A-Level Maclaurin series: worked solution

Question

For y=f(x) with ln y=π/4−arctan(eˣ) and (1+e^{2x}) dy/dx + y eˣ = 0, determine the Maclaurin expansion of y as far as the x² term. [5]

Two pieces of information are supplied, and each does a different job: the logarithmic relation fixes the value of y at x = 0, while the differential equation generates the derivatives at x = 0. Step 1: find y(0) from ln y = π/4 - arctan(eˣ). At x = 0: eˣ = e⁰ = 1, and arctan(1) = π/4. ln y(0) = π/4 - π/4 = 0 y(0) = e⁰ = 1 Step 2: find y'(0) from the differential equation, by substituting x = 0. (1 + e^{2x}) dy/dx + y eˣ = 0 At x = 0: e^{2x} = 1 and eˣ = 1, so (1 + 1) y'(0) + y(0)(1) = 0 2 y'(0) + 1 = 0 y'(0) = -1/2 Step 3: to get y''(0), differentiate the whole equation with respect to x. Both terms are products, so the product rule is needed on each. d/dx[(1 + e^{2x}) y'] = 2e^{2x} y' + (1 + e^{2x}) y'' d/dx[y eˣ] = y' eˣ + y eˣ so 2e^{2x} y' + (1 + e^{2x}) y'' + y' eˣ + y eˣ = 0 Step 4: put x = 0 and substitute the values already found, y(0) = 1 and y'(0) = -1/2. 2(1)(-1/2) + (1 + 1) y''(0) + (-1/2)(1) + (1)(1) = 0 -1 + 2 y''(0) - 1/2 + 1 = 0 2 y''(0) - 1/2 = 0 y''(0) = 1/4 Step 5: apply the Maclaurin formula y = y(0) + y'(0)x + y''(0)x²/2! + ... y = 1 + (-1/2)x + (1/4)(x²/2) y = 1 - (1/2)x + (1/8)x² + ...