It is given that $y = \sqrt{1 + \tan^{-1} 2x}$ where $-\frac{1}{2} < x < \frac{1}{2}$. Show that $y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \frac{8x}{(1 + 4x^2)^2} = 0.$ Hence find the Maclaurin series for $y$, up to and including the term in $x^3$.

1. Eliminate the square root to make implicit differentiation easier.

$y = \sqrt{1 + \tan^{-1} 2x}$
$y^2 = 1 + \tan^{-1} 2x$

2. Differentiate implicitly with respect to $x$ for the first time.

Differentiating both sides with respect to $x$:
$2y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + (2x)^2} \cdot 2$
$2y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2}{1 + 4x^2}$
$y \frac{\mathrm{d}y}{\mathrm{d}x} = (1 + 4x^2)^{-1}$

3. Differentiate implicitly a second time to obtain the given differential equation.

Using the product rule on the left hand side and the chain rule on the right hand side:
$y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -1(1 + 4x^2)^{-2} \cdot (8x)$
$y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{-8x}{(1 + 4x^2)^2}$
$y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \frac{8x}{(1 + 4x^2)^2} = 0 \quad \text{(Shown)}$

4. Evaluate $y$, $\frac{\mathrm{d}y}{\mathrm{d}x}$, and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ at $x = 0$.

When $x = 0$:
$y = \sqrt{1 + \tan^{-1} 0} = \sqrt{1 + 0} = 1$
Substitute into the first derivative equation $y \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + 4x^2}$:
$(1) \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + 0} \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 1$
Substitute into the second derivative equation:
$(1)\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (1)^2 + 0 = 0 \implies \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -1$

5. Differentiate the given equation to find the third derivative at $x = 0$.

Differentiating $y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 8x(1 + 4x^2)^{-2} = 0$ with respect to $x$:
$\left[ y \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + \frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} \right] + 2\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left[ 8(1 + 4x^2)^{-2} + 8x(-2)(1 + 4x^2)^{-3}(8x) \right] = 0$
$y \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 3\frac{\mathrm{d}y}{\mathrm{d}x}\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \frac{8}{(1 + 4x^2)^2} - \frac{128x^2}{(1 + 4x^2)^3} = 0$
Substitute $x = 0$, $y = 1$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 1$, $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -1$:
$(1)\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 3(1)(-1) + \frac{8}{1} - 0 = 0$
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} - 3 + 8 = 0 \implies \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = -5$

6. Form the Maclaurin series for $y$ up to the term in $x^3$.

Using the Maclaurin series formula:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 1 + 1(x) + \frac{-1}{2}x^2 + \frac{-5}{6}x^3$
$y = 1 + x - \frac{1}{2}x^2 - \frac{5}{6}x^3$