WrittenHW02-Solutions

WrittenHW02-Solutions - Math 216 (Section 50) Written...

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Unformatted text preview: 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 = ⇒ Z 1 x- 1- 1 x dx =- Z 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 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)...
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This homework help was uploaded on 04/02/2008 for the course MATH 216 taught by Professor Stenstones? during the Winter '07 term at University of Michigan.

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WrittenHW02-Solutions - Math 216 (Section 50) Written...

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