The discussion revolves around a non-linear ordinary differential equation (ODE) with an initial value problem. The equation is piecewise defined, with different expressions for the derivative based on the value of the dependent variable y. The original poster expresses uncertainty about how to approach the problem, particularly due to the piecewise nature of the ODE.
The discussion is ongoing, with participants providing guidance on how to interpret the piecewise conditions. There is a focus on clarifying the definitions and implications of the conditions for y and x, and the potential for different types of solutions based on those interpretations.
Participants note the importance of understanding the correct interpretation of the piecewise conditions, as this affects the type of solutions that can be derived. There is mention of continuity and differentiability of solutions depending on the chosen conditions.
mistereko said:Thanks very much. That should be ok. :)
mistereko said:So, will I have two solutions in the end?
mistereko said: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) [itex]\in[/itex] C0 [0,[itex]\infty[/itex])
I'm trying to find the non-trivial solution.
Homework Equations
The Attempt at a Solution
Ray Vickson said:It is important to know if you mean that
[tex]y' = \frac{y}{3} - 3, \; y > 0, \text{ and } y' = -\frac{y}{3} - 3, \;<br /> y \leq 0,[/tex]
or
[tex]y' = \frac{y}{3} - 3, \; x > 0, \text{ and } y' = -\frac{y}{3} -3, \; x \leq 0.[/tex]
Both systems have [itex]C^0[/itex] solutions; one of them has [itex]C^1[/itex] solutions, but the other does not (except for one very particular choice of initial conditions).
RGV