Hi all,

I have the following D.E.

$\displaystyle \frac{dy}{dt} = \sqrt{y^2+1},\\ y(t_0) = y_0$

How can i find a value of $\displaystyle y_0$ and a value of $\displaystyle t_0$ such that there is a unique solution to the initial-value problem?

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- Mar 20th 2011, 11:48 PMOilerUniqueness Theorem
Hi all,

I have the following D.E.

$\displaystyle \frac{dy}{dt} = \sqrt{y^2+1},\\ y(t_0) = y_0$

How can i find a value of $\displaystyle y_0$ and a value of $\displaystyle t_0$ such that there is a unique solution to the initial-value problem? - Mar 21st 2011, 12:24 AMFernandoRevilla
The functions

$\displaystyle f(t,y)=\sqrt{y^2+1},\;\;\dfrac{\partial f}{\partial y}$

are continuous on $\displaystyle \mathbb{R}^2$ , so for every $\displaystyle (t_0,y_0)\in \mathbb{R}^2$ there exists a unique solution satisfying $\displaystyle y(t_0)=y_0$ .