- #1

roam

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## Homework Statement

http://img694.imageshack.us/img694/6672/37517439.jpg

## The Attempt at a Solution

For part

**(a)**, according to the uniqueness theorem, if f(t,y) and ∂f/∂y are continious in a given interval in which (t

_{0}, y

_{0}) exists, and if y

_{1}(t) and y

_{2}(t) are two functions that solve the initial value problem then

y

_{1}(t) = y

_{2}(t)

So... in my problem the function √y+1 is continious for values greater than or equal to -1. And the partial derivative

[itex]\frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y+1}}[/itex]

is continious for y>-1.

So, can I use y(0)=3 (i.e. t

_{0}=0, and y

_{0}=3)? What else do I need to do in order to show that the solution is unique to the IVP?

For part

**(b)**, are there any values of t

_{0}and y

_{0}such that the initial value problem has more than one solution? The uniqueness theorem says if both f(t,y) and ∂f/∂y are continious, then the uniquesness of the solution is guaranteed. So how is it possible to find an example where the uniqueness fails?

Any help is greatly appreciated.

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