= Solution
Substitute the <power series> and compare coefficients. The constant coefficient gives $-a_1-a_0=0$. For $k\geq1$, the coefficient of $x^k$ gives
$$
(k+1)(k-1)a_{k+1}+(k-1)a_k=0.
$$
Thus $a_1=-a_0$, while the $k=1$ equation places no restriction on $a_2$. For $k\geq2$,
$$
a_{k+1}=-\frac{a_k}{k+1},\qquad
a_k=\frac{2a_2(-1)^k}{k!}.
$$
The <power series> therefore sums to
$$
y=a_0(1-x)+2a_2(e^{-x}-1+x).
$$
Renaming the two arbitrary constants gives \b[$\boxed{y(x)=C(1-x)+De^{-x}}$].
Both functions solve the equation by direct substitution. Their <Wronskian>, in the order $1-x,e^{-x}$, is $xe^{-x}$, which is nonzero for $x\ne0$. Hence they give the full two-dimensional solution space on either interval away from the singular point. Both are entire, and the displayed formula also gives every twice continuously differentiable solution through zero: matching $y(0)$ and $y''(0)$ fixes the same $C,D$ on the two sides.
Back to article page
