It is given that $y={{e}^{{{ an }^{-1}}4x}}$. Show that $\left( 1+16{{x}^{2}} ight)\frac{{{\mathr

Question

It is given that $y={{e}^{{{	an }^{-1}}4x}}$. Show that $\left( 1+16{{x}^{2}} ight)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+k\frac{\mathrm{d}y}{\mathrm{d}x}\left( 8x-1 ight)=0$, where $k$ is a constant to be found. [3] By further differentiation of the result in part (a), find the Maclaurin series for $y$ up to and including the term in ${{x}^{3}}$. [3] The first three non-zero terms of the Maclaurin expansion of ${{e}^{{{	an }^{-1}}4x}}$ are equal to the first three non-zero terms of the series expansion of $\ln \left( \frac{a}{1-bx} ight)$, where $a$ and $b$ are constants. Using standard series from the List of Formulae (MF27), find the exact values of $a$ and $b$. [2]

Accepted Answer

(a) (k=4) (b) (1+4x+8x^2-\dfrac{32}{3}x^3+\cdots) (c) (a=e,\ b=4)

2025 DHS Promo Q6

2025 DHS Promo Q6

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