Series Solutions — Question 3

Question 3 – Solution

Strategy. Combine a highest-degree argument with the exact termination mechanism in the recurrence.

Step 1: Identify the only possible parameters. Coefficient matching gives $$\boxed{a_{n+4}=\frac{(n-\lambda)a_n}{(n+4)(n+3)(n+2)(n+1)}.}$$ If a nonzero polynomial has degree $m$, its leading term in $-xy'+\lambda y$ is $(\lambda-m)a_mx^m$; $y^{(4)}$ has smaller degree. Hence $\lambda=m$ must be a nonnegative integer.

Step 2: Construct and count the terminating chain. For $\lambda=m$, normalize $a_m=1$. Descending the recurrence gives $$\boxed{p_m(x)=\sum_{j=0}^{\lfloor m/4\rfloor} \frac{(-1)^j m!}{4^j j!(m-4j)!}\,x^{m-4j}.}$$ At $n=m$ the next coefficient is zero and the chain terminates. The displayed finite sum satisfies every coefficient equation, so it is a solution. All its coefficients are fixed by the leading coefficient; thus the polynomial solution space, including zero, is one-dimensional.

Step 3: Check a concrete eighth-degree solution. For $m=8$, $$\boxed{p_8=x^8-420x^4+1260.}$$ We have $p_8^{(4)}=1680x^4-10080$ and $-xp_8'+8p_8=-1680x^4+10080$, so the residual is identically zero.

Step 4: Rule out hidden terminating chains. The recurrence preserves the index modulo four. A nonzero chain can terminate only when its numerator $n-m$ vanishes, which occurs solely in the class containing $m$. Any nonzero seed in another class continues forever and cannot be part of a polynomial. Equivalently, a second polynomial of lower degree would require that degree to equal $m$, a contradiction. For negative or nonintegral $\lambda$, only the zero polynomial solves the equation.