Series Solutions — Question 5

Question 5 – Solution

Strategy. Use coefficient termination to distinguish polynomial solutions from general local series.

Step 1: Derive the parity recurrence. Matching $x^n$ gives $$\boxed{a_{n+2}=\frac{n(n+1)-\lambda}{(n+2)(n+1)}a_n\quad(n\ge 0).}$$ Even and odd indices evolve independently from $a_0$ and $a_1$.

Step 2: Characterize termination completely. If a nonzero polynomial has degree $m$, its highest coefficient obeys $[\lambda-m(m+1)]a_m=0$, so $\lambda=m(m+1)$ for an integer $m\ge 0$. Conversely, choose the starting coefficient of parity $m$ nonzero and the other zero. For indices of that parity below $m$, no numerator vanishes; at $n=m$ it does, terminating the chain. Thus these and only these parameter values work, including $m=0$, $\lambda=0$ (constants).

Step 3: Construct and verify the requested polynomial. For $\lambda=20$, the recurrence yields $a_2=-15/4$, $a_4=35/8$, and $a_6=0$. Hence $$\boxed{y(x)=\frac{35x^4-30x^2+3}{8}.}$$ Here $y'=(35x^3-15x)/2$ and $y''=(105x^2-15)/2$. Substitution in $(1-x^2)y''-2xy'+20y$ cancels every coefficient. Initial data also hold, so uniqueness near $0$ identifies the solution.

Step 4: Locate roots and extend the solution. Solving the quadratic in $x^2$ gives four distinct zeros: $$\boxed{x=\pm\sqrt{\frac{15-2\sqrt{30}}{35}},\quad x=\pm\sqrt{\frac{15+2\sqrt{30}}{35}}.}$$ They lie in $(-1,1)$, and $y(-1)=y(1)=1$. The polynomial satisfies the original equation identically for every real $x$, including $\pm 1$. Singular normalized coefficients do not forbid this special solution from extending through them.