= Solution
Substitute a <Frobenius series> $y=\sum_{n\geq0}a_nz^{n+\rho}$ into the transformed equation. For $n\geq1$, equating the coefficient of $z^{n+\rho-1}$ gives the <recurrence relation>
$$
(n+\rho)(n+\rho-2)a_n+(n+\rho-1+c)a_{n-1}=0.
$$
At the larger <indicial root> $\rho=2$, the coefficient of $a_n$ is $n(n+2)\ne0$, so a <Frobenius solution> always exists. At the smaller <indicial root> $\rho=0$, the first two equations are
$$
a_1=ca_0,\qquad0\cdot a_2+(1+c)a_1=0.
$$
Since $a_0\ne0$, the exceptional equation is consistent exactly when
$$
\boxed{c(c+1)=0,\qquad c=0\ \text{or}\ c=-1.}
$$
These are \b[both admissible values], including the zero value; the singular wording of the question does not remove either. This is the <undetermined coefficient at Frobenius resonance>: at either value $a_2$ is free, whereas otherwise the second independent solution requires a <Logarithmic Frobenius solution>.
Explicit pairs demonstrate sufficiency and <linear independence>. In the variable $z=x+1$ one may take
$$
\begin{array}{c|cc}
c&u_0(z)&u_2(z)\\\hline
0&1&1-(1+z)e^{-z}\\
-1&1-z&e^{-z}+z-1.
\end{array}
$$
Both displayed $u_2$ have leading term $z^2/2$, while $u_0(0)=1$. They therefore give independent <Frobenius solutions> with the two <indicial exponents>, and direct <differentiation> verifies the equation. The point is an <apparent regular singular point> in these two cases.
For the requested closed forms, particularly short choices are
$$
\boxed{y(x)=1\quad(c=0),\qquad y(x)=x\quad(c=-1).}
$$
In addition, $(x+2)e^{-x}$ solves the zero-parameter equation and $e^{-x}$ solves the equation with $c=-1$.
Back to article page
