Power-Series Solution of a Second-Order ODE: 04-BS-5, May 2013, Q1

Question 1: Power-Series Solution of a Second-Order ODE (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 $(x+4)y''+2xy'+2y=0$; at $x=0$ the leading coefficient $(x+4)$ equals $4\ne0$, so $x=0$ is an ordinary point.

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

Approach. Substitute $y=\sum_{n=0}^{\infty}a_nx^n$, shift the summation index of every term so all carry the same power $x^m$, match coefficients to obtain a recurrence for $a_{m+2}$, then generate the even-start solution ($a_0=1,a_1=0$) and the odd-start solution ($a_0=0,a_1=1$).

Substitute the series and align every term on the power $x^m$. With $y=\sum a_nx^n$, $y'=\sum na_nx^{n-1}$, $y''=\sum n(n-1)a_nx^{n-2}$: $$xy''=\sum_{m=1}^{\infty}(m+1)m\,a_{m+1}x^{m},\qquad 4y''=\sum_{m=0}^{\infty}4(m+2)(m+1)a_{m+2}x^{m},$$ $$2xy'=\sum_{m=1}^{\infty}2m\,a_mx^{m},\qquad 2y=\sum_{m=0}^{\infty}2a_mx^{m}.$$
Match the coefficient of $x^m$ and isolate $a_{m+2}$. Summing all four series and setting the total coefficient of $x^m$ to zero, $$(m+1)m\,a_{m+1}+4(m+2)(m+1)a_{m+2}+2m\,a_m+2a_m=0.$$ Dividing by $(m+1)$, $$\boxed{a_{m+2}=-\dfrac{m\,a_{m+1}+2a_m}{4(m+2)}}\qquad(m=0,1,2,\dots)$$ (the $x^0$ balance $8a_2+2a_0=0$ is just this formula at $m=0$, so one recurrence generates every coefficient from $a_0,a_1$.)
Build $y_1$ from $a_0=1,\ a_1=0$. The recurrence gives $a_2=-\tfrac14,\ a_3=\tfrac1{48},\ a_4=\tfrac{11}{384},\ a_5=-\tfrac{49}{7680}$, so $$y_1(x)=\boxed{1-\dfrac14x^2+\dfrac1{48}x^3+\dfrac{11}{384}x^4-\dfrac{49}{7680}x^5+\cdots}$$
Build $y_2$ from $a_0=0,\ a_1=1$. The same recurrence gives $a_2=0,\ a_3=-\tfrac16,\ a_4=\tfrac1{48},\ a_5=\tfrac{13}{960}$, so $$y_2(x)=\boxed{x-\dfrac16x^3+\dfrac1{48}x^4+\dfrac{13}{960}x^5+\cdots}$$ Since $y_1(0)=1,\,y_1'(0)=0$ and $y_2(0)=0,\,y_2'(0)=1$, the Wronskian $W(0)=y_1(0)y_2'(0)-y_1'(0)y_2(0)=1\ne0$, so $y_1,y_2$ are linearly independent; the general solution is $y(x)=C_1y_1(x)+C_2y_2(x)$, convergent for $|x|\lt 4$ (the distance to the nearest singular point $x=-4$).

Quantity	Result
Recurrence	$a_{m+2}=-\dfrac{m\,a_{m+1}+2a_m}{4(m+2)}$
$y_1(x)$	$1-\tfrac14x^2+\tfrac1{48}x^3+\tfrac{11}{384}x^4-\tfrac{49}{7680}x^5+\cdots$
$y_2(x)$	$x-\tfrac16x^3+\tfrac1{48}x^4+\tfrac{13}{960}x^5+\cdots$
Radius of convergence	$\|x\|\lt 4$

Concept & Theory

Topic. Power-series (Taylor) solutions of a linear ODE about an ordinary point.

Key relations. For $p(x)y''+q(x)y'+r(x)y=0$ with $p,q,r$ polynomials and $p(x_0)\ne0$, substituting $y=\sum a_n(x-x_0)^n$ and matching powers of $(x-x_0)^m$ always yields a recurrence expressing $a_{m+2}$ in terms of earlier coefficients.

Why this works. Fuchs' theorem guarantees a genuine convergent power-series solution about any ordinary point, with radius of convergence at least the distance to the nearest singularity of $q/p$ or $r/p$ in the complex plane; here $p(x)=x+4$ vanishes only at $x=-4$, so both series converge for $|x|\lt 4$.

Common pitfall. Misaligning the index shifts — $xy''$ effectively starts at $m=1$ while $4y''$ and $2y$ start at $m=0$ — is the single most common slip and silently corrupts every coefficient from $a_2$ onward if not handled term by term.

Source. Kreyszig, Advanced Engineering Mathematics, Ch. 5 (Power Series Solutions of ODEs).

Question 1 of 7: Power-Series Solution of a Second-Order ODE

Question 1: Power-Series Solution of a Second-Order ODE (20 marks)