- #1
Gridvvk
- 56
- 1
dy/dt = 2 - 2ty
y(0) = 1
I am not asked to solve this (I know it's not easy to solve), but what I am asked is,
"for large values of t is the solution y(t) greater than, less than, or equal to 1/t"?
I would think less than because 1/e^(t^2) converges faster than 1/t, but at the same time I solved dy/dt = 0 and got y = 1/t, so I'm not sure what that signifies.
y(0) = 1
I am not asked to solve this (I know it's not easy to solve), but what I am asked is,
"for large values of t is the solution y(t) greater than, less than, or equal to 1/t"?
I would think less than because 1/e^(t^2) converges faster than 1/t, but at the same time I solved dy/dt = 0 and got y = 1/t, so I'm not sure what that signifies.