Given that $y = \tfrac{1}{2}\ln\left(1 + \tan x\right)$, show that $2e^{2y}\dfrac{dy}{dx} = \sec^2 x.$ Hence write down the first two non-zero terms in the Maclaurin series for $y$. Given that the first two non-zero terms in the Maclaurin series for $y$ are equal to the first two non-zero terms in the series expansion of $\dfrac{x}{a + bx}$, where $a$ and $b$ are constants, find $a$ and $b$.

1. Show the given differential equation by expressing $y$ in exponential form and differentiating implicitly.

$y = \frac{1}{2}\ln(1+\tan x)$
$2y = \ln(1+\tan x)$
$e^{2y} = 1 + \tan x$

Differentiating implicitly with respect to $x$:
$e^{2y} \cdot 2\frac{dy}{dx} = \sec^2 x$
$2e^{2y}\frac{dy}{dx} = \sec^2 x \quad \text{(shown)}$

2. Evaluate $y$ and $\frac{dy}{dx}$ at $x=0$.

At $x = 0$:
$e^{2y} = 1 + \tan(0) = 1 \implies 2y = 0 \implies y = 0$

Substitute into the differential equation to find the first derivative:
$2e^{0}\frac{dy}{dx} = \sec^2(0)$
$2(1)\frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2}$

3. Differentiate implicitly a second time to find $\frac{d^2y}{dx^2}$ at $x=0$.

Differentiating $2e^{2y}\frac{dy}{dx} = \sec^2 x$ with respect to $x$ using the product and chain rules:
$2 \left[ \frac{d}{dx}(e^{2y}) \cdot \frac{dy}{dx} + e^{2y} \cdot \frac{d^2y}{dx^2} \right] = 2\sec^2 x \tan x$
$2 \left[ \left(2e^{2y} \frac{dy}{dx}\right) \frac{dy}{dx} + e^{2y} \frac{d^2y}{dx^2} \right] = 2\sec^2 x \tan x$
$4e^{2y}\left(\frac{dy}{dx}\right)^2 + 2e^{2y}\frac{d^2y}{dx^2} = 2\sec^2 x \tan x$

Substitute $x=0, y=0, \frac{dy}{dx}=\frac{1}{2}$:
$4(1)\left(\frac{1}{2}\right)^2 + 2(1)\frac{d^2y}{dx^2} = 2(1)(0)$
$1 + 2\frac{d^2y}{dx^2} = 0 \implies \frac{d^2y}{dx^2} = -\frac{1}{2}$

4. Write down the first two non-zero terms of the Maclaurin series.

The Maclaurin series is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$
$y \approx 0 + \left(\frac{1}{2}\right)x + \frac{-1/2}{2}x^2$
$y \approx \frac{1}{2}x - \frac{1}{4}x^2$

1. Concisely obtain the first two non-zero terms of the Maclaurin series for $y$ as directed by part (i).

From part (i), we have $2e^{2y}\frac{dy}{dx} = \sec^2 x$. When $x = 0$, $y = \frac{1}{2}\ln(1) = 0$.
Substituting these into the derivative equation:
$2e^0 \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2}$
Differentiating $2e^{2y}\frac{dy}{dx} = \sec^2 x$ with respect to $x$:
$4e^{2y}\left(\frac{dy}{dx}\right)^2 + 2e^{2y}\frac{d^2y}{dx^2} = 2\sec^2 x \tan x$
At $x = 0, y = 0, \frac{dy}{dx} = \frac{1}{2}$:
$4(1)\left(\frac{1}{2}\right)^2 + 2(1)\frac{d^2y}{dx^2} = 0 \implies 1 + 2\frac{d^2y}{dx^2} = 0 \implies \frac{d^2y}{dx^2} = -\frac{1}{2}$
Therefore, the Maclaurin series for $y$ is:
$y \approx y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 = \frac{1}{2}x - \frac{1}{4}x^2$

2. Expand the given rational expression using the standard binomial series.

$\frac{x}{a+bx} = x(a+bx)^{-1}$
$= x\left[a\left(1+\frac{b}{a}x\right)\right]^{-1}$
$= \frac{x}{a}\left(1+\frac{b}{a}x\right)^{-1}$
Using the standard binomial series expansion $(1+u)^{-1} = 1 - u + \dots$:
$= \frac{x}{a}\left(1 - \frac{b}{a}x + \dots\right)$
$= \frac{1}{a}x - \frac{b}{a^2}x^2 + \dots$

3. Equate the corresponding coefficients to find $a$ and $b$.

Given that the first two non-zero terms are equal, we compare the coefficients of $x$ and $x^2$:
$\frac{1}{a}x - \frac{b}{a^2}x^2 = \frac{1}{2}x - \frac{1}{4}x^2$
Comparing the coefficients of $x$:
$\frac{1}{a} = \frac{1}{2} \implies a = 2$
Comparing the coefficients of $x^2$ and substituting $a = 2$:
$-\frac{b}{a^2} = -\frac{1}{4}$
$-\frac{b}{2^2} = -\frac{1}{4}$
$\frac{b}{4} = \frac{1}{4} \implies b = 1$