<p>It is given that $y={{e}^{{{\tan }^{-1}}4x}}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show that $\left( 1+16{{x}^{2}} \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+k\frac{\mathrm{d}y}{\mathrm{d}x}\left( 8x-1 \right)=0$, where $k$ is a constant to be found.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">By further differentiation of the result in part <strong>(a)</strong>, find the Maclaurin series for $y$ up to and including the term in ${{x}^{3}}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The first three non-zero terms of the Maclaurin expansion of ${{e}^{{{\tan }^{-1}}4x}}$ are equal to the first three non-zero terms of the series expansion of $\ln \left( \frac{a}{1-bx} \right)$, where $a$ and $b$ are constants. Using standard series from the List of Formulae (MF27), find the exact values of $a$ and $b$.</div><p style="text-align: right;">[2]</p></li></ol>

It is given that $y={{e}^{{{\tan }^{-1}}4x}}$.

Show that $\left( 1+16{{x}^{2}} \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+k\frac{\mathrm{d}y}{\mathrm{d}x}\left( 8x-1 \right)=0$, where $k$ is a constant to be found.

By further differentiation of the result in part **(a)**, find the Maclaurin series for $y$ up to and including the term in ${{x}^{3}}$.

The first three non-zero terms of the Maclaurin expansion of ${{e}^{{{\tan }^{-1}}4x}}$ are equal to the first three non-zero terms of the series expansion of $\ln \left( \frac{a}{1-bx} \right)$, where $a$ and $b$ are constants. Using standard series from the List of Formulae (MF27), find the exact values of $a$ and $b$.

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