It is given that $(1+9x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 3y$, and the curve \(y = \mathrm{f}(x)

Question

It is given that $(1+9x^2)\frac{\mathrm{d}y}{\mathrm{d}x} = 3y$, and the curve $y = \mathrm{f}(x)$ passes through the $y$-axis at $(0, \mathrm{e}^2)$. Show that $(1+9x^2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + (18x-3)\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.[2] Find the Maclaurin series for $y$, up to and including the term in $x^3$, giving the exact coefficients for each term.[4] Given that $\ln y = 2 + \tan^{-1} 3x$, find the Maclaurin series for $\mathrm{e}^{\tan^{-1} 3x}$, up to and including the term in $x^3$.[2]

Accepted Answer

(a) $(1+9x^2)y^{\prime\prime}+(18x-3)y^{\prime}=0$ (shown). (b) $y=\mathrm e^2+3\mathrm e^2x+\frac{9\mathrm e^2}{2}x^2-\frac{9\mathrm e^2}{2}x^3+\cdots$. (c) $\mathrm e^{\tan^{-1}(3x)}=1+3x+\frac92x^2-\frac92x^3+\cdots$.

2025 RVHS P1 Q8

2025 RVHS P1 Q8

Solution:

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