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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.