solnswk2 - MATH 2930 Practice Problems - Week 2, Summer...

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Unformatted text preview: 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 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 + 1 which yields A = 1 / 9. 2 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. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 4 6 8 10 12 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. Ill address the latter one first. We are given the differential equation dP/dt = aP- bP 2 and some information. Lets look at that information now. The initial population is given by P (0) = P , 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 and D , which means that these are true for time t = 0, the same time at which P ( t ) = P . Thus, we can immediately say 3 that B = aP (0) and D = b [ P (0)] 2 . Solving for a and b , we get a = B P and b = D P 2 . Put these aside for now....
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This note was uploaded on 01/21/2011 for the course MATH 2930 taught by Professor Terrell,r during the Spring '07 term at Cornell University (Engineering School).

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solnswk2 - MATH 2930 Practice Problems - Week 2, Summer...

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