变量 $x$ and $y$ 满足关系式 $(x+y)\frac{dy}{dx} + ky = 2\quad\text{and}\quad y = 1\text{ 在 }x = 0\text{ 时},$ 其中 $k$ 为常数。证明 $(x+y)\frac{d^{2}y}{dx^{2}} + (1+k)\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^{2} = 0.$ 已知 $x$ 很小，求 $g(x)=\dfrac{1}{\sin^{2}\left(2x+\dfrac{\pi}{2}\right)}$ 关于 $x$ 的升幂级数展开式，直到 $x^{2}$ 项（含该项）。 若 $g(x)$ 展开式中 $x^{2}$ 的系数等于 (i) 中 $y$ 的麦克劳林级数中 $x^{2}$ 系数的两倍，求 $k$ 的值。 通过对 (i) 中所得结果进行进一步微分，并设 $k=1$，求 $y$ 的麦克劳林级数，直到 $x^{3}$ 项（含该项）。

1. Differentiate implicitly with respect to $x$ using the product rule on the first term.

Differentiating $(x+y)\frac{dy}{dx} + ky = 2$ with respect to $x$:
$\frac{d}{dx}(x+y) \cdot \frac{dy}{dx} + (x+y) \cdot \frac{d}{dx}\left(\frac{dy}{dx}\right) + \frac{d}{dx}(ky) = \frac{d}{dx}(2)$
$\left(1 + \frac{dy}{dx}\right)\frac{dy}{dx} + (x+y)\frac{d^2y}{dx^2} + k\frac{dy}{dx} = 0$

2. Expand and group the terms to arrive at the required expression.

$\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 + (x+y)\frac{d^2y}{dx^2} + k\frac{dy}{dx} = 0$
Grouping the $\frac{dy}{dx}$ terms:
$(x+y)\frac{d^2y}{dx^2} + (1+k)\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = 0 \quad \text{(Shown)}$

1. Simplify $g(x)$ using a trigonometric identity and approximate its denominator for small $x$.

Since $\sin(2x + \frac{\pi}{2}) = \cos(2x)$,
$g(x) = \frac{1}{\cos^2(2x)} = (\cos 2x)^{-2}$
Using the standard expansion $\cos x \approx 1 - \frac{x^2}{2}$ for small $x$:
$\cos 2x \approx 1 - \frac{(2x)^2}{2} = 1 - 2x^2$
Thus, $g(x) \approx (1 - 2x^2)^{-2}$.

2. Expand $g(x)$ using the binomial expansion to find the coefficient of $x^2$.

Using the binomial expansion $(1+u)^n \approx 1 + nu$:
$g(x) \approx 1 + (-2)(-2x^2) = 1 + 4x^2$
Therefore, the series expansion of $g(x)$ up to the $x^2$ term is $1 + 4x^2$, and the coefficient of $x^2$ is $4$.

3. Find the value of $\frac{dy}{dx}$ at $x=0$.

From the first equation in (i):
$(x+y)\frac{dy}{dx} + ky = 2$
Substitute $x=0$ and $y=1$:
$(0+1)\frac{dy}{dx} + k(1) = 2 \implies \frac{dy}{dx} = 2 - k$

4. Find the value of $\frac{d^2y}{dx^2}$ at $x=0$ to determine the coefficient of $x^2$ in the Maclaurin series for $y$.

Using the differential equation established in (i):
$(x+y)\frac{d^2y}{dx^2} + (1+k)\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = 0$
Substitute $x=0$, $y=1$, and $\frac{dy}{dx} = 2-k$:
$(1)\frac{d^2y}{dx^2} + (1+k)(2-k) + (2-k)^2 = 0$
$\frac{d^2y}{dx^2} + (2 - k + 2k - k^2) + (4 - 4k + k^2) = 0$
$\frac{d^2y}{dx^2} + 6 - 3k = 0 \implies \frac{d^2y}{dx^2} = 3k - 6$

The coefficient of $x^2$ in the Maclaurin series for $y$ is given by $\frac{1}{2!} \frac{d^2y}{dx^2}$, which equals $\frac{3k - 6}{2}$.

5. Equate the coefficients based on the given condition to solve for $k$.

The problem states that the coefficient of $x^2$ in $g(x)$ is equal to twice the coefficient of $x^2$ in $y$:
$4 = 2 \left( \frac{3k - 6}{2} \right)$
$4 = 3k - 6$
$3k = 10 \implies k = \frac{10}{3}$

1. Evaluate the first and second derivatives at $x=0$ with $k=1$.

Given $(x+y)\frac{dy}{dx} + ky = 2$ and $y = 1$ at $x = 0$.
With $k=1$, $(x+y)\frac{dy}{dx} + y = 2$.
Substitute $x=0, y=1$:
$(0+1)\frac{dy}{dx} + 1 = 2 \implies \frac{dy}{dx} = 1$

From part (i), $(x+y)\frac{d^2y}{dx^2} + (1+k)\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = 0$.
With $k=1$, $(x+y)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = 0$.
Substitute $x=0, y=1, \frac{dy}{dx}=1$:
$(0+1)\frac{d^2y}{dx^2} + 2(1) + (1)^2 = 0 \implies \frac{d^2y}{dx^2} = -3$

2. Differentiate the result from part (i) to find an expression for the third derivative.

Differentiating $(x+y)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = 0$ implicitly with respect to $x$:

$\frac{d}{dx}\left[ (x+y)\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 \right] = 0$
$(x+y)\frac{d^3y}{dx^3} + \left(1 + \frac{dy}{dx}\right)\frac{d^2y}{dx^2} + 2\frac{d^2y}{dx^2} + 2\frac{dy}{dx}\frac{d^2y}{dx^2} = 0$

3. Substitute the known values to evaluate the third derivative and construct the Maclaurin series.

Substitute $x=0, y=1, \frac{dy}{dx}=1, \frac{d^2y}{dx^2}=-3$ into the third derivative equation:
$(0+1)\frac{d^3y}{dx^3} + (1+1)(-3) + 2(-3) + 2(1)(-3) = 0$
$\frac{d^3y}{dx^3} - 6 - 6 - 6 = 0 \implies \frac{d^3y}{dx^3} = 18$

The Maclaurin series for $y$ is:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 1 + (1)x + \frac{-3}{2}x^2 + \frac{18}{6}x^3$
$y = 1 + x - \frac{3}{2}x^2 + 3x^3$