Differential Equation - Uniqueness Theroem

[cse63146]

Homework Statement

The differential equation that models the volume of a raindrop is $$\frac{dv}{dt} = kv^{2/3}$$ where $$k = 3^{2/3}(4 \pi)^{1/3}$$

A) Why doesn't this equation satisfy the hypothesis of the Uniqueness Theroem?
B) Give a physical interpertation of the fact that solution to this equation with the initial condition v(0) = 0 are not unique. Does this model say anything about the way raindrops begin to form?

Homework Equations

The Attempt at a Solution

A) The equation doesn't satisfy the hypothesis Uniqueness Theroem because when v = 0, the equation's derivative does not exist.

B) At time t = 0, the raindrop does not have volume, but as t increases, it's volume increases as well.

Am I correct for both parts

 

[HallsofIvy]

Homework Statement

The differential equation that models the volume of a raindrop is $$\frac{dy}{dt} = kv^{2/3}$$ where $$k = 3^{2/3}(4 \pi)^{1/3}$$

Do you mean "dv/dt", rather than "dy/dt"?

A) Why doesn't this equation satisfy the hypothesis of the Uniqueness Theroem?
B) Give a physical interpertation of the fact that solution to this equation with the initial condition v(0) = 0 are not unique. Does this model say anything about the way raindrops begin to form?

Homework Equations

The Attempt at a Solution

A) The equation doesn't satisfy the hypothesis Uniqueness Theroem because when v = 0, the equation's derivative does not exist.

Strictly speaking an "equation" doesn't have a derivative. What you mean is that the function $kv^{2/3}$ has no derivative at v= 0. That is true and is a reason why the uniqueness theorem does not hold.

B) At time t = 0, the raindrop does not have volume, but as t increases, it's volume increases as well.

How do you conclude that "its volume increases"? Certainly v(t)= 0 for all t satisfies $dv/dt= kv^{2/3}$ as well as v(0)= 0.

Am I correct for both parts

 

[cse63146]

so to prove that the volume does increase, I would need to find it's general solution?