It is given that $e^y = 1 + e^x$. Show that \(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \le

Question

It is given that $e^y = 1 + e^x$. Show that $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = \frac{\mathrm{d}y}{\mathrm{d}x}.$[2] Hence find the first four non-zero terms of the Maclaurin series for $y$.[5] Hence find the Maclaurin series for $\frac{e^{2x}}{1 + e^{2x}}$, up to and including the term in $x^3$.[3]

Accepted Answer

(a) $y''+(y')^2=y'$ (b) $y=\ln2+\dfrac12x+\dfrac18x^2-\dfrac1{192}x^4+\cdots$ (c) $\dfrac12+\dfrac12x-\dfrac16x^3+\cdots$

2025 YIJC P2 Q5

2025 YIJC P2 Q5

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