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5 votes
If
(dy)/(dt) =ky, and k is a nonzero constant, then y could be

(A)
2e^(kty)

(B)
2e^(kt)

(C)
e^(kt) +3

(D)
kty+5

(E)
(1)/(2) ky^2+ (1)/(2)

User DasBeasto
by
8.6k points

1 Answer

6 votes

This ODE is separable; we have


(\mathrm dy)/(\mathrm dt)=ky\implies\frac{\mathrm dy}y=k\,\mathrm dt

Integrating both sides gives a general solution of


\ln|y|=kt+C\implies y=e^(kt+C)=Ce^(kt)

B is the only choice that is applicable.

User Henry Lee
by
8.3k points
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