H3_130 - M y = M ( x ) y = 0 and N x = 0. Since they are...

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Math 415 Homework 3 2.5.15 Suppose that a certain population obeys the logistic equation dy/dt = ry [1 - ( y/K )]. In that case we can separate the equation to get Z dy y [1 - ( y/K )] = r Z dt, a.k.a., using partial fractions, ln | y | - ln | K - y | = rt + C , or y K - y = De rt . (1) The initial condition (not explicitly stated) y (0) = y 0 gives us that the constant D is equal to y 0 K - y 0 . We can solve for y to get y ( t ) = Ky 0 e rt y 0 e rt + y 0 - K . (a) If y 0 = K/ 3, find the time τ at which the initial population has doubled. Find the value of τ corresponding to r = 0 . 025 per year. If we plug this specific y 0 into equation (1) we get y K - y = 1 2 e rt . The population has doubled when y = 2 K/ 3, and plugging this in to the left side, we find that it happens when 2 = 1 2 e . or τ = 1 r ln 4. When r = 0 . 025, τ 55 . 4518. 2.6.18 Any separable equation, M ( x ) + N ( y ) y 0 = 0 , is also exact. If we write this in the standard form for exact (or near exact) equations, we get M ( x ) dx + N ( y ) dy = 0. To check exactness, we take a partial derivative of each part of this, namely the partial derivative with respect to the variable that is not in the differential next to the term. In this case we compare
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Unformatted text preview: M y = M ( x ) y = 0 and N x = 0. Since they are the same, any separable equation is also exact (since it can be written in the above form). 2.5.17 (a) Solve the Gompertz equation dy dt = ry ln K y , subject to the initial condition y (0) = y . Separating gives us Z dy y ln K y = Z r dt. For the integral on the left, if we let u = ln( K/y ) as suggested, then du =-dy/y , so the integral becomes-R u-1 du =-ln | ln( K/y ) | . So ln | ln( K/y ) | =-rt + C or ln( K/y ) = De-rt . 1 Finally, solving for the constant D , and then solving for y we get y ( t ) = Ke-ln( K/y ) e-rt = Ke ln( y /K ) e-rt . (b) If r = 0 . 71 per year, K = 80 . 5 10 6 kg, and y /K = 0 . 25, then y (2) = 57 . 580 10 6 . (c) y ( ) = 0 . 75 K when e ln(0 . 25) e-. 71 = 0 . 75 , i.e. when =-1 . 71 ln ln 0 . 75 ln 0 . 25 2 . 2148 . 2...
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This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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H3_130 - M y = M ( x ) y = 0 and N x = 0. Since they are...

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