Series Solutions — Question 7

Question 7 – Solution

Strategy. Build an entire basis from the recurrence, then examine its exponentially amplified coefficients at each end of the real line.

Step 1: Follow the even and odd chains. Matching $x^n$ gives $a_{n+3}=2a_{n+1}/(n+3)$ for $n\geq 0$. The independent seeds are $a_0=a$, $a_1=c$, $a_2=b$. Ratio tests along each chain give infinite radius. A normalized basis is $$1,\quad e^{x^2}-1,\quad H(x)=x+\frac{2x^3}{3}+\frac{4x^5}{15}+\cdots.$$

Step 2: Identify and verify the odd basis. The function $H=e^{x^2}\int_0^x e^{-t^2}\,dt$ satisfies $H'=2xH+1$; differentiating twice proves the third-order equation and its initial triple $(0,1,0)$. The other two basis functions also satisfy the equation. Thus $$\boxed{y=a+b(e^{x^2}-1)+ce^{x^2}\int_0^x e^{-t^2}\,dt.}$$ Its initial coefficients are exactly $(a,c,b)$ in degrees zero, one and two.

Step 3: Classify both ends. Put $K=\sqrt\pi/2$. At $+\infty$, a nonzero $b+cK$ forces exponential growth. If $b+cK=0$, the remaining tail tends to zero because, for $x>0$, $$0\leq e^{x^2}\int_x^\infty e^{-t^2}dt\leq\frac 1{2x}.$$ The inequality follows by replacing $1$ with $t/x$ in the integral. Therefore boundedness at $+\infty$ is equivalent to $b+cK=0$, with limit $a-b$. Oddness of the integral gives the condition $b-cK=0$ at $-\infty$, also with limit $a-b$. Both conditions hold exactly when $b=c=0$, leaving constants.

Step 4: Check the selected response. Here $\boxed{y=-e^{x^2}\int_x^\infty e^{-t^2}dt<0}$ for $x\geq 0$, and the tail estimate gives $y\to 0$ as $x\to+\infty$.