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.
No one has answered this question yet.
scroll down through this:
http://hyperphysics.phy-astr.gsu.edu/hbase/diff.html
To show that the constant function y1(t) = 0 is a solution to the differential equation dy/dt = y/t^2, we need to substitute y1(t) = 0 into the equation and show that it satisfies the equation.
First, let's substitute y1(t) = 0 into the differential equation:
dy/dt = y/t^2
d(0)/dt = 0/t^2
0 = 0
As we can see, when y1(t) = 0, the differential equation is satisfied. Therefore, y1(t) = 0 is a solution.
Now, let's proceed with part b. We need to find infinitely many other functions that satisfy the differential equation, agree with the solution y1(t) = 0 when t≤0, but are nonzero when t>0.
To find such functions, we can take advantage of the fact that the constant of integration, C, can be different for different intervals. Let's split the function into two cases:
Case 1: When t ≤ 0
In this case, we want our function to be y(t) = y1(t) = 0, as given in part a. Therefore, we can define y(t) = 0 when t ≤ 0.
Case 2: When t > 0
For t > 0, we need our function to be nonzero. Let's consider the equation ln|y| = (-1/t) + C that you obtained after integrating:
ln|y| = (-1/t) + C
Now, since we want our function to be nonzero when t > 0, we can consider the exponential of both sides:
e^(ln|y|) = e^((-1/t) + C)
Since e^(ln|y|) = |y|, the equation becomes:
|y| = e^((-1/t) + C)
Finally, let's consider the absolute value of y separately:
For t > 0:
y(t) = e^((-1/t) + C) if y > 0
y(t) = -e^((-1/t) + C) if y < 0
These functions satisfy the differential equation dy/dt = y/t^2, agree with the solution y1(t) = 0 when t ≤ 0, and are nonzero when t > 0. You can choose different values of C to get infinitely many solutions.
For part c, this example does not contradict the Uniqueness Theorem because although there are infinitely many solutions, they are distinct and defined by different values of C for the interval when t > 0. The Uniqueness Theorem states that if a differential equation has a unique solution on a specific interval with given initial conditions, then no other solution can exist on the same interval with the same initial conditions. In this case, the solutions obtained have different values for C, and therefore they are different solutions, not contradicting the theorem.