Power-Series Solution about an Ordinary Point: 04-BS-5, May 2017, 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 $(1-x^2)y''-5xy'-3y=0$, to be solved about $x=0$.

Find. Two linearly independent solutions expanded in powers of $x$ about the ordinary point $x=0$, together with the interval on which the expansions are valid.

Approach. Confirm that $x=0$ is an ordinary point, substitute $y=\sum a_nx^n$, shift indices so every sum runs over the same power of $x$, extract a single two-term recurrence, and then run that recurrence separately on the even branch ($a_0=1,a_1=0$) and the odd branch ($a_0=0,a_1=1$).

Verify that $x=0$ is an ordinary point. Writing the equation in standard form, $$y''-\dfrac{5x}{1-x^2}\,y'-\dfrac{3}{1-x^2}\,y=0$$ the coefficient functions $p(x)=-5x/(1-x^2)$ and $q(x)=-3/(1-x^2)$ are analytic at $x=0$ because the leading coefficient $1-x^2$ equals $1$ there. The nearest singular points are $x=\pm1$, so the theory guarantees two independent power-series solutions valid at least for $|x|\lt1$.
Substitute the series and its derivatives. $$y=\sum_{n=0}^{\infty}a_nx^n,\qquad y'=\sum_{n=1}^{\infty}na_nx^{n-1},\qquad y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$ Inserting these and distributing the factor $(1-x^2)$ across $y''$ gives four sums: $$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}-\sum_{n=2}^{\infty}n(n-1)a_nx^{n}-5\sum_{n=1}^{\infty}na_nx^{n}-3\sum_{n=0}^{\infty}a_nx^{n}=0$$
Re-index the first sum and collect one recurrence. Replacing $n$ by $n+2$ in the first sum makes every series run over $x^n$ from $n=0$ (the extra $n=0,1$ terms of the second and third sums vanish identically), so matching the coefficient of $x^n$ gives $$(n+2)(n+1)a_{n+2}-\big[n(n-1)+5n+3\big]a_n=0$$ The bracket factors cleanly, $n^2+4n+3=(n+1)(n+3)$, and one factor of $(n+1)$ cancels: $$\boxed{a_{n+2}=\dfrac{n+3}{n+2}\,a_n},\qquad n=0,1,2,\dots$$ Because the recurrence links $a_{n+2}$ to $a_n$ only, the even-indexed and odd-indexed coefficients form two completely independent families — which is exactly what supplies the two required solutions.
Even branch: $a_0=1,\ a_1=0$. Iterating $a_{2m}=\dfrac{2m+1}{2m}\,a_{2m-2}$ gives $$a_2=\tfrac32,\quad a_4=\tfrac54\cdot\tfrac32=\tfrac{15}{8},\quad a_6=\tfrac76\cdot\tfrac{15}{8}=\tfrac{35}{16},\quad a_8=\tfrac98\cdot\tfrac{35}{16}=\tfrac{315}{128}$$ The telescoping product $a_{2m}=\dfrac{3\cdot5\cdot7\cdots(2m+1)}{2\cdot4\cdot6\cdots(2m)}$ collapses to a closed coefficient, so $$\boxed{y_1(x)=\sum_{m=0}^{\infty}\dfrac{(2m+1)!}{4^m\,(m!)^2}\,x^{2m}=1+\dfrac32x^2+\dfrac{15}{8}x^4+\dfrac{35}{16}x^6+\dfrac{315}{128}x^8+\cdots}$$
Odd branch: $a_0=0,\ a_1=1$. Iterating $a_{2m+1}=\dfrac{2m+2}{2m+1}\,a_{2m-1}$ gives $$a_3=\tfrac43,\quad a_5=\tfrac65\cdot\tfrac43=\tfrac85,\quad a_7=\tfrac87\cdot\tfrac85=\tfrac{64}{35},\quad a_9=\tfrac{10}{9}\cdot\tfrac{64}{35}=\tfrac{128}{63}$$ so that $$\boxed{y_2(x)=\sum_{m=0}^{\infty}\dfrac{4^m\,m!\,(m+1)!}{(2m+1)!}\,x^{2m+1}=x+\dfrac43x^3+\dfrac85x^5+\dfrac{64}{35}x^7+\dfrac{128}{63}x^9+\cdots}$$
Confirm linear independence and the radius of convergence. $y_1$ is even and $y_2$ is odd, and the Wronskian evaluated at the origin is $$W(0)=y_1(0)y_2'(0)-y_1'(0)y_2(0)=(1)(1)-(0)(0)=1\ne0$$ so the pair is linearly independent and $y=c_1y_1+c_2y_2$ is the general solution. Applying the ratio test to either series, $\left|a_{n+2}x^{n+2}/(a_nx^n)\right|=\dfrac{n+3}{n+2}|x|^2\to|x|^2$, so both converge for $|x|\lt1$ and diverge for $|x|\gt1$: the radius of convergence is $R=1$, set by the singular points $x=\pm1$ found in Step 1.
Closed forms (a useful check, not required by the question). Both series sum to elementary functions, which is a convenient way to confirm the recurrence: $$y_1(x)=(1-x^2)^{-3/2},\qquad y_2(x)=\dfrac{x\sqrt{1-x^2}+\arcsin x}{2\,(1-x^2)^{3/2}}$$ Substituting each of these into the original ODE by computer algebra returns an exact residual of zero, and their Maclaurin expansions reproduce the coefficients boxed in Steps 4 and 5 term for term.

Quantity	Result
Recurrence	$a_{n+2}=\dfrac{n+3}{n+2}a_n$
$y_1(x)$ (even solution)	$1+\tfrac32x^2+\tfrac{15}{8}x^4+\tfrac{35}{16}x^6+\cdots=(1-x^2)^{-3/2}$
$y_2(x)$ (odd solution)	$x+\tfrac43x^3+\tfrac85x^5+\tfrac{64}{35}x^7+\cdots$
Wronskian at $x=0$	$W(0)=1\ne0$ (independent)
Radius of convergence	$R=1$ (singular points $x=\pm1$)

Topic: Power-series solution of a linear second-order ODE at an ordinary point.

Key relations: ordinary-point test (leading coefficient non-zero); index shift $\sum n(n-1)a_nx^{n-2}=\sum(n+2)(n+1)a_{n+2}x^n$; the two-term recurrence $a_{n+2}=\frac{n+3}{n+2}a_n$; ratio test for $R$.

Why this works: when the leading coefficient does not vanish at the expansion point, Fuchs’ theorem guarantees two analytic solutions there, so an undetermined power series is certain to succeed. Because the ODE contains only even-order couplings between the coefficient index and the power of $x$, the recurrence steps by two, which splits the coefficients into non-interacting even and odd families — each family generates one solution, and the two are automatically independent by parity. The radius of convergence never has to be computed from the series itself: it equals the distance from the expansion point to the nearest zero of the leading coefficient in the complex plane, here $|{\pm}1-0|=1$.

Common pitfall: forgetting to distribute $(1-x^2)$ before re-indexing, which drops the $-n(n-1)a_n$ contribution and yields the wrong bracket; and quoting an infinite radius of convergence out of habit — the polynomial coefficient $1-x^2$ does vanish, at $x=\pm1$, so the series here genuinely stop converging at $|x|=1$.

Source: Kreyszig, Advanced Engineering Mathematics, 10th ed., Ch. 5 §5.1 (Power Series Method) and §5.2 (Legendre’s Equation).

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

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