1.
2.
3.
4.
9 Differential Equations
(C) v(t) = 2t^2 − t + C; v(1) = 3; so C = 2.
(B) If a(t) = 20t^3 − 6t, then
v(t) = 5t^4 − 3t^2 + C 1 ,
s(t) = t^5 − t^3 + C 1 t + C 2 ,
Since
s(−1) = −1 + 1 − C 1 + C 2 = 2
and
s(1) = 1 − 1 + C 1 + C 2 = 4,
therefore
2 C 2 = 6, C 2 = 3,
C 1 = 1.
So
v(t) = 5t^4 − 3t^2 + 1.
(D) From Answer 2, s(t) = t^5 − t^3 + t + 3, so s(0) = C 2 = 3.
(B) Since a(t) = −32, v(t) = −32t + 40, and the height of the stone s(t) =
−16t^2 + 40t + C. When the stone hits the ground, 4 sec later, s(t) = 0, so
