110_pdfsam_math 54 differential equation solutions odd

110_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 we have input rate = 5 (gal / min) · 0 . 001% 100% = 5 · 10 5 (gal / min) , output rate = 5 (gal / min) · x ( t ) (gal) 10 , 000 (gal) = 5 · 10 4 x ( t ) (gal / min) , and the equation for x ( t ) becomes dx dt = 5 · 10 5 5 · 10 4 x dx dt + 5 · 10 4 x = 5 · 10 5 . This is a linear equation. Solving yields x ( t ) = 0 . 1 + Ce 5 · 10 4 t = 0 . 1 + Ce 0 . 0005 t . Using the initial condition, x (0) = 10 , 000 (gal) · 0 . 01% 100% = 1 (gal) , we find the value of C : 1 = 0 . 1 + Ce 0 . 0005 · 0 C = 0 . 9 . Therefore, x ( t ) = 0 . 1 + 0 . 9 e 0 . 0005 t and the concentration of chlorine, say, c ( t ), in the pool at time t is c ( t ) = x ( t ) (gal) 10 , 000 (gal) · 100% = x ( t ) 100 % = 0 . 001 + 0 . 009 e 0 . 0005 t % . After 1 hour (i.e., t = 60 min), c (60) = 0 . 001 + 0
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Unformatted text preview: . 009 e − . 0005 · 60 = 0 . 001 + 0 . 009 e − . 03 ≈ . 0097 % . To answer the second question, we solve the equation c ( t ) = 0 . 001 + 0 . 009 e − . 0005 t = 0 . 002 ⇒ t = ln(1 / 9) − . 0005 ≈ 4394 . 45 (min) ≈ 73 . 24 (h) . 7. Let x ( t ) denote the mass of salt in the Frst tank at time t . Assuming that the initial mass is x (0) = x , we use the mathematical model described by equation (1) on page 90 of the text to 106...
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