$V_s = 5e^{-3t}$ $V_c(0) = 1V$ $V_c(t) = ?$ $\cdot y(t) = e^{-pt} \int qe^{pt}dt + Ke^{-pt}$
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We are given that Vc(0) = 1V. This means that at t = 0, the voltage across the capacitor is 1V. Show more…
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