344_pdfsam_math 54 differential equation solutions odd

344_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 For the top juncture, all the currents flow out, and the Kirchhof’s current law gives I 1 I 2 I 3 =0 I 1 + I 2 + I 3 =0 . There±ore, the system, describing the current in RLC ,is q C = R 2 I 2 , R 2 I 2 = R 1 I 3 + LI 0 3 , I 1 + I 2 + I 3 =0 . With given data, R 1 = R 2 =1Ω , L =1H ,and C = 1 F, and the relation I 1 = dq/dt ,th is system becomes q = I 2 , I 2 = I 3 + I 0 3 , q 0 + I 2 + I 3 =0 . Replacing in the last two equations I 2 by q ,weget I 0 3 + I 3 q =0 , q 0 + q + I 3 =0 . We eliminate q by substituting q = I 0 3 + I 3 into the second equation and obtain I 0 3 +2 I 0 3 +2 I 3 =0 . The characteristic equation, r 2 +2 r + 2 = 0, has roots r = 1
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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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