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)

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- Mar 18th 2006, 12:09 PMfrozenflamesDerivative Question
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) - Mar 18th 2006, 03:25 PMThePerfectHackerQuote:

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.