Consider the differential equation: dy/dt=y/t^2
a) Show that the constant function y1(t)=0 is a solution.
b)Show that there are infinitely many other functions that satisfy the differential equation, that agree with this solution when t<=0, but that are nonzero when t>0 [Hint: you need to define these functions using language like " y(t)=...when t<=0 and y(t)=...when t>0 and "]
c) Why doesn't this example contradict the Uniqueness Theorem?
I'm trying to do part b and after I separated and integrated I got
ln|y|=(-1/t)+C
I'm not sure if I can get C with the solution they gave in part a)y1(t)=0.
Anyways, I get y(t)=Ce^-(1/t). I don't know where to go from there.
dy/y = dt/t^2
ln y = -1/t + c
y = e^(-1/t+c) = e^c e^(-1/t)
=C e^(-1/t) agree
C can be anything so
what if C = 0 ?
then y(t) = 0 for all t
if C = 0 for t</= 0 then C can be anything at all for t>0
If y(t) depends on what C is, then how this equation doesn't contradict the uniqueness theorem if it has many solutions?
Because the general solution contains an an arbitrary constant C. The value of C depends on your boundary conditions, for example if y =5 at t = 2
then
5 = C e^-(1/2)
5 = C (.606)
C = 8.24
NOW your solution is unique.
scroll down through this:
http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html
To solve part (b) of the problem, let's start from where you left off with the equation ln|y| = (-1/t) + C.
First, we need to determine the value of the constant C that satisfies the condition y(t) = 0 when t ≤ 0. Since ln|y| is not defined when y = 0, we can't directly substitute t = 0 to find C. However, we can use the limit:
lim(t->0-) ln|y| = lim(t->0-) (-1/t) + C
Since the right-hand side is a constant (C), the left-hand side must also be a constant. As we take the limit, y approaches zero (since y(t) = 0 for t ≤ 0), and ln|y| approaches negative infinity. Therefore, the only way for the left-hand side to be a constant is if that constant is negative infinity. So, we have:
-∞ = lim(t->0-) (-1/t) + C
Solving for C, we get C = -∞.
Now, let's express the solution y(t) in two separate pieces:
For t ≤ 0, we have y(t) = 0, as given in part (a).
For t > 0, we have y(t) = Ce^(-1/t), as you correctly found. Since C = -∞ in this case, we can write:
y(t) = (-∞)e^(-1/t)
However, it is important to note that (-∞) is not a well-defined value. One way to interpret this is to consider (-∞) as approaching negative infinity as t approaches 0 from the positive side.
Therefore, when t > 0, we can rewrite the solution as:
y(t) = lim(C->-∞) Ce^(-1/t)
So, there are infinitely many solutions to the differential equation that agree with y1(t) = 0 when t ≤ 0 but become nonzero when t > 0. These solutions can be expressed as y(t) = lim(C->-∞) Ce^(-1/t).
Now, let's discuss part (c) and why this example doesn't contradict the Uniqueness Theorem.
The Uniqueness Theorem states that if a differential equation has a unique solution on an interval, then the solution is unique on that interval.
In this example, y(t) = 0 and y(t) = lim(C->-∞) Ce^(-1/t) are both solutions on their respective intervals. However, they are different solutions on different intervals. The theorem does not require the solutions to be unique on the entire real line, just on each specific interval.
In this case, the solution y(t) = 0 is valid for t ≤ 0, while y(t) = lim(C->-∞) Ce^(-1/t) is valid for t > 0. These two solutions only overlap at t = 0, but they are different functions on their respective intervals. Therefore, the Uniqueness Theorem is not violated.