Question

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)`

Answer 1

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.