Question 1 of 7: Power-Series Solution of a Variable-Coefficient ODE

Question 1: Power-Series Solution of a Variable-Coefficient 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 ODE $(x^2+1)y''+2xy'+2y=0$. Writing it in standard form $y''+P(x)y'+Q(x)y=0$ gives $P(x)=2x/(x^2+1)$ and $Q(x)=2/(x^2+1)$, both analytic at $x=0$ (the point $x^2+1=0$ never touches the real axis), so $x=0$ is an ordinary point.

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

Approach. Substitute the series $y=\sum a_nx^n$ directly into the ODE, collect the coefficient of each power of $x$ to obtain a three-term recurrence for $a_n$, then generate the even series (from $a_0=1,a_1=0$) and the odd series (from $a_0=0,a_1=1$) separately — these two choices of initial data are automatically linearly independent.

1. Set up the series and its derivatives. $$y=\sum_{n=0}^{\infty}a_nx^n,\qquad y'=\sum_{n=1}^{\infty}na_nx^{n-1},\qquad y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$
2. Substitute into the ODE and re-index every sum to a common power $x^n$. Splitting $(x^2+1)y''=x^2y''+y''$ and shifting the $y''$ term (which starts at $x^0$ once re-indexed) gives four series in $x^n$: $$\sum_{n=2}^{\infty}n(n-1)a_nx^n+\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=1}^{\infty}2na_nx^n+\sum_{n=0}^{\infty}2a_nx^n=0$$ Because the $n=0,1$ terms of the first sum vanish identically ($n(n-1)=0$), every sum can be taken to start at $n=0$ and combined term by term.
3. Read off the recurrence. Collecting the coefficient of $x^n$: $$(n+2)(n+1)a_{n+2}+\big[n(n-1)+2n+2\big]a_n=0\ \Longrightarrow\ (n+2)(n+1)a_{n+2}+(n^2+n+2)a_n=0$$ $$\boxed{a_{n+2}=-\dfrac{n^2+n+2}{(n+2)(n+1)}\,a_n},\qquad n=0,1,2,\dots$$ This single recurrence generates two independent series once $a_0,a_1$ are each free — even-indexed coefficients depend only on $a_0$, odd-indexed only on $a_1$.
4. Even solution $y_1$: set $a_0=1,\,a_1=0$. Applying the recurrence at $n=0,2,4,6$: $$a_2=-1,\quad a_4=\tfrac{2}{3},\quad a_6=-\tfrac{22}{45},\quad a_8=\tfrac{121}{315}$$ $$\boxed{y_1(x)=1-x^2+\dfrac{2}{3}x^4-\dfrac{22}{45}x^6+\dfrac{121}{315}x^8-\cdots}$$
5. Odd solution $y_2$: set $a_0=0,\,a_1=1$. Applying the recurrence at $n=1,3,5$: $$a_3=-\tfrac{2}{3},\quad a_5=\tfrac{7}{15},\quad a_7=-\tfrac{16}{45}$$ $$\boxed{y_2(x)=x-\dfrac{2}{3}x^3+\dfrac{7}{15}x^5-\dfrac{16}{45}x^7+\cdots}$$ $y_1$ and $y_2$ are linearly independent (one is purely even, the other purely odd, and neither is identically zero), so the general solution is $y(x)=c_1y_1(x)+c_2y_2(x)$.