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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 Kirchhoff’s current law gives I 1 I 2 I 3 = 0 I 1 + I 2 + I 3 = 0 . Therefore, the system, describing the current in RLC , is q C = R 2 I 2 , R 2 I 2 = R 1 I 3 + LI 3 , I 1 + I 2 + I 3 = 0 . With given data, R 1 = R 2 = 1 Ω, L = 1 H, and C = 1 F, and the relation I 1 = dq/dt , this system becomes q = I 2 , I 2 = I 3 + I 3 , q + I 2 + I 3 = 0 . Replacing in the last two equations I 2 by q , we get I 3 + I 3 q = 0 , q + q + I 3 = 0 . We eliminate q by substituting q = I 3 + I 3 into the second equation and obtain I 3 + 2 I 3 + 2 I 3 = 0 . The characteristic equation, r 2 + 2 r + 2 = 0, has roots r = 1 ± i and so, a general solution
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