 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/8/i/solution

y = \sum_{j \geq 0} a_{j} x^{j}

and equate coefficients in the power series. The constant coefficient gives

a_{2} = 0

. For

m \geq 1

x^{m}

gives

(m + 2) (m + 1) a_{m + 2} - [(m - 1) (m - 2) + 6 (m - 1) + 6] a_{m - 1} = 0.

(m + 1) (m + 2)

a_{j + 3} = a_{j}

for

j \geq 0

a_{0} = 1

a_{1} = 0

a_{2} = 0

. Hence

y (x) = 1 + x^{3} + x^{6} + x^{9} + \dots, ∣ x ∣ < 1.

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