Given that $y= e^{\sin^{-1}2x}$, show that $\left(1-4x^{2}\right)\dfrac{d^{2}y}{dx^{2}}=4y+4x\dfrac{dy}{dx}$.

1. Find the first derivative of $y$ with respect to $x$.

$y = e^{\sin^{-1}2x}$

Using the chain rule:
$\dfrac{dy}{dx} = e^{\sin^{-1}2x} \cdot \dfrac{1}{\sqrt{1-(2x)^2}} \cdot 2$

Since $y = e^{\sin^{-1}2x}$:
$\dfrac{dy}{dx} = \dfrac{2y}{\sqrt{1-4x^2}}$

2. Rearrange the equation and square both sides to avoid fractions and roots.

Multiply both sides by $\sqrt{1-4x^2}$:
$\sqrt{1-4x^2} \dfrac{dy}{dx} = 2y$

Squaring both sides:
$(1-4x^2) \left(\dfrac{dy}{dx}\right)^2 = 4y^2$

3. Differentiate implicitly with respect to $x$ to obtain the final second-order differential equation.

Differentiating both sides with respect to $x$ using the product and chain rules:
$(1-4x^2) \left(2 \dfrac{dy}{dx} \dfrac{d^2y}{dx^2}\right) + \left(\dfrac{dy}{dx}\right)^2 (-8x) = 8y \dfrac{dy}{dx}$

Dividing the entire equation by $2\dfrac{dy}{dx}$:
$(1-4x^2) \dfrac{d^2y}{dx^2} - 4x \dfrac{dy}{dx} = 4y$

Rearranging to match the required form:
$\left(1-4x^{2}\right)\dfrac{d^{2}y}{dx^{2}} = 4y + 4x\dfrac{dy}{dx}$

1. Evaluate the function and its first derivative at $x=0$.

When $x=0$,
$y = e^{\sin^{-1}(0)} = e^0 = 1$

Differentiating $y=e^{\sin^{-1}2x}$ with respect to $x$ using the chain rule:
$\frac{dy}{dx} = e^{\sin^{-1}2x} \cdot \frac{2}{\sqrt{1-4x^2}}$

When $x=0$,
$\frac{dy}{dx} = 1 \cdot \frac{2}{\sqrt{1-0}} = 2$

2. Evaluate the second derivative at $x=0$ using the given identity from part (a).

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

3. Differentiate the identity with respect to $x$ to find a relationship for the third derivative.

Applying the product rule to differentiate $\left(1-4x^{2}\right)\frac{d^{2}y}{dx^{2}} = 4y + 4x\frac{dy}{dx}$ implicitly with respect to $x$:
$-8x\frac{d^{2}y}{dx^{2}} + \left(1-4x^{2}\right)\frac{d^{3}y}{dx^{3}} = 4\frac{dy}{dx} + \left( 4\frac{dy}{dx} + 4x\frac{d^{2}y}{dx^{2}} \right)$
$-8x\frac{d^{2}y}{dx^{2}} + \left(1-4x^{2}\right)\frac{d^{3}y}{dx^{3}} = 8\frac{dy}{dx} + 4x\frac{d^{2}y}{dx^{2}}$

4. Evaluate the third derivative at $x=0$ and formulate the Maclaurin series.

Substitute $x=0$, $\frac{dy}{dx}=2$, and $\frac{d^{2}y}{dx^{2}}=4$ into the differentiated equation:
$(1-0)\frac{d^{3}y}{dx^{3}} = 8(2) + 0$
$\frac{d^{3}y}{dx^{3}} = 16$

Constructing the series expansion using Maclaurin's formula:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 1 + 2x + \frac{4}{2!}x^2 + \frac{16}{3!}x^3 + \dots$
$y = 1 + 2x + 2x^2 + \frac{8}{3}x^3$

1. Derive the Maclaurin series expansion of $y$ up to $x^3$ as required by part (b).

We first find the values of $y$ and its derivatives at $x = 0$.
At $x = 0$, $y = e^0 = 1$.
$\frac{dy}{dx} = \frac{2 e^{\sin^{-1}2x}}{\sqrt{1-4x^2}} \implies y'(0) = 2$.
Substituting $x = 0$ into $(1-4x^2)\frac{d^2y}{dx^2} = 4y + 4x\frac{dy}{dx}$ gives $y''(0) = 4(1) = 4$.
Differentiating the given differential equation implicitly:
$-8xy'' + (1-4x^2)y''' = 4y' + 4y' + 4xy''$
Substituting $x = 0$ gives $y'''(0) = 8(2) = 16$.
Thus, the Maclaurin series is:
$y \approx 1 + 2x + \frac{4}{2!}x^2 + \frac{16}{3!}x^3 = 1 + 2x + 2x^2 + \frac{8}{3}x^3$

2. Determine the value of $x$ needed to approximate $e^{\pi/6}$.

We need $e^{\sin^{-1}2x} = e^{\pi/6}$.
Equating the exponents:
$\sin^{-1}2x = \frac{\pi}{6}$
$2x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
$x = \frac{1}{4}$

3. Substitute $x = 1/4$ into the expansion to find the approximation.

Substitute $x = \frac{1}{4}$ into the series expansion:
$e^{\pi/6} \approx 1 + 2\left(\frac{1}{4}\right) + 2\left(\frac{1}{4}\right)^2 + \frac{8}{3}\left(\frac{1}{4}\right)^3$
$= 1 + \frac{1}{2} + 2\left(\frac{1}{16}\right) + \frac{8}{3}\left(\frac{1}{64}\right)$
$= 1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{24}$
$= \frac{24 + 12 + 3 + 1}{24}$
$= \frac{40}{24} = \frac{5}{3}$