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

E

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)

Answer 2

The equation is separable, so solving it is trivial:

`(\mathrm dy)/(\mathrm dt)=ky\implies\frac{\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=\frac34\implies k=ln\frac34`

so the answer is E.