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Homework Statement
[itex]\frac{dy}{dt}[/itex]=y(y-1)(y+3), y(0)=4
What does the exist/unique theorem say about the solution?
Homework Equations
Exist Theorem:
Suppose some continuous function f(t,y) goes through a rectangle. If (to,yo)is a point in the rectanglein the ty-plane, then their exist a [itex]\epsilon[/itex]>0 and func. y(t)
defined for to-[itex]\epsilon[/itex]<t<to+[itex]\epsilon[/itex]
that solves the init. val. prob.
[itex]\frac{dy}{dt}[/itex]=f(t,y), y(to)=yo
Unique Theorem:
Suppose f(t,y) and ∂f/∂y are continuous func. in a rectangle in the ty-plane. If (to,yo) is a point in the rectangle and if y1(t) and y2(t) are two functions that solve the initial value prob.
[itex]\frac{dy}{dt}[/itex]=f(t,y), y(to)=yo
for all t in to-[itex]\epsilon[/itex]<t<to+[itex]\epsilon[/itex] then
y1(t)=y2(t) for that interval.
The Attempt at a Solution
Their are 3 equilibrium solutions namely y=0,1,-3
y(t)>3 because why? I know given my initial condition that if I am starting at y=4 when t=0 I can only approach y=3. Also in the book the solution includes that
y(t) goes to 3 as t decreases and y(t) goes to infinity as t increases. Why can't it be that y(t) goes to infinity as t increases and y(t) goes to 3 as t decreases. How do I know from which way my functions going if I just know the equilib. soln. and the initial cond.?
Also If I have an initial cond. of y(0)=2 how do I know which equilibrium soln it is approaching as t increases? Thanks
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