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

For the differential equation dy/dt=ky, k is a constant, y(0)=24, and y(1)=18. What-example-1
User Corey Ross
by
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2 Answers

6 votes

Answer:

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)

User FosterZ
by
7.6k points
3 votes

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.

User DaveE
by
7.5k points
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