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Differential equation word problem?
I have a question that I could do with some help with:
Doctors are studying the spread of measles. Using the variable y to represent the proportion
of a population af fected by measles, and using t to represent time (days) they fi nd the rate
of spread of measles through a city can be modeled by the differential equation dy/dt = 0.2(0.6-y)
Suppose initially ten percent of a cities population has measles.
i) Solve initial value problem.?
ii) How long will it take before half the population are affected by measles?
iii) What is the long term proportion that will be affected by measles?
1 Answer
- gp4rtsLv 78 years agoFavorite Answer
The equation is dy/dt = 0.2*(0.6 - y)
the variables are separable, and the equation can be written
dy/(0.6 - y) = 0.2*dt
the solution of which is -ln|0.6 - y| = 0.2*t + C
0.6 - y = e^(-0.2*t - C) = e^(0.2*t)*e^-C let k = e^-C
y = 0.6 - k*e^-0.2*t; use the initial conditions to find k
The initial condition is y = 0.1 at t = 0:
0.1 = 0.6 - k
k = 0.6 - 0.1 = 0.5
y = 0.6 - 0.5*e^-0.2*t
half the population affected means y = 0.5
0.5 = 0.6 - 0.5*e^-0.2*t solve for t
-0.1 = - 0.5*e^-0.2*t
0.2 = e^-0.2*t
ln(0.2) = -0.2*t
-1.6 = -0.2*t
t = 8.1 days
y = 0.6 - 0.5*e^-0.2*t at t ---> ∞ e^-0.2*t ---> 0
y∞ = 0.6
