An LR circuit, modeled by the initial value problem
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has the solution
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which is made up of a transient response, −(l/5)e_t, and a steady-state response, (2/5) sin 2t + (1/5) cos 2t.
(a) On one plot, sketch the graph of the solution in black, the transient response in red, and the steady-state response in blue.
(b) Create a second plot that contains the steady-state response in blue. Use your numerical solver to superimpose the solutions of I' + I = cos 2t on your plot for the initial conditions 7(0) = −1, −0.5,…, 1. Explain the behavior of the solutions in terms of the steady-state and transient response.
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