Power-Series Solution About an Ordinary Point: 04-BS-5, May 2018, Q1

Question 1: Power-Series Solution About an Ordinary Point (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The linear second-order ODE $(x^2+5)y''+3xy'+3y=0$. The leading coefficient $x^2+5$ vanishes only at the complex points $x=\pm i\sqrt5$, so $x=0$ is an ordinary point and a power-series solution $y=\sum_{n=0}^{\infty}a_nx^n$ is guaranteed, with radius of convergence at least $\sqrt5$ (the distance from $0$ to the nearest singularity in the complex plane).

Find. Two linearly independent power-series solutions $y_1(x)$, $y_2(x)$, valid about $x=0$.

Approach. Substitute the series for $y,y',y''$, re-index the $x^2y''$ and $5y''$ pieces onto a common power $x^n$, collect the coefficient of each $x^n$ to get a two-term recurrence, then generate the even solution ($a_0=1,a_1=0$) and the odd solution ($a_0=0,a_1=1$) separately.

Set up the series and its derivatives. With $y=\sum_{n=0}^{\infty}a_nx^n$, $$y'=\sum_{n=1}^{\infty}na_nx^{n-1},\qquad y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$
Re-index every term onto $x^n$. $x^2y''=\sum_{n=0}^{\infty}n(n-1)a_nx^n$ is already aligned (the $n=0,1$ terms vanish automatically). For $5y''$, shift the index $m=n-2$: $$5y''=5\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}=5\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}$$ The remaining terms are already in $x^n$ form: $3xy'=\sum_{n=0}^{\infty}3na_nx^n$, $3y=\sum_{n=0}^{\infty}3a_nx^n$.
Collect the coefficient of $x^n$ and set it to zero. $$5(n+2)(n+1)a_{n+2}+\big[n(n-1)+3n+3\big]a_n=0\ \Rightarrow\ 5(n+2)(n+1)a_{n+2}+(n^2+2n+3)a_n=0$$ $$\boxed{a_{n+2}=-\dfrac{n^2+2n+3}{5(n+1)(n+2)}\,a_n},\qquad n=0,1,2,\dots$$
Even solution: $a_0=1,\ a_1=0$. Repeated application of the recurrence gives $a_2=-\tfrac{3}{10}$, $a_4=\tfrac{11}{200}$, $a_6=-\tfrac{99}{10000}$, $a_8=\tfrac{5049}{2800000}$, so $$\boxed{y_1(x)=1-\dfrac{3}{10}x^2+\dfrac{11}{200}x^4-\dfrac{99}{10000}x^6+\dfrac{5049}{2800000}x^8-\cdots}$$
Odd solution: $a_0=0,\ a_1=1$. The same recurrence, now driven by $a_1$, gives $a_3=-\tfrac15$, $a_5=\tfrac{9}{250}$, $a_7=-\tfrac{57}{8750}$, so $$\boxed{y_2(x)=x-\dfrac15x^3+\dfrac{9}{250}x^5-\dfrac{57}{8750}x^7+\cdots}$$ $y_1$ and $y_2$ are linearly independent: one is even and the other odd, and $y_1(0)=1,\ y_1'(0)=0$ while $y_2(0)=0,\ y_2'(0)=1$, so the Wronskian at $x=0$ is $\begin{vmatrix}1&0\\0&1\end{vmatrix}=1\ne0$. The general solution is $y=C_1y_1(x)+C_2y_2(x)$.

Quantity	Result
Recurrence	$a_{n+2}=-\dfrac{n^2+2n+3}{5(n+1)(n+2)}a_n$
$y_1(x)$ (even, $a_0=1$)	$1-\tfrac{3}{10}x^2+\tfrac{11}{200}x^4-\tfrac{99}{10000}x^6+\tfrac{5049}{2800000}x^8-\cdots$
$y_2(x)$ (odd, $a_1=1$)	$x-\tfrac15x^3+\tfrac{9}{250}x^5-\tfrac{57}{8750}x^7+\cdots$
General solution	$y=C_1y_1(x)+C_2y_2(x)$

Topic: Power-series solutions of a linear ODE about an ordinary point.

Key relations: series substitution $y=\sum a_nx^n$; re-indexing a derivative series so every term lines up on the same power of $x$; a two-term recurrence $a_{n+2}=\phi(n)a_n$, which decouples into independent even- and odd-indexed solution families from the two free initial coefficients $a_0,a_1$.

Why this works: a linear ODE $p_2(x)y''+p_1(x)y'+p_0(x)y=0$ has an ordinary point at $x_0$ whenever $p_2(x_0)\ne0$; the existence theorem for ordinary points then guarantees two independent analytic solutions, and the coefficients of every power series solution are forced (up to the two free constants $a_0,a_1$) by matching powers of $x$ term by term.

Common pitfall: forgetting to re-index the $5y''$ term (leaving its sum starting at $n=2$ instead of $n=0$) silently drops the recurrence's $n=0,1$ equations; also mixing up which initial pair $(a_0,a_1)$ generates the even vs. odd branch, or stopping the recurrence at the wrong parity.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 5.1–5.2 (Power Series Method, ordinary points).

Question 1 of 7: Power-Series Solution About an Ordinary Point

Question 1: Power-Series Solution About an Ordinary Point (20 marks)