280_pdfsam_math 54 differential equation solutions odd

# 280_pdfsam_math 54 differential equation solutions odd - c...

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Chapter 5 Tank B: y 0 =1L / min · x 100 kg / L 1L / min · y 100 kg / L . Hence, we obtain the system x 0 =0 . 8 x 20 + y 100 , y 0 = x 100 y 100 . From the second equation, we fnd that x = 100 y 0 + y . Substitution into the frst equation yields (100 y 0 + y ) 0 =0 . 8 100 y 0 + y 20 + y 100 100 y 0 +6 y 0 + 1 25 y =0 . 8 y 0 +0 . 06 y 0 +0 . 0004 y =0 . 008 . (5.20) The characteristic equation r 2 +0 . 06 r +0 . 0004 = 0 o± the corresponding homogeneous equation has roots r = 0 . 06 ± p (0 . 06) 2 4(1)(0 . 0004) 2 = 3 ± 5 100 , and so y h ( t )= c 1 e ( 3 5) t/ 100 + c 2 e ( 3+ 5) t/ 100 is a general solution to the homogeneous equation. We now look ±or a particular solution o± the ±orm y p ( t )= c . Substitution into (5.20) gives 0 . 0004
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Unformatted text preview: c = 0 . 008 c = . 008 . 0004 = 20 . Thus y ( t ) = y p ( t ) + y h ( t ) = 20 + c 1 e ( 3 5) t/ 100 + c 2 e ( 3+ 5) t/ 100 (5.21) is a general solution to (5.20). Then x ( t ) = y + 100 y = 20 + (1 3 5) c 1 e ( 3 5) t/ 100 + (1 3 + 5) c 2 e ( 3+ 5) t/ 100 = 20 (2 + 5) c 1 e ( 3 5) t/ 100 + ( 2 + 5) c 2 e ( 3+ 5) t/ 100 . (5.22) 276...
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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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