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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.
 Winter '07
 Stenstones?
 Math

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