Non-linear ODE with IV problem

[mistereko]

Homework Statement

I just need to know how to start this. I've never seen a piece wise ODE before and I don't really know where to start to be honest. I know it's non-linear and it's of order one.

dy/dx = (1/3)y - 3, y > 0 and dy/dx = -(1/3)y - 3 ≤ 0. y(0) = 1 with y(x) $\in$ C₀ [0,$\infty$)

I'm trying to find the non-trivial solution.

Homework Equations

The Attempt at a Solution

 

[Dick]

Just solve the two parts separately.

 

[mistereko]

Thanks very much. That should be ok. :)

 

[mistereko]

Thanks very much. That should be ok. :)

So, will I have two solutions in the end?

 

[Dick]

So, will I have two solutions in the end?

You'll have a single solution defined in two pieces. Are you sure that the pieces are y>0 and y<=0? Not x?

 

[HallsofIvy]

No. You will have a single solution- a single function given by one formula for x< 0 and second formula for x> 0- a "piecewise function".

I clearly type too slowly!

 

[mistereko]

Thanks guys.

 

[mistereko]

It's definitely a linear ODE right?

 

[Ray Vickson]

Homework Statement

I just need to know how to start this. I've never seen a piece wise ODE before and I don't really know where to start to be honest. I know it's non-linear and it's of order one.

dy/dx = (1/3)y - 3, y > 0 and dy/dx = -(1/3)y - 3 ≤ 0. y(0) = 1 with y(x) $\in$ C₀ [0,$\infty$)

I'm trying to find the non-trivial solution.

Homework Equations

The Attempt at a Solution

It is important to know if you mean that
$$y' = \frac{y}{3} - 3, \; y > 0, \text{ and } y' = -\frac{y}{3} - 3, \;<br /> y \leq 0,$$
or
$$y' = \frac{y}{3} - 3, \; x > 0, \text{ and } y' = -\frac{y}{3} -3, \; x \leq 0.$$
Both systems have $C^0$ solutions; one of them has $C^1$ solutions, but the other does not (except for one very particular choice of initial conditions).

RGV

 

[mistereko]

It is important to know if you mean that
$$y' = \frac{y}{3} - 3, \; y > 0, \text{ and } y' = -\frac{y}{3} - 3, \;<br /> y \leq 0,$$
or
$$y' = \frac{y}{3} - 3, \; x > 0, \text{ and } y' = -\frac{y}{3} -3, \; x \leq 0.$$
Both systems have $C^0$ solutions; one of them has $C^1$ solutions, but the other does not (except for one very particular choice of initial conditions).

RGV

The first one you wrote. Cheers.