Power-series solution of a second-order ODE: 04-BS-5, May 2015, 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 ODE $(x^2+2)y''+3xy'-y=0$; the point $x=0$ is ordinary because the leading coefficient $x^2+2$ is nonzero there.

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

Approach. Substitute $y=\sum_{n=0}^{\infty}a_nx^n$, collect the coefficient of $x^n$ into a recurrence relating $a_{n+2}$ to $a_n$, then generate the even series ($a_0=1,a_1=0$) and the odd series ($a_0=0,a_1=1$) separately.

Differentiate the series and shift indices. With $y=\sum a_nx^n$, $y'=\sum na_nx^{n-1}$, $y''=\sum n(n-1)a_nx^{n-2}$. Re-indexing the $2y''$ term so every series is a power of $x^n$: $$x^2y''=\sum_{n=0}^{\infty}n(n-1)a_nx^n,\qquad 2y''=2\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n$$
Collect the coefficient of $x^n$ and set it to zero. Substituting into $(x^2+2)y''+3xy'-y=0$: $$n(n-1)a_n+2(n+2)(n+1)a_{n+2}+3na_n-a_n=0$$ Grouping the $a_n$ terms, $\big[n(n-1)+3n-1\big]a_n=-\big(n^2+2n-1\big)a_n$, so the recurrence is $$\boxed{a_{n+2}=-\dfrac{n^2+2n-1}{2(n+2)(n+1)}\,a_n}$$
Even series $y_1$ ($a_0=1,\ a_1=0$). Applying the recurrence for $n=0,2,4,\dots$: $$a_2=\tfrac14,\quad a_4=-\tfrac{7}{96},\quad a_6=\tfrac{161}{5760}$$ $$\boxed{y_1(x)=1+\tfrac14x^2-\tfrac{7}{96}x^4+\tfrac{161}{5760}x^6-\cdots}$$
Odd series $y_2$ ($a_0=0,\ a_1=1$). Applying the recurrence for $n=1,3,5,\dots$: $$a_3=-\tfrac16,\quad a_5=\tfrac{7}{120},\quad a_7=-\tfrac{17}{720}$$ $$\boxed{y_2(x)=x-\tfrac16x^3+\tfrac{7}{120}x^5-\tfrac{17}{720}x^7+\cdots}$$ Since $y_1$ starts with $1$ (even powers only) and $y_2$ starts with $x$ (odd powers only), their Wronskian at $x=0$ is $y_1(0)y_2'(0)-y_1'(0)y_2(0)=1\ne0$, so they are linearly independent and $y=C_1y_1(x)+C_2y_2(x)$ is the general solution.

Final results — Question 1
Solution	Series (through $x^6$/$x^7$)
$y_1$ (even, $a_0=1$)	$1+\tfrac14x^2-\tfrac{7}{96}x^4+\tfrac{161}{5760}x^6-\cdots$
$y_2$ (odd, $a_1=1$)	$x-\tfrac16x^3+\tfrac{7}{120}x^5-\tfrac{17}{720}x^7+\cdots$

Topic: Power-series solutions about an ordinary point.

Key relations: $y=\sum a_nx^n$; a linear ODE with analytic coefficients has an ordinary point wherever the leading coefficient is nonzero, and the series solution converges at least out to the nearest singularity of $\{x:x^2+2=0\}$ in the complex plane, i.e. radius $\sqrt2$.

Why this works: Substituting the series turns the ODE into a linear recurrence between coefficients two apart in index; picking $(a_0,a_1)=(1,0)$ and $(0,1)$ generates two solutions that are automatically independent because one is purely even and the other purely odd in $x$.

Common pitfall: forgetting to re-index the $2y''$ term (it must shift by $-2$ in $n$ before it lines up with the other $x^n$ series) and mixing up the sign of the $-y$ term when collecting coefficients.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., §5.1–5.3 (Power Series Method).

Question 1 of 7: Power-series solution of a second-order ODE

Question 1: Power-series solution of a second-order ODE (20 marks)