Theorem of existence and unicity.

[Lengalicious]

Homework Statement

In the following case: x'(t) = log (3t (x(t) - 2)) does the theorem of existence and unicity guarantee a unique solution for the initial value problem x(3) = 5, justify your answer?

Homework Equations

x'(t) = log (3t (x(t) - 2))

The Attempt at a Solution

Ok what I would do is take the domain of f(x,t) and the domain of ∂f(x,t)/∂x and then find the common interval and if the initial value problem is within this interval then there is a unique solution. However there is x(t) and t in the function so I don't understand how to find the domain?

 

[HallsofIvy]

For what x and t is that function, log(3t(x- 2), continuous and differentiable? What is the largest interval containing (3, 5) on which the function is continuous and differentiable?

 

[Lengalicious]

Yeh, how would I find that out? Thats what I'm confused about.

EDIT: Would I just plug numbers into the x and t? Until I got an x and t that came up with 'Math error'? Because if that's the case then when t = 0 the function is discontinuous and when x = 2? Am I right?
and then the partial derivative is 3t/(3t(x-2)) = 1/(x-2) in which case x cannot equal 2 and t is eqal to all real numbers? So the common intervals are x cannot = 2 and t cannot = 0, with intial values t_0 = 3 and x_0 = 5 which are within the common intervals, therefore there is a unique solution? On a side note, should i have taken the partial derivative with respect to t or x? Or do I take partial derivative with respect to both and then compare all 3 intervals.