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solnswk2 - MATH 2930 Practice Problems Week 2 Summer 2009 S...

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MATH 2930 Practice Problems - Week 2, Summer 2009 S. Bhat June 17, 2009 § 2.1 #2: We have to solve the initial value problem dx/dt = 10 x - x 2 , x (0) = 1. We first factor the right hand side of the equation, so it reads dx dt = x (10 - x ) . Then, via separation of variables, we can rewrite this as dx x (10 - x ) = dt. Now, if we just look at the left hand side, we want to separate this into two simpler- to-integrate terms using partial fractions. Thus, leaving out the dx for now, we want to find some A and B such that: 1 x (10 - x ) = A x + B 10 - x . Now, we simply add the fractions on the right side as normal, and end up with 1 x (10 - x ) = A (10 - x ) + Bx x (10 - x ) . Since the denominators are the same, the numerators must also be equal, and thus, we can write 1 = A (10 - x ) + Bx. 1
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Now, we combine like terms, and obtain 10 A + ( B - A ) x = 1 . Thus, since there is only a constant term on the right hand side, it must be the same as that on the left-hand side, and the coefficient of the x term on the left hand side must be zero. So: 10 A = 1 = A = 1 10 and B - A = 0 = A = B. Thus, we return to the separated equation, and rewrite 1 / 10 x dx + 1 / 10 10 - x dx = Z dt or, equivalently, Z 1 x dx + Z 1 10 - x dx = Z 10 t dt. This yields 1 10 ln | x | - 1 10 ln | 10 - x | = 10 t + C, C R . This simplifies to ln x 10 - x = 10 t + C. We can take the exponential function of both sides, and simplify this to x 10 - x = Ae 10 t , A R . Omitting the rest of the algebra, the general solution then becomes x ( t ) = 10 1 A e - 10 t + 1 . Now, we plug in the initial value, x (0) = 1, to obtain the particular solution. We get 1 = 10 1 A e 0 + 1 which yields A = 1 / 9. 2
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Thus, our particular solution is simply: x ( t ) = 10 9 e - 10 t +1 . Several solution curves are given below, with our particular solution highlighted. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 2 4 6 8 10 12 t x #15: This problem is a bit more involved than the last one. There are really two ways of doing this problem: using the formulas the book gives, or — much better for your understanding — solving the problem generally from scratch. I’ll address the latter one first. We are given the differential equation dP/dt = aP - bP 2 and some information. Let’s look at that information now. The initial population is given by P (0) = P 0 , the birth rate is given by B = aP , and the death rate is D = bP 2 . Now, The initial birth and death rates are, respectively, B 0 and D 0 , which means that these are true for time t = 0, the same time at which P ( t ) = P 0 . Thus, we can immediately say 3
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that B 0 = aP (0) and D 0 = b [ P (0)] 2 . Solving for a and b , we get a = B 0 P 0 and b = D 0 P 0 2 . Put these aside for now. Now, we solve the differential equation. Like problem 2 above, we factor the right hand side and separate variables, to obtain dP P ( a - bP ) = dt. Using partial fractions again, we obtain the integral equation Z 1 /a P dP + Z b/a a - bP dP = Z dt. Integrating this, we end up with 1 a ln | P | - 1 a ln | a - bP | = t + C, C R .
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