Maclaurin Series: implicit differentiation of a differential equation & series for \(e^{\tan^{-1}3x}\) — RVHS 2025 H2 Math Prelim P1

(a) Differentiate the given equation

Differentiate \((1+9x^2)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3y\) with respect to \(x\) using the product rule on the left.

\[ \begin{aligned} (1+9x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 18x\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 3\frac{\mathrm{d}y}{\mathrm{d}x}\\[4pt] (1+9x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (18x-3)\frac{\mathrm{d}y}{\mathrm{d}x} &= 0 \end{aligned} \]

(b) Build the Maclaurin series for \(y\)

Differentiate the equation in (a) once more, then evaluate every derivative at \(x=0\).

\[ (1+9x^2)\frac{\mathrm{d}^3y}{\mathrm{d}x^3} + 18x\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (18x-3)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 18\,\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \]

When \(x = 0\):

\[ y = e^2, \qquad \frac{\mathrm{d}y}{\mathrm{d}x} = 3e^2, \qquad \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 9e^2, \qquad \frac{\mathrm{d}^3y}{\mathrm{d}x^3} = -27e^2 \] \[ y = e^2 + 3e^2\,x + \frac{9e^2}{2}\,x^2 - \frac{27e^2}{6}\,x^3 + \cdots = e^2 + 3e^2\,x + \frac{9e^2}{2}\,x^2 - \frac{9e^2}{2}\,x^3 + \cdots \]

(c) Deduce the series for \(e^{\tan^{-1}3x}\)

Rewrite \(y\) using the given logarithmic relation, then divide the series for \(y\) by \(e^2\).

Since \(\ln y = 2 + \tan^{-1}3x\), we have \(y = e^{2+\tan^{-1}3x} = e^2 \cdot e^{\tan^{-1}3x}\). Therefore

\[ e^{\tan^{-1}3x} = \frac{y}{e^2} = 1 + 3x + \frac{9}{2}\,x^2 - \frac{9}{2}\,x^3 + \cdots \]