WrittenHW02-Solutions

# WrittenHW02-Solutions - Math 216(Section 50 1(2.11 Written...

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Math 216 (Section 50) Written Homework #2 – Solutions Fall 2007 1 ( 2.1–1 ) dx dt = x - x 2 = x (1 - x ) = - x ( x - 1) = 1 x ( x - 1) dx dt = - 1 = 1 ( x - 1)( x + 1) dx = - dt. However, 1 x ( x - 1) = 1 x - 1 - 1 x , and thus 1 x - 1 - 1 x dx = - dt = 1 x - 1 - 1 x dx = - dt = 1 x - 1 - 1 x dx = - dt = ln | x - 1 | - ln | x | = - t + C = ln x - 1 x = - t + C. = x - 1 x = e C e - t = A e 2 t . = x - 1 x = ± A e 2 t = B e 2 t . Plugging ( t, x ) = (0 , 2) into the above equation, 1 2 = B. Therefore now the equation becomes x - 1 x = 1 2 e - t = 1 - 1 x = 1 2 e - t = 1 x = 1 - 1 2 e - t = 2 - e - t 2 x = 2 2 - e - t . 2 ( 2.1–10 ) The general equation for the population model: dP dt = β P - δ P, 1

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where β and δ are birth and death rates per unit Based on the information given, β = 0 , δ = k 1 P for some constant k . Then the differential equation for P ( t ) is given by dP dt = - k P, (1) where k is a constant. The initial condition is given by P (0) = 900 , (2) and we also have P (6) = 441 . (3) Now let us solve the differential equation. 1 P dP dt = - k = 1 P dP = - k dt = 1 P dP = - k dt = 2 P = - kt + C.
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