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Series R-L with exponential voltage: particular solution

Short answer

The forcing exponent equals the natural exponent, so the particular solution takes the form K t e^(-4t). Substituting into the differential equation yields K = 8. Thus the particular solution is i_p(t) = 8 t e^(-4t) A.

An exponential voltage v(t) = 2e^(−4t) is applied at time t = 0 to a series R-L circuit comprising a resistor R = 1 Ω and an inductor L = 0.25 H. Obtain the particular solution for current i(t) through the circuit. Assume zero initial current through the inductor. Use classical approach.

Circuit equation. Applying KVL to the series R-L loop:

L di/dt + R i = v(t) → 0.25 di/dt + i = 2e^(−4t) → di/dt + 4i = 8e^(−4t)

Complementary (natural) solution

Characteristic equation: m + 4 = 0, so m = −4 and i_c = A e^(−4t).

Particular solution

The forcing term e^(−4t) has the same exponent as the natural response (−R/L = −4), so the usual guess K e^(−4t) fails (it gives 0 = 8). This is the resonance case, so try

i_p = K t e^(−4t)

di_p/dt = K e^(−4t) − 4K t e^(−4t)
di_p/dt + 4 i_p = K e^(−4t) = 8 e^(−4t)   →   K = 8

Particular solution: i_p(t) = 8t e^(−4t) A

Complete solution

i(t) = A e^(−4t) + 8t e^(−4t)

Initial condition i(0⁺) = 0 (inductor current cannot jump) gives A = 0.

Answer: i(t) = 8t e^(−4t) A for t ≥ 0. Check: at t = 0.25 s, i = 2e^(−1) = 0.736 A and di/dt + 4i = 8e^(−1), which satisfies the equation. The current peaks at t = 1/4 s with i = 0.736 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

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