If dy/dt = ky and k is a nonzero constant, then y could be
a. 2e^kry
b. 2e^kt
c. e^kt +3
d. kty + 5
e. (1/2)ky^2 + (1/2)
All solutions to this diffrencial equation are,Originally Posted by frozenflames
$\displaystyle y=Ce^{kt}$
Notice that #2:
$\displaystyle 2e^{kt}$ works because $\displaystyle C=2$. The easiest way is to check which one makes the differencial equation true.