A-Level Maclaurin series: worked solution

Question

By differentiating that result further, obtain the Maclaurin series of y up to and including the x³ term.

The result being differentiated further is the relation from the previous part, (25 + x²)(dy/dx) = 5 Differentiate both sides with respect to x. The left-hand side is a product, so use the product rule: d/dx(25 + x²) × dy/dx + (25 + x²) × d/dx(dy/dx) = 0 2x(dy/dx) + (25 + x²)(d²y/dx²) = 0 Differentiate again, using the product rule on each of the two products: [2(dy/dx) + 2x(d²y/dx²)] + [2x(d²y/dx²) + (25 + x²)(d³y/dx³)] = 0 2(dy/dx) + 4x(d²y/dx²) + (25 + x²)(d³y/dx³) = 0 Now substitute x = 0 into each relation in turn, since Maclaurin's series needs the values of the derivatives at x = 0. The curve passes through the origin, so y = 0 when x = 0. From the first relation: (25 + 0)(dy/dx) = 5, so dy/dx = 5/25 = 1/5 at x = 0. From the second: 2(0)(1/5) + (25 + 0)(d²y/dx²) = 0, so d²y/dx² = 0 at x = 0. From the third: 2(1/5) + 4(0)(0) + (25 + 0)(d³y/dx³) = 0 2/5 + 25(d³y/dx³) = 0 d³y/dx³ = -(2/5)/25 = -2/125 at x = 0. Maclaurin's series is y = y(0) + y'(0)x + y''(0)x²/2! + y'''(0)x³/3! + ... y = 0 + (1/5)x + 0 × x²/2 + (-2/125)(x³/6) + ... The last term is -2x³/750 = -x³/375. So y = (1/5)x - (1/375)x³ + ...