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 $$x=1$$?
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.