0
$\begingroup$

We have that $|t|\frac{dy}{dx}=\sqrt{|y|}$ with $y(t_0)=y_0$ and I am asked to state the values of $y_0$ and $t_0$ such that we are guaranteed a solution exists. Now I understand if $f(y,t)=\frac{\sqrt{|y|}}{|t|}$ is continuous on an interval containing $t_0$ then we have a solution, but $y(t)=0$ with $y(0)=0$ is a solution to the differential equation but $f(y,t)$ is clearly not continuous at $t=0$, so what in what I have said/understanding is wrong, or

$\endgroup$
1
$\begingroup$

The existence theorem is not an " if and only if" theorem.

If $f(t,y)$ is continuous then we have a solution for $ y'=f(t,y)$ but the continuity is not necessary for having a solution.

$\endgroup$

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.