consider the following ODE which can be solved with separation of variables:
$ x' = 2t (1+x^2), x(0) = 0 $
The solution is:
$ \lambda(t) = \tan(t^2) $ But what can I say about the uniqueness of the solution?
Thank you
consider the following ODE which can be solved with separation of variables:
$ x' = 2t (1+x^2), x(0) = 0 $
The solution is:
$ \lambda(t) = \tan(t^2) $ But what can I say about the uniqueness of the solution?
Thank you
The solution is unique, by the Picard–Lindelöf theorem.
The solution is unique because $1+x^2\neq 0$ for any $x$. Hence the variables can be separated and the equation can be integrated.
In general, the uniqueness of the autonomous equations $\dot x=f(x)$ gets broken if for $\hat{x}$ such that $f(\hat{x})=0$ the improper integral $ \int_{x_0}^\hat{x}\frac{dx}{f(x)} $ converges.