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A homework problem I assigned said: The Wronskian of a second order linear ODE is given by: $$ (x-1)e^x. $$ It then asks whether the functions are linearly independent on the whole real line: yes, it is not $0$ for all $x$.

Then it asks whether the functions are solutions on the whole real line to a second order linear ODE: The answers say no because it is $0$ at $x=1$ and Abel's theorem says they must be nonzero for all $x$ to be solutions.

What am I missing? This seems contradictory. Abel's theorem implies that the Wronskian is $0$ for all $x$ or never $0$. How can that function ever be a Wronskian since it is $0$ at $x=1$, but nonzero elsewhere? Also, how can the functions be linearly independent if their Wronskian is $0$ at $x=1$?

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Abel's theorem is true for the domain where the coefficients of $y''(x)+p(x)y'(x)+q(x)y(x)=0$ exist and are continuous.

What you can conclude from the root of the Wronskian at $x=1$ is that these assumptions do not hold there, the domain can not contain the vertical line $\{(x,y):x=1\}$. Any solution has to completely lie inside the domain, so it can not cross this line, its domain is either completely left or completely right of it.

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