- #1

BrainHurts

- 102

- 0

## Homework Statement

Find a solution of the IVP

[itex]\frac{dy}{dt}[/itex] = t(1-y

^{2})

^{[itex]\frac{1}{2}[/itex]}and y(0)=0 (*)

other than y(t) = 1. Does this violate the uniqueness part of the Existence/Uniqueness Theorem. Explain.

## Homework Equations

Initial Value Problem [itex]\frac{dy}{dt}[/itex]=f(t,y) y(t

_{0})=y

_{0}has a solution if f is continuous on B = [t

_{0},t

_{0}+ a] x [y

_{0}-b,y

_{0}+b]

## The Attempt at a Solution

So when I solved * by separation of variables, my solution

y(t)= sin(t

^{2}+[itex]\frac{πk}{2}[/itex]) where k = ±1,±2,...

f(t,y) = t(1-y

^{2})

^{[itex]\frac{1}{2}[/itex]}

So [itex]\frac{∂f}{∂y}[/itex] = ty(1-y

^{2})[itex]\frac{-1}{2}[/itex]

I want to say that that as long as |y|≤ 1, f(t,y) is continuous and because the set B is closed, there exists a Maximum value M = max

_{(t,y)[itex]\inB[/itex]}|[itex]\frac{∂f}{∂y}[/itex]|, this is the Lipschitz condition and f satisfies this condition.