Series Solutions — Question 3

Question 3 – Solution

Strategy. Distinguish a solution’s real domain from the disk reached by a particular Taylor center.

Step 1: Determine the Maclaurin coefficients. For $n\ge 0$, coefficient matching gives $(n+2)(n+1)a_{n+2}-(n+1)^2a_{n+1}=0$. With $a_0=0,a_1=1$, this yields $$a_{n+2}=\frac{n+1}{n+2}a_{n+1},\qquad \boxed{y(x)=\sum_{n=1}^{\infty}\frac{x^n}{n}}.$$

Step 2: Identify two different domains. The differential equation says $[(1-x)y']'=0$. Initial data give $y'=1/(1-x)$ and $\boxed{y=-\ln(1-x)}$ on $(-\infty,1)$, its maximal real interval through $0$. The series has radius $1$. It diverges at $1$ (harmonic series) and converges at $-1$ (alternating series), so its interval is $[-1,1)$. Integrating the geometric series identifies its sum on $(-1,1)$; at $-1$, the alternating harmonic value is $-\ln 2$.

Step 3: Diagnose the failed evaluation. At $-2$, the terms $(-2)^n/n$ do not tend to zero. The series therefore diverges although $y(-2)=-\ln 3$ is finite. The obstruction is the center’s distance to the singularity at $1$, not a singularity at $-2$.

Step 4: Move the expansion center. Let $h=x+1$ and $y=\sum b_nh^n$. The equation is $(2-h)y''-y'=0$, giving $b_{n+2}=(n+1)b_{n+1}/[2(n+2)]$. Since $b_0=-\ln 2$, $b_1=1/2$, $$\boxed{y=-\ln 2+\sum_{n\ge 1}\frac{h^n}{n2^n} =-\ln 2+\frac h2+\frac{h^2}{8}+\frac{h^3}{24}+\frac{h^4}{64}+\cdots.}$$ The radius in $h$ is $2$, with convergence at $h=-2$ and divergence at $h=2$; in $x$ the interval is $[-3,1)$. Now $x=-2$ is interior, and the sum is $-\ln 2-\ln(3/2)=\boxed{-\ln 3}$.