Series Solutions — Question 5

Question 5 – Solution

Strategy. Separate formal coefficient solvability from convergence and from one-sided smooth existence.

Step 1: Compute the formal coefficients. For $z=\sum b_nx^n$, coefficient matching gives $b_0=0$, $b_1=1$, and $b_n=(n-1)b_{n-1}$ for $n\geq 2$. Thus $b_n=(n-1)!$ for $n\geq 1$. Two integrations with zero constants give $$\boxed{y\sim\sum_{n=1}^\infty\frac{(n-1)!}{(n+1)(n+2)}x^{n+2} =\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{10}+\frac{x^6}{5}+\cdots.}$$

Step 2: Prove divergence. The ratio of consecutive displayed coefficients is $n(n+1)/(n+3)\to\infty$. For every fixed nonzero $x$, the terms eventually fail to tend to zero. The radius is zero. Any analytic solution would have exactly these forced coefficients, so no analytic solution at zero exists.

Step 3: Construct and verify a smooth solution. For $x\leq 0$, $1-xt\geq 1$. For $n\geq 1$, $$\partial_x^n\frac{x}{1-xt}=\frac{n!t^{n-1}}{(1-xt)^{n+1}}.$$ An integrable majorant permits differentiation of every order up to the left endpoint, giving $z(0)=0$ and $z^{(n)}(0-)=n!(n-1)!$. Also $$x^2z'-z=x\int_0^\infty\frac{d}{dt}\left(\frac{e^{-t}}{1-xt}\right)dt=-x.$$ Hence $\boxed{y(x)=\int_0^x(x-v)z(v)\,dv}$, with oriented integration, is smooth on $(-\infty,0]$, satisfies the equation there and has the zero initial triple. The equality at zero follows by continuity.

Step 4: Resolve the apparent contradiction. Its derivatives give exactly the formal coefficients above, but smoothness does not require the Taylor series to converge or represent the function. The leading coefficient $x^2$ vanishes at zero, so this is not an ordinary point. The analytic existence theorem for a normalized regular equation does not apply.