Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

The solution is unique, by the Picard–Lindelöf theorem.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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