Question 1 of 7: Power-Series Solution of a Legendre-Type ODE

Question 1: Power-Series Solution of a Legendre-Type 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 $(1-x^2)y''-2xy'+12y=0$; at $x=0$ the leading coefficient $(1-x^2)$ equals $1\ne0$, so $x=0$ is an ordinary point. Writing the equation as $(1-x^2)y''-2xy'+n(n+1)y=0$ with $n(n+1)=12$ identifies it as Legendre's equation with $n=3$.

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

Approach. Substitute $y=\sum_{k=0}^{\infty}a_kx^k$, match coefficients of $x^k$ to obtain a two-term recurrence for $a_{k+2}$ in terms of $a_k$, then generate the even-start solution ($a_0=1,a_1=0$) and the odd-start solution ($a_0=0,a_1=1$) separately.

1. Substitute the series and match the coefficient of $x^k$. With $y=\sum a_kx^k$, $y'=\sum ka_kx^{k-1}$, $y''=\sum k(k-1)a_kx^{k-2}$, re-index $y''$ as $\sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^{k}$ and substitute into $(1-x^2)y''-2xy'+12y=0$: $$\sum_{k=0}^{\infty}\Big[(k+2)(k+1)a_{k+2}-k(k-1)a_k-2ka_k+12a_k\Big]x^{k}=0.$$
2. Isolate the recurrence. Setting the bracket to zero for every $k$ and simplifying $-k(k-1)-2k+12=-[k(k+1)-12]$, $$\boxed{a_{k+2}=\dfrac{k(k+1)-12}{(k+1)(k+2)}\,a_k}\qquad(k=0,1,2,\dots)$$
3. Build $y_1$ from $a_0=1,\ a_1=0$ (even series). The recurrence gives $a_2=-6,\ a_4=3,\ a_6=\tfrac45,\ a_8=\tfrac37,\ a_{10}=\tfrac27,\dots$, so $$y_1(x)=\boxed{1-6x^2+3x^4+\tfrac45x^6+\tfrac37x^8+\tfrac27x^{10}+\cdots}$$ Every odd-index coefficient of $y_1$ is zero because $a_1=0$ and the recurrence only ever multiplies by two steps in $k$. None of the numerators $k(k+1)-12$ vanish for even $k\ge0$ except $k=0$ trivially through $a_2$’s own definition, so this series never terminates.
4. Build $y_2$ from $a_0=0,\ a_1=1$ (odd series). $a_3=\dfrac{1\cdot2-12}{2\cdot3}(1)=-\dfrac53$. At $k=3$ the numerator $k(k+1)-12=3\cdot4-12=0$, so $a_5=0$ and every later odd coefficient inherits that zero. The series terminates exactly: $$y_2(x)=\boxed{x-\tfrac53x^3}$$ This is a genuine polynomial solution: $y_2(x)=-\tfrac23P_3(x)$ where $P_3(x)=\tfrac12(5x^3-3x)$ is the degree-3 Legendre polynomial, and direct substitution confirms $(1-x^2)y_2''-2xy_2'+12y_2\equiv0$.
5. Confirm linear independence. $y_1(0)=1,\ y_1'(0)=0$ and $y_2(0)=0,\ y_2'(0)=1$, so the Wronskian $W(0)=y_1(0)y_2'(0)-y_1'(0)y_2(0)=1\ne0$. Hence $y_1,y_2$ are linearly independent and the general solution is $y(x)=C_1y_1(x)+C_2y_2(x)$. $y_1$ converges for $|x|\lt1$ (the nearest singular points of the ODE, where $1-x^2=0$, are $x=\pm1$); $y_2$ is a polynomial and converges everywhere.