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Year 3 of the mathematics course of the University of Oxford

Differential equations and applications course of the University of Oxford

Past exam of the differential equations and applications course of the University of Oxford

2010

Substitute $y=\sum_{k\geq0}a_kx^k$ and compare the coefficient of $x^k$. The two <derivative> terms give $(k+1)^2a_{k+1}$, while the remaining terms give $(n-k)a_k$. Thus
$$
\boxed{a_{k+1}=\frac{k-n}{(k+1)^2}a_k,\qquad a_0=1.}
$$
Iterating the recurrence gives
$$
\boxed{a_k=\frac{(-1)^kn!}{(n-k)!(k!)^2}=\frac{(-1)^k}{k!}\binom nk\quad(0\leq k\leq n),\qquad a_k=0\quad(k>n).}
$$
Indeed the recurrence first vanishes at $k=n$ and then remains zero. Consequently the <Laguerre polynomial> is
$$
\boxed{L_n(x)=\sum_{k=0}^n\frac{(-1)^k}{k!}\binom nk x^k,}
$$
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 function>[analytic] solution with this normalization.

Solution

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

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

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