Given a differential equation
$$ x y''(x) - (1 - x^2) y'(x) - (1 + x) y(x) = 0 $$
and a solution $ y_1(x) = 1 - x $, we were asked to compute the Wronskian for a second independent $ y_2(x) $ which satisfies $ y(0) = 0, y''(0) = 1 $.
My attempt: When I try to plug in the given conditions $ y_2(x) $ back into the differential equation I get:
$$ 0 \cdot y_2''(0) - (1 - 0^2) y_2'(0) - (1 + 0) y(0) = 0 $$
which simplifies to $ y_2'(0) = 0 $. This means then that if both $ y_2(0) = y_2'(0) = 0 $ then the Wronskian should be zero, which I thought would mean that the solutions are not independent. Am I missing something?
Obviously the second solution, if it exists at all, is not the zero solution, behaves like $\frac{x^2}2$ close to zero. This implies for the Wronski matrix $W[y_1,y_2](x)=(1-x)x+\frac{x^2}2+...=x-\frac{x^2}2+...$
On the other hand we know from Abels theorem $W'/W=-p=\frac1x-x$, so that $$ W[y_1,y_2](x)=Ce^{\ln x-\frac{x^2}2}=Cx e^{-\frac{x^2}2} $$ which would give $C=1$ from the linear term, but is at first glance incompatible with the second term. One would need more terms in the power series expansion of the second solution to see if that resolves.