已知 $y = \tan^{-1} \dfrac{1}{5} x$，证明 $\left(25 + x^2\right) \dfrac{d^2 y}{dx^2} + 2x \dfrac{dy}{dx} = 0$。 通过对此结果的进一步求导，求 $y$ 的麦克劳林级数，直到包含 $x^3$ 的项为止。

1. Find the first derivative $\frac{dy}{dx}$ using the chain rule and simplify the expression.

$y = \tan^{-1} \left(\frac{1}{5} x\right)$
$\frac{dy}{dx} = \frac{1}{1 + \left(\frac{x}{5}\right)^2} \cdot \frac{d}{dx}\left(\frac{x}{5}\right)$
$\frac{dy}{dx} = \frac{1}{1 + \frac{x^2}{25}} \cdot \frac{1}{5}$
$\frac{dy}{dx} = \frac{25}{25 + x^2} \cdot \frac{1}{5}$
$\frac{dy}{dx} = \frac{5}{25 + x^2}$

2. Rearrange the equation and differentiate implicitly using the product rule to show the target result.

$(25 + x^2) \frac{dy}{dx} = 5$

Differentiating both sides with respect to $x$:
$(25 + x^2) \frac{d}{dx}\left(\frac{dy}{dx}\right) + \frac{dy}{dx} \frac{d}{dx}(25 + x^2) = 0$
$\left(25 + x^2\right) \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} = 0 \quad \text{(Shown)}$

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

Given $y = \tan^{-1}\left(\frac{1}{5}x\right)$, when $x = 0$, $y(0) = \tan^{-1}(0) = 0$.

From the derivation in part (a), $\frac{dy}{dx} = \frac{1/5}{1 + (x/5)^2} = \frac{5}{25 + x^2}$.
When $x = 0$, $\frac{dy}{dx} = \frac{5}{25} = \frac{1}{5}$.

Substitute $x = 0$ into the equation from part (a), $(25 + x^2) \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} = 0$:
$25 \frac{d^2 y}{dx^2} + 0 = 0 \implies \frac{d^2 y}{dx^2} = 0$.

2. Differentiate the given second-order differential equation with respect to $x$.

$\frac{d}{dx} \left[ (25 + x^2) \frac{d^2 y}{dx^2} + 2x \frac{dy}{dx} \right] = 0$

Using the product rule on both terms:
$\left[ (25 + x^2) \frac{d^3 y}{dx^3} + 2x \frac{d^2 y}{dx^2} \right] + \left[ 2x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} \right] = 0$

$(25 + x^2) \frac{d^3 y}{dx^3} + 4x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = 0$

3. Evaluate the third derivative at $x = 0$.

Substitute $x = 0$, $\frac{dy}{dx} = \frac{1}{5}$, and $\frac{d^2 y}{dx^2} = 0$ into the differentiated equation:

$25 \frac{d^3 y}{dx^3} + 4(0)(0) + 2\left(\frac{1}{5}\right) = 0$
$25 \frac{d^3 y}{dx^3} = -\frac{2}{5}$
$\frac{d^3 y}{dx^3} = -\frac{2}{125}$

4. Formulate the Maclaurin's series up to the term in $x^3$.

The Maclaurin's series is given by:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$

$y = 0 + \left(\frac{1}{5}\right)x + \frac{0}{2!}x^2 + \frac{-2/125}{3!}x^3 + \dots$

$y = \frac{1}{5}x - \frac{2/125}{6}x^3$
$y = \frac{1}{5}x - \frac{1}{375}x^3$