3
$\begingroup$

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?

$\endgroup$
0
5
$\begingroup$

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.

$\endgroup$

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.