HW3 - 293 HW 3 Geoffrey Recktenwald 1.4-2 Solve the ODE dy...

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293 HW 3 Geoffrey Recktenwald 9-13-2005 1.4-2 Solve the ODE dy dx = - 2 xy 2 [Method - Separate] - dy y 2 = 2 xdx - dy y 2 = (2 xdx ) 1 y = x 2 + c y = 1 x 2 + c 1.4-7 Solve the ODE dy dx = (64 xy ) 1 / 3 dy dx = 4 3 x 3 y [Method - Separate] dy 3 y = 4 3 xdx dy 3 y = 4 3 xdx 3 2 y 2 / 3 = 4 3 4 x 4 / 3 + c y = 2 x 4 / 3 + C 3 / 2 1.4-25 Solve the ODE with IC x dy dx - y = 2 x 2 y dy dx = 2 x 2 y + y x [Method - Separate] dy y = 2 x 2 + 1 x dx dy y = 2 x + 1 x dx ln | y | = x 2 + ln | x | + c y = Cxe x 2 [Apply Initial condition y (1) = 1] 1 = Ce 1 2 C = e - 1 y = xe x 2 - 1 1.4-40 Generic exponential decay equation A ( t ) = A 0 e - kt For our example 1 2 = e - 5 . 27 k Solving for k yields k = ln(2) / 5 . 27 0 . 1315 The generic equation becomes A ( t ) = A 0 e - 0 . 1315 t 100 = 1 e - 0 . 1315 t t = ln(100) / 0 . 1315 35 . 01 The area is inhabitable in about 35 years. 1.4-50 Tripling every 7.5 years implies exponential growth. The Generic exponential growth equation is A ( t ) = A 0 e kt For our example 3 = e 7 . 5 k Solve for k k = ln(3) / 7 . 5 = 2 15 ln(3) = ln(3 2 / 15 ) Our growth equation becomes A ( t ) = A 0 e ln(3 2 / 15 ) t = A 0 3 2 t/ 15 a). The initial amount A 0 = 10 so our equation becomes A ( t ) = 10 · 3 2 t/ 15 1
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b). Given A 0 = 10 how much ( A ( t )) will be present in 5 years? A ( t ) = 10 · 3 2(5) / 15 = 10 · 3 2 / 3 20 . 80 pu c). How long ( t ) will it take to reach A ( t ) = 100 ? 100 = 10 · 3 2 t/ 15 t = 15 2 · ln 10 ln 3 15 . 72 years 1.4-57 Torricelli’s law Equation (1.4-24) A ( y ) dy dx = - k y For a Cylinder, A(y) is a constant so we absorb it into k dy dt = - K y Solve for y dy y = - Kdt 2 y = - Kt + c y = - 1 2 Kt + C 2 Use the initial and final conditions to solve for c y (0) = h y ( T ) = 0 y (0) = - 1 2 K · 0 + C 2 = h C = h y ( T ) = - 1 2 K · T + h 2 = 0 K = 2 h T
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