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
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.
Differential Equations - Damon, Thursday, January 31, 2013 at 3:52pm
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
Differential Equations - Erica, Thursday, January 31, 2013 at 3:58pm
If y(t) depends on what C is, then how this equation doesn't contradict the uniqueness theorem if it has many solutions?
Differential Equations - Damon, Thursday, January 31, 2013 at 4:20pm
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
5 = C e^-(1/2)
5 = C (.606)
C = 8.24
NOW your solution is unique.
Differential Equations - Damon, Thursday, January 31, 2013 at 4:34pm
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