Question

If {dy}{dt} =ky" align="absmiddle" class="latex-formula">, and k is a nonzero constant, then y could be
(A) $2e^{kty}$

(B) $2e^{kt}$

(C) $e^{kt} +3$

(D) $kty+5$

(E) $\frac{1}{2} ky^2+ \frac{1}{2}$

Answer 1

This ODE is separable; we have

$\dfrac{\mathrm dy}{\mathrm dt}=ky\implies\dfrac{\mathrm dy}y=k\,\mathrm dt$

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.