contestada

If [tex] \frac{dy}{dt} =ky[/tex], and k is a nonzero constant, then y could be

(A) [tex]2e^{kty} [/tex]

(B) [tex]2e^{kt} [/tex]

(C) [tex]e^{kt} +3[/tex]

(D) [tex]kty+5[/tex]

(E) [tex] \frac{1}{2} ky^2+ \frac{1}{2} [/tex]

Respuesta :

LammettHash

This ODE is separable; we have

[tex]\dfrac{\mathrm dy}{\mathrm dt}=ky\implies\dfrac{\mathrm dy}y=k\,\mathrm dt[/tex]

Integrating both sides gives a general solution of

[tex]\ln|y|=kt+C\implies y=e^{kt+C}=Ce^{kt}[/tex]

B is the only choice that is applicable.

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