Series Solutions — Question 6

Question 6 – Solution

Strategy. Apply the original equation at its singular center before invoking any ordinary-point intuition.

Step 1: Retain the exceptional equation. The coefficient of $x^0$ is $2a_1=0$. For $n\ge 1$, the coefficient of $x^n$ is $(n+1)(n+2)a_{n+1}+a_{n-1}=0$. Equivalently, $$\boxed{a_1=0,\qquad a_{m+2}=-\frac{a_m}{(m+2)(m+3)}\quad(m\ge 0).}$$ Thus $a_0$ is free but $a_1$ is not.

Step 2: Identify all analytic solutions. Every odd coefficient vanishes. The even chain gives $$\boxed{y(x)=a_0\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k}}{(2k+1)!} =a_0\frac{\sin x}{x}\quad(x\ne 0),\qquad y(0)=a_0.}$$ The ratio test gives infinite radius. Termwise substitution verifies the original equation, including $x=0$. Conversely, every analytic solution must obey the recurrence, so the displayed family is exhaustive.

Step 3: Test the proposed initial values. The data $y(0)=1$, $y'(0)=1$ are incompatible with $a_1=0$. More generally, a $C^2$ solution satisfies the equation at $0$, where it reduces to $2y'(0)=0$. Even if the equation were initially imposed only at nearby nonzero points, continuity of $y,y',y''$ would give the same constraint by taking the limit. Such a $C^2$ solution cannot have the proposed data.

Step 4: Explain the lost second parameter. With $v=xy$, $v''=xy''+2y'$, so $v''+v=0$ for $x>0$. Consequently $$\boxed{y(x)=A\frac{\sin x}{x}+B\frac{\cos x}{x}\quad(x>0).}$$ The second term behaves like $B/x$ as $x\to 0^+$; boundedness forces $B=0$. The remaining term extends analytically with value $A$ and slope $0$. The ordinary-point theorem does not guarantee two arbitrary analytic initial coefficients at a center where the leading coefficient is zero.