MATB44 Assignment 1 Question 3: solution (3) Determine a model for the population P (t) if both the birth and the death rate are proportional to the population present at time t. Solution: The birth rate is βP (t). The death rate is δP (t). The birth rate is the number of births per unit time per unit population. So dP = (β − δ)P 2 (t) dt This equation is separable: 1 dP == β − δ P 2 dt so − 1 = (β − δ)t − C P (t) or P (t) = 1 . C + (δ − β)t Here C is a constant. 1 Solution to Assignment 1 questions 7 and 8 Question 7: y ′ = −1 − y/x This equation is homogeneous. Set v = y/x So dy/dx = v + xdv/dx = −1 − v dv = −1 − 2v dx dv dx = −1 − 2v x x Integrating, we obtain 1 − ln |1 + 2v| = ln |x| + C1 2 Taking the exponential, |1 + 2v|−1/2 = C2 x Squaring and taking the reciprocal, 1 + 2v = C3 (1/x2 ) or x + 2y C3 = 2 x x C3 x + 2y = x or C4 x − . x 2 Here C1 , . . . , C4 are constants. (C4 = C3 /2, C3 = (C2 )−2 and C2 = eC1 .) y= 1 Question 8: (x2 − 3y 2 )dx + 2xydy = 0, y(2) = 1 This equation is homogeneous. Substitute v = y/x dy/dx = −(x2 − 3y 2)/2xy = −(1 − 3v 2 )/2v dv v+x = −1/v + 3/2v dx dv x = −1/v + 1/2v dx dx 2vdv = 2 −2 + v x Sub w = v 2 dx dw = −2 + w x ln |w − 2| = ln |x| + C ln |v 2 − 2| = ln |x| + C ln |y 2/x2 − 2| = ln |x| + C This is the general solution (in implicit form). Taking the exponential, y 2 /x2 − 2 = |x|eC y 2 /x2 = 2 + |x|eC y 2 = (2 + |x|eC )x2 Introducing the initial condition y(2) = 1, ln |1/4 − 2| = ln 2 + C so C = ln(7/4) − ln 2. eC = (7/4) × (1/2) = 7/8. y 2 = (2 + 7/8|x|)x2 2
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