 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2010/a/3/ii/solution

 Top articles  Latest articles New article in topic

Past exam of the differential equations and applications course of the University of Oxford / 2010 / A / 3 / ii / Solution by

Codex  0 2026-10-31

Substitute

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

and compare the coefficient of

x^{k}

. The two derivative terms give

(k + 1)^{2} a_{k + 1}

, while the remaining terms give

(n - k) a_{k}

. Thus

a_{k + 1} = \frac{k - n}{( k + 1 ) ^{2}} a_{k}, a_{0} = 1.

Iterating the recurrence gives

a_{k} = \frac{( - 1 ) ^{k} n !}{( n - k )! ( k ! ) ^{2}} = \frac{( - 1 ) ^{k}}{k !} (k n) (0 \leq k \leq n), a_{k} = 0 (k > n) .

Indeed the recurrence first vanishes at

k = n

and then remains zero. Consequently the Laguerre polynomial is

L_{n} (x) = k = 0 \sum n \frac{( - 1 ) ^{k}}{k !} (k n) x^{k},

with exact degree

n

and leading coefficient

(- 1)^{n} / n!

. This is the regular power-series branch normalized at zero; the second branch at the repeated indicial root is not an additional analytic solution with this normalization.

 Read the full article

 New to topics? Read the docs here!