 Solution

ID: past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2021/2/b/solution

r = a

and normalize the first coefficient to one:

y_{+} (x) = x^{a} \sum_{n = 0}^{\infty} c_{n} x^{n}, c_{0} = 1, c_{j} = 0 (j < 0) .

Substitution and comparison of the coefficient of

x^{n + a}

give

[(n + a)^{2} - a^{2}] c_{n} + b c_{n - β} = 0, c_{n} = - \frac{b c _{n - β}}{n ( n + 2 a )} (n \geq 1) .

Every denominator is nonzero because

a \geq 0

For

β = 1

c_{n} = - b c_{n - 1} / [n (n + 2 a)]

, so

y_{+} (x) = x^{a} [1 - \frac{b}{1 + 2 a} x + \frac{b ^{2}}{2 ( 1 + 2 a ) ( 2 + 2 a )} x^{2} + \dots], c_{n} = \frac{( - b ) ^{n}}{\prod _{j = 1}^{n} j ( j + 2 a )} .

For

β = 2

c_{1} = 0

and all odd coefficients vanish; the even coefficients satisfy

c_{2 k} = - b c_{2 k - 2} / [2 k (2 k + 2 a)]

. Hence

y_{+} (x) = x^{a} [1 - \frac{b}{4 ( 1 + a )} x^{2} + \frac{b ^{2}}{32 ( 1 + a ) ( 2 + a )} x^{4} + \dots] .

β

, induction in the recurrence relation shows that only indices divisible by

β

survive:

y_{+} (x) = x^{a} k = 0 \sum \infty \frac{( - b ) ^{k} x ^{k β}}{\prod _{j = 1}^{k} j β ( j β + 2 a )} .

k = 0

equals one. Successive terms in the power series factor have ratio of magnitudes

\frac{∣ b ∣ ∣ x ∣ ^{β}}{( k + 1 ) β (( k + 1 ) β + 2 a )} ⟶ 0.

Thus the ratio test gives convergence for every finite

x

; a nonintegral

a

still requires a chosen side of zero or a complex branch for the factor

x^{a}

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