Question

For the differential equation dy/dt=ky, k is a constant, y(0)=24, and y(1)=18. What is the value of k?

Answer 1

The equation is separable, so solving it is trivial:

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

Integrating both sides gives

$\ln|y|=kt+C\implies y=e^{kt+C}=Ce^{kt}$

Given $y(0)=24$ and $y(1)=18$, we find

$24=C$

$18=Ce^k=24e^k\implies e^k=\dfrac34\implies k=ln\dfrac34$

so the answer is E.

Answer 2

Answer:

E

Step-by-step explanation:

dy/dt = ky

1/y .dy = k .dt

ln(y) = kt + c

t = 0, y = 24

ln(24) = 0 + c

c = ln(24)

ln(y) = kt + ln(24)

t = 1, y = 18

ln(18) = k + ln(24)

k = ln(18) - ln(24)

k = ln(18/24) = ln(3/4)