Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 6A / a / Solution

Substitute the power series and compare coefficients. The constant coefficient gives $-a_1-a_0=0$. For $k\ge 1$, the coefficient of $x^k$ givesThus $a_1=-a_0$, while the $k=1$ equation places no restriction on $a_2$. For $k\ge 2$,The power series therefore sums toRenaming the two arbitrary constants gives $\boxed{y(x)=C(1-x)+D{e}^{-x}}$.

Both functions solve the equation by direct substitution. Their Wronskian, in the order $1-x,e^{-x}$, is $x{e}^{-x}$, which is nonzero for $x\ne 0$. 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 $y^{\prime \prime }(0)$ fixes the same $C,D$ on the two sides.