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# week05su - Math 205 Summer II 2007 B Dodson Week 3 2.5...

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Math 205, Summer II 2007 B. Dodson Week 3: 2.5 Wronskian 3.1 Intro, Slope Fields, verify solution 3.2 Separable DE 3.4 Linear Equations ————– Problem: Verify that the function y = c 1 x is a solution of y = y 2 x Solution: Compute y and check. y = c 1 ( 1 2 ) x - 1 2 . y 2 x = c 1 x 2 x = c 1 ( 1 2 ) x ( x ) 2 = c 1 ( 1 2 ) 1 x = y . —————–

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2 Problem: Inflamable substance, temp T 0 = 50 (F), placed in hot oven, temp T m = 450 (F). After 20 min substance temp T = 150 . Find temp at 40 min. If substance ignites at 350, find time of combustion. Solution: Newton’s Law of cooling: T = T ( t ) temp at time t, dT dt = - k ( T - T m ) , T m = 450 , t in min. ————– dT dt = - k ( T - T m ) , T m = 450 , T (0) = 50 , T (20) = 150 , find T (40) and t c so T ( t c ) = 350 . Method: separation. dT T - T m = - k dt, ( T = T m ). integral: ln | T - T m | = - k t + c, T m = 450 . Initial Data: ln (450 - T ) = - k t + c, T (0) = 50 , T (20) = 150 . When t = 0 , ln 400 = c, e c = 400 . So 450 - T = e - kt e c = 400 e - kt , and T = T ( t ) = 450 - 400 ( e - k ) t . Next, when t = 20 , 150 = 450 - 400 ( e - k ) 20 , so - 300 = - 400 ( e - k ) 20 , e - k = ( 3 4 ) 1 20 , and T ( t ) = 450 - 400 ( 3 4 ) t 20 . Finally, T ( t ) = 450 - 400 ( 3 4 ) t 20 , gives T (40) = 225 , and T ( t c ) = 350 gives t c = 96 . 4 minutes. (why?)
3 For t = 40 , t 20 = 2 , then ( 3 4 ) 2 = 9 16 and 400 * 9 16 = 25 · 9 = 225 , so T (40) = 450 - 225 = 225 . Next, for T := t

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