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.
To answer part (b), you are on the right track with your integration of the differential equation. After separating the variables and integrating, you obtained ln|y| = (-1/t) + C.
To determine the value of the constant C, you need to consider the solution they gave in part (a), which is y1(t) = 0. For this function, y1(t) = 0 for all values of t.
Substituting this into the equation ln|y| = (-1/t) + C, we get ln|0| = (-1/t) + C. However, ln|0| is undefined since the natural logarithm is only defined for positive values. Therefore, we need to treat the case of y1(t) = 0 separately.
For the nonzero solutions that satisfy the differential equation when t > 0, we can use the condition y(t) = 0 when t ≤ 0 to determine the value of C. For t ≤ 0, we know that y(t) = 0. Substituting this into ln|y| = (-1/t) + C, we get ln|0| = (-1/t) + C, which simplifies to ln|0| = C. Again, ln|0| is undefined, so we cannot determine the value of C using this condition.
However, despite not being able to determine the value of C, we can still describe the behavior of the solutions. We can define the functions as follows:
y(t) = 0 when t ≤ 0 (to satisfy the condition of y1(t) = 0)
y(t) = Ce^(-1/t) when t > 0
Here, C represents any nonzero constant. Note that for t ≤ 0, the solution is y(t) = 0, which corresponds to y1(t) = 0. For t > 0, the solution is y(t) = Ce^(-1/t), which is nonzero.
Finally, to answer part (c), the example does not contradict the Uniqueness Theorem. The Uniqueness Theorem states that if a differential equation has a unique solution on an interval, then no other solution can exist on that interval. In this case, we have shown that there are infinitely many solutions that satisfy the given differential equation, but they have different behavior for t ≤ 0 and t > 0. Therefore, the uniqueness of the solutions is preserved because they differ on different intervals.