The solution of a differential equation to the system is then given by I(t) = (3/4)e-4.48t + (-1/4)e4.48t and V(t) = (3/4)e-4.48t – (2/4)e4.48t.
a. When L = 4R2C, the eigenvalues of the system of differential equations are found by solving the characteristic equation, which is -4±2√(4R2C) = -4±8RC. Since R, C, and L are all positive constants, the eigenvalues are real and equal, and they are both equal to -4RC.
b. Setting R = 1.12, C = 1 F, and L = 4 H, we find that the eigenvalues of the system of differential equations are -4·1.12·1 = -4.48. The solution to the system is then given by I(t) = Ae-4.48t + Be4.48t and V(t) = -4RAe-4.48t + 4RB e4.48t.
Substituting the initial conditions I(0) = 1 A and V(0) = 2 V, we get A + B = 1 and -4RA + 4RB = 2. Solving for A and B, we find that A = 3/4 and B = -1/4.
The solution to the system is then given by I(t) = (3/4)e-4.48t + (-1/4)e4.48t and V(t) = (3/4)e-4.48t – (2/4)e4.48t.
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