Series Solutions — Question 2

Question 2 – Solution

Strategy. Keep derivative values separate from coefficients divided by factorials.

Step 1: Derive periodic derivative data. Differentiating the equation $n$ times at zero gives $$b_{n+4}+b_{n+2}+b_n=1\quad(n\geq 0).$$ The first six values are $(0,0,0,0,1,1)$. Subtracting this equation from its version at $n+2$ gives $b_{n+6}=b_n$. Thus $b_n=1$ precisely when $n\equiv 4,5\pmod 6$, and otherwise $b_n=0$.

Step 2: Compare the two generating functions. The ordinary generating function of the derivatives is $$\boxed{B(z)=\frac{z^4+z^5}{1-z^6},\qquad R_B=1.}$$ There is a genuine pole at $z=1$. In contrast, the solution is $$\boxed{y(x)=\sum_{k=0}^\infty\left(\frac{x^{6k+4}}{(6k+4)!} +\frac{x^{6k+5}}{(6k+5)!}\right),\qquad R_y=\infty.}$$ Absolute convergence follows by comparison with $e^{|x|}$.

Step 3: Verify the equation and normalization. The coefficient of $x^n/n!$ in $y^{(4)}+y''+y$ is $b_{n+4}+b_{n+2}+b_n=1$, giving $e^x$. The first four derivative values vanish. Using $b_n$ as ordinary coefficients would make the fourth derivative at zero $24$, whereas the equation and initial values require $y^{(4)}(0)=1$.

Step 4: Find the dominant exponential. A particular solution is $e^x/3$, since $1^4+1^2+1=3$. The characteristic polynomial factors as $$r^4+r^2+1=(r^2+r+1)(r^2-r+1).$$ Its distinct roots have real parts $-1/2$ or $1/2$. Every homogeneous term is therefore $O(e^{x/2})$ as $x\to+\infty$, regardless of its constants. It follows that $\boxed{\lim_{x\to+\infty}e^{-x}y(x)=1/3}$.