279_pdfsam_math 54 differential equation solutions odd

279_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.2 and the general solution, x ( t ), is x ( t )= x h ( t )+ x p ( t C 1 e r 1 t + C 2 e r 2 t +20 . From the frst equation in (5.18) we fnd y ( t ) = 100 ·{ ( D +0 . 07)[ x ] 1 . 2 } = 100 dx dt +7 x ( t ) 120 = 100 ± r 1 C 1 e r 1 t + r 2 C 2 e r 2 t ² ± C 1 e r 1 t + C 2 e r 2 t ² 120 = ³ 2 7 ´ C 1 e r 1 t + ³ 2+ 7 ´ C 2 e r 2 t . The initial conditions imply 0= x (0) = C 1 + C 2 , 20 = y (0) = ( 2 7 ) C 1 + ( 7 ) C 2 C 1 + C 2 = 20 , ( 2 7 ) C 1 + ( 7 ) C 2 =0 C 1 = µ 10 + 20 7 ,C 2 = µ 10 20 7 . Thus the solution to the problem is x ( t µ 10 + 20 7 e r 1 t µ 10 20 7 e r 2 t +20(kg) , y ( t 30 7 e r 1 t 30 7 e r 2 t . 33. Since no solution flows in or out o± the system ±rom the tank B, we conclude that the solution flows ±rom the tank B to the tank A with the same rate as it does ±rom A to B, that is, 1 L/min. Furthermore, the solution flows in and out o± the tank A with the same rate, 4 L/min, and so
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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