Wronskian which is zero at one point

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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?

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I assume you're referring to Abel's Identity and its obvious corollary. If we write:

$$ y^{\prime\prime} - \frac{x+2}{x} y^\prime + \frac{2}{x}y = 0 $$

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.