Respuesta :
Answer:
a) This was proved to have a solution
b) This was proved not to have any unique solution
c) This was also proved not to have any unique solution
Step-by-step explanation:
The step by step explanation is as attached below
Answer:
(a) Proved
(b) discussed
(c) There are infinite number of solutions.
Step-by-step explanation:
It will be easier just to give a solution that satisfies the differential equation, but that will not suffice.
These are first order Nonlinear Differential Equations whose solutions are not as straightforward as they might seem. Two questions must be asked:
1. Does the solution to the differential equation exist?
2. If it exists, is it unique?
I will explain the general case, and then explain how they correlate with your work.
EXISTENCE
Suppose F(t, x) is a continuous function. Then the initial value problem
x'= F(t, x), x(t_0) = a
has a solution x = f(t) that is, at least, defined for some δ > 0.
This guarantees the existence of solution to the initial value problem, at
least for infinitesimal times (t). In some cases, this is the most that can be said, although in many cases the maximal interval α < t < β of the existence of solution might be much larger, possibly infinite, −∞ < t < ∞, resulting in a general solution.
The interval of existence of a solution strongly depends upon both the equation and the particular initial values. For instance, even though its right hand side is defined everywhere, the solutions to the scalar initial value problem x' = x^⅓ only exist up until time 1/(x_0) (1/0 in this case, which is infinity), and so, the larger the initial value, the shorter the time of existence.
UNIQUENESS
having talked about the importance of existence of solution, we need to ask ourselves, does the initial value problem
have more than one solution? If it does, changes will happen everytime, and we cannot use the differential equation to predict the future state of the system. The continuity of the right hand side of the differential equation will ensure the existence of a solution, but it is not sufficient to guarantee uniqueness of the solution to the initial value problem. The difficulty can be appreciated by looking at the first differential equation you gave.
x' = x^⅓ , x(0) = 0
From the explanation above, since the right hand side is a continuous function, there exists a solution, at least for t close to 0. This equation can be easily solved by the method of integration:
dx/dt = x^⅓
dx/(x^⅓) = dt
Int{x^(-⅓)dx} = dt
(x^⅔)/(⅔) = t + c
(3/2)x^⅔ = t + c
x = (⅔t + c1)^(3/2)
Applying the initial condition x(0) = 0
implies that c1 = 0, and hence,
x = ⅔t^(3/2) is a solution to the initial value problem.
But again, since the right hand side of the differential equation vanishes at x = 0, the constant function x(t) ≡ 0 is an equilibrium solution to the differential equation. Moreover, the equilibrium solution has the same initial value x(0) = 0. Therefore, we have two different solutions to the initial value problem, which invalidates its uniqueness. In fact, there is an infinite number of solutions to the initial value problem. For any positive a, the function
x(t) = 0 for 0 ≤ t ≤ a,
= (⅔t − a)^(3/2) for 2t ≥ 3a,
is differentiable at every point.
This explains the situation of questions (a) and (b).
For question (c) x' = x/t² for x(0) = 0.1
This is quite different
Solving by integration, we have
dx/x = t^(-2) dt
ln x = -1/t + c
x = kexp(-1/t)
Applying the initial condition, we realise that as n approaches 0, the lim n approaches negative infinity.
Which also means there are infinitely many solutions.
I hope this helps
