The following problem deals with logistic growth. Let y be the population of a b
ID: 3139623 • Letter: T
Question
The following problem deals with logistic growth. Let y be the population of a bacterial colony counted in thousands. Suppose that the initial population is 5.000: that is, y = 5 when t = 0. Assume that y can be modeled using the differential equation dy/dt = .025 y(10 - y), where t is counted in weeks. This differential equation means that the carrying capacity of the population's environment is 10,000. Use separation of variables to deduce the differential equation 1 / y(10-y) dy = .025 dt Verify the following identity. 1 / y(10-y) = 1/10 (1/y + 1/10-y) Use the identity from part (b) to integrate both sides, and use the initial value of the population to get ln(y/10-y) = .25t. Solve for y and use this equation to predict the population after 5 weeks. After 15 weeks? This problem deals with Euler's method for approximating solutions to differential equations. More details can be found in Section 6.7 of your text. Given y' = f (x,y) . a starting point of (x0, y0), and a step-size h, we repeat the following steps for n = 0 ,1 ,2 ,3 , 1. Set x n+1 = xn + h. Set yn+1 = y n + h middot f (x n, yn). Consider the differential equation dy/dx = 5(x + y) with initial point (x0, y0) = (0.1). Use Euler's method with step size h = .1 to approximate the y-value corresponding to x = .5. In other words, y(.5) y5.Explanation / Answer
1)dy/dt = 0.025 y(10-y)
dy/y(10-y) = 0.025 dt
[1/10 (1/y + 1/10-y ) ] dy = 0.025 dt
dy/y + dy/10-y =0.25dt
lny - ln(10-y) =0.25t + c
y=5 at t=0
ln5-ln5=c=0
ln(y/(10-y)) =0.25t
t=5
ln(y/(10-y)) =0.25 x5 = 1.25
y/(10-y) =3.49
y=7.77359
t=15
ln(y/(10-y)) =0.25 x 15 = 3.75
y/(10-y) = 42.52
y=9.7702
2)
x1=0+0.1
y1= 1+ 0.1x[5(0+1)] =1.5
x2=0.1+0.1 =0.2
y2=1.5+0.1x[5(0.1+1.5)]=2.3
x3=0.2+0.1 =0.3
y3=2.3+0.1x[5(0.2+2.3)]=3.55
x4=0.4
y4=3.55+0.1[5(0.3+3.55)]= 5.475
x5=0.5
y5 = 5.475 +0.1(5(0.4+5.475)) = 8.4125
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