Are there two maximal solutions to this differential equation?

[Susanne217]

Homework Statement

The binary relation $$\mathbb{R} \times (0, \infty)$$ which is identitical to

(as I understand it) $$\mathbb{R} \times \mathbb{R}_{+}$$

this is supposedly the set on which the solution of the differential equation

y' = f(t,y) is defined upon. Where y' = y/t + y^2 (no initial condition given).

I find the solution to be

y(t) = (2t)/(t^2 + c)

Then I am suppose to find all maximale solution on the interval above.

the solution is defined as follows on -infinity to infinity for an initial condition

y(t_0) = x_0 where t_0 > 0.

Because as I see it in Maple the solution phase portrait doesn't pass through zero if x_0 > 0.

Then I draw the phase portrait with the above condition the it look asymptotic?

But if I make x_0 then it I get what looks to be the real line.

So please bare with me.

With these two differential initial conditions I get different looking phase portrait. Does this mean that I have two maximal solutions? for this eqn?

Sincerely
Susanne

 

[lippyka]

Homework Equations y' = f(t,y) The Attempt at a Solution Since the initial condition is y(t_0) = x_0 where t_0 > 0, then the solution of the differential equation will be y(t) = (2t)/(t^2 + c) where c is a constant that depends on the initial condition. The phase portrait for this equation with the given initial condition will look asymptotic, meaning that it will not pass through zero. However, if you change the initial condition to x_0 < 0, then the phase portrait will pass through zero and hence have a maximal solution. Therefore, it can be concluded that there are two maximal solutions to this differential equation, depending on the initial condition.