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