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112_pdfsam_math 54 differential equation solutions odd

112_pdfsam_math 54 differential equation solutions odd - 20...

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Chapter 3 This is a linear equation in standard form. Using the method given on page 51 of the text we find the general solution to be y ( t ) = x 0 20 te t/ 20 + Ce t/ 20 . The constant C can be found from the initial condition: 0 = y (0) = x 0 20 · 0 · e 0 / 20 + Ce 0 / 20 C = 0 . Therefore, y ( t ) = ( x 0 / 20) te t/ 20 . To investigate y ( t ) for maximum value we calculate dy dt = x 0 20 e t/ 20 y ( t ) 20 = x 0 20 e t/ 20 1 t 20 . Thus dy dt = 0 1 t 20 = 0 t = 20 , which is the point of global maximum (notice that dy/dt > 0 for t < 20 and dy/dt < 0 for t > 20). In other words, at this moment the water in the second tank will taste saltiest, and comparing concentrations, it will be y (20) / 60 x 0 / 60 = y (20) x 0 = 1 20 · 20 · e
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Unformatted text preview: 20 / 20 = e − 1 times as salty as the original brine. 9. Let p ( t ) be the population of splake in the lake at time t . We start counting the population in 1980. Thus, we let t = 0 correspond to the year 1980. By the Malthusian law stated on page 93 of the text, we have p ( t ) = p e kt . Since p = p (0) = 1000, we see that p ( t ) = 1000 e kt . To Fnd k we use the fact that the population of splake was 3000 in 1987. Therefore, p (7) = 3000 = 1000 e k · 7 ⇒ 3 = e k · 7 ⇒ k = ln 3 7 . 108...
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