# Wronskian which is zero at one point

For the ODE $$xy''-(x+2)y'+2y=0~,$$ which has solutions $$y_1 = e^x$$ and $$y_2 = x^2+2x+2$$, the Wronskian is $$W=-e^x x^2$$.

As per the known theorem, Wronskian is either identically zero (i.e. zero for all $$x$$) or is never zero.

But $$-e^x x^2$$ is zero at $$x=0$$, which appears to be in the domain of solutions of this equation (since $$y(0) = y'(0)$$.

Does this somehow contradict the theorem?

$$y^{\prime\prime} - \frac{x+2}{x} y^\prime + \frac{2}{x}y = 0$$
Then Abel's identity applies only if the functions in front of $y^\prime$ and $y$ are continuous on the corresponding open interval. $\frac{2}{x}$ is not continuous/defined at $x=0$. As expected, any open interval excluding $0$ satisfies the theorem. There is no contradiction.