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Series R-C with exponential voltage: particular solution

Short answer

The forcing exponent matches the natural exponent, so the particular solution is of the form K t e^(-4t). Substitution gives K = -80. Hence i_p(t) = -80 t e^(-4t) A.

An exponential voltage v(t) = 20e^(−4t) is suddenly applied at time t = 0 to a series RC circuit with R = 1 Ω, C = 0.25 F. Obtain the particular solution i(t) in the circuit. Assume zero initial charge across capacitor. [Use classical method.]

Circuit equation. KVL around the series loop: R i + (1/C)∫i dt = v(t). Differentiating:

R di/dt + i/C = dv/dt
di/dt + 4i = −80 e^(−4t)    (R = 1, 1/RC = 4, dv/dt = −80e^(−4t))

Complementary solution

m + 4 = 0 → m = −4, so i_c = A e^(−4t).

Particular solution

The forcing exponent (−4) equals the natural exponent (−1/RC = −4), so the trial function must be multiplied by t:

i_p = K t e^(−4t)

di_p/dt + 4 i_p = K e^(−4t) = −80 e^(−4t)   →   K = −80

Particular solution: i_p(t) = −80 t e^(−4t) A

Complete solution

i(t) = (A − 80t) e^(−4t)

Initial condition: with the capacitor uncharged, v_C(0) = 0, so all the source voltage appears across R at t = 0⁺: i(0⁺) = v(0)/R = 20 A. Hence A = 20.

Answer: i(t) = (20 − 80t) e^(−4t) A, t ≥ 0.

Check: v_C = v − R i = 20e^(−4t) − (20 − 80t)e^(−4t) = 80t e^(−4t), which is 0 at t = 0 and satisfies i = C dv_C/dt = 0.25(80 − 320t)e^(−4t) = (20 − 80t)e^(−4t) ✓. The current changes sign at t = 0.25 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

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