^2$. Hence find the Maclaurin series for $\frac{1}{2 + y}$ up to and including the term in $x^2$.

1. Find the value of $y$ when $x = 0$.

Substituting $x = 0$ into the given equation:
$y^3 + y^2 + 2y = (0)^2 - 3(0)$
$y(y^2 + y + 2) = 0$
Since $y^2 + y + 2 = (y + \frac{1}{2})^2 + \frac{7}{4} > 0$ for all real $y$, the quadratic factor has no real roots. Therefore, the only real root is:
$y = 0$

2. Differentiate the equation implicitly to find the value of the first derivative at $x = 0$.

Differentiating implicitly with respect to $x$:
$3y^2 \frac{dy}{dx} + 2y \frac{dy}{dx} + 2 \frac{dy}{dx} = 2x - 3$
$(3y^2 + 2y + 2)\frac{dy}{dx} = 2x - 3$
Substitute $x = 0$ and $y = 0$:
$(0 + 0 + 2)\frac{dy}{dx} = 2(0) - 3$
$2\frac{dy}{dx} = -3 \implies \frac{dy}{dx} = -\frac{3}{2}$

3. Differentiate implicitly a second time to find the value of the second derivative at $x = 0$.

Differentiating $(3y^2 + 2y + 2)\frac{dy}{dx} = 2x - 3$ with respect to $x$ using the product rule:
$\left(6y \frac{dy}{dx} + 2 \frac{dy}{dx}\right)\frac{dy}{dx} + (3y^2 + 2y + 2)\frac{d^2y}{dx^2} = 2$
Substitute $x = 0$, $y = 0$, and $\frac{dy}{dx} = -\frac{3}{2}$:
$\left( 0 + 2\left(-\frac{3}{2}\right) \right)\left(-\frac{3}{2}\right) + (0 + 0 + 2)\frac{d^2y}{dx^2} = 2$
$2\left(-\frac{3}{2}\right)^2 + 2\frac{d^2y}{dx^2} = 2$
$\frac{9}{2} + 2\frac{d^2y}{dx^2} = 2$
$2\frac{d^2y}{dx^2} = 2 - \frac{9}{2} = -\frac{5}{2}$
$\frac{d^2y}{dx^2} = -\frac{5}{4}$

4. Construct the Maclaurin series for $y$ up to the term in $x^2$.

The Maclaurin series expansion is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$
Substituting the evaluated derivatives:
$y = 0 + \left(-\frac{3}{2}\right)x + \frac{-\frac{5}{4}}{2}x^2$
$y = -\frac{3}{2}x - \frac{5}{8}x^2$

1. Derive the Maclaurin series for $y$ up to the term in $x^2$ (from part (i)).

Substitute $x = 0$ into $y^3 + y^2 + 2y = x^2 - 3x$:
$y(y^2 + y + 2) = 0 \implies y = 0$.

Differentiate implicitly with respect to $x$:
$3y^2 \frac{dy}{dx} + 2y \frac{dy}{dx} + 2 \frac{dy}{dx} = 2x - 3$.
Substitute $x=0, y=0 \implies 2 \frac{dy}{dx} = -3 \implies \frac{dy}{dx} = -\frac{3}{2}$.

Differentiate again:
$6y \left(\frac{dy}{dx}\right)^2 + 3y^2 \frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^2 + 2y \frac{d^2y}{dx^2} + 2 \frac{d^2y}{dx^2} = 2$.
Substitute $x=0, y=0, \frac{dy}{dx}=-\frac{3}{2}$:
$0 + 0 + 2\left(-\frac{3}{2}\right)^2 + 0 + 2 \frac{d^2y}{dx^2} = 2$
$\frac{9}{2} + 2 \frac{d^2y}{dx^2} = 2 \implies 2 \frac{d^2y}{dx^2} = -\frac{5}{2} \implies \frac{d^2y}{dx^2} = -\frac{5}{4}$.

The Maclaurin series for $y$ is:
$y \approx y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 = -\frac{3}{2}x - \frac{5}{8}x^2$.

2. Expand $\frac{1}{2 + y}$ using the standard binomial series up to the term in $y^2$.

Rewrite the expression to use the standard series $(1+x)^n$:
$\frac{1}{2 + y} = (2 + y)^{-1} = 2^{-1} \left(1 + \frac{y}{2}\right)^{-1}$

Expand using the binomial theorem:
$= \frac{1}{2} \left[ 1 + (-1)\left(\frac{y}{2}\right) + \frac{(-1)(-2)}{2!} \left(\frac{y}{2}\right)^2 + \dots \right]$
$= \frac{1}{2} \left( 1 - \frac{y}{2} + \frac{y^2}{4} - \dots \right)$
$= \frac{1}{2} - \frac{1}{4}y + \frac{1}{8}y^2 - \dots$

3. Substitute the Maclaurin series for $y$ into the expansion and simplify up to $x^2$.

Substitute $y = -\frac{3}{2}x - \frac{5}{8}x^2$ into the expanded series:
$\frac{1}{2+y} \approx \frac{1}{2} - \frac{1}{4}\left(-\frac{3}{2}x - \frac{5}{8}x^2\right) + \frac{1}{8}\left(-\frac{3}{2}x - \frac{5}{8}x^2\right)^2$

Expand and ignore terms of degree higher than 2:
$\approx \frac{1}{2} + \frac{3}{8}x + \frac{5}{32}x^2 + \frac{1}{8}\left( -\frac{3}{2}x \right)^2$
$= \frac{1}{2} + \frac{3}{8}x + \frac{5}{32}x^2 + \frac{1}{8}\left( \frac{9}{4}x^2 \right)$
$= \frac{1}{2} + \frac{3}{8}x + \frac{5}{32}x^2 + \frac{9}{32}x^2$
$= \frac{1}{2} + \frac{3}{8}x + \frac{14}{32}x^2$
$= \frac{1}{2} + \frac{3}{8}x + \frac{7}{16}x^2$