ECE 211 · Circuit Analysis · Interactive Notes

Lecture 11: Solving Second-Order DC Switched Circuits

This page develops the complete procedure for a circuit that contains both an inductor and a capacitor: the three boundary values, the second-order differential equation, the characteristic equation, and the three damping cases with their solution templates. Three examples are worked, one per damping case. The material is reconstructed from the Lecture 11 board notes as interactive figures.

Teal indicates linked content. Hovering over (or tabbing to) a boxed term such as vcv_c highlights the corresponding circuit element; the linkage also operates in the reverse direction, from the circuit to the equations.

§1 · The procedure

Two storage elements produce a second-order equation

A circuit that stores energy in one capacitor and one inductor is governed by a second-order differential equation, because two independent state variables are present. A DC switched circuit is one whose sources change value exactly once, at t=0t = 0, which is written with the unit step u(t)u(t). The response is obtained by the same five steps every time:

  1. Collect the three boundary values of the chosen unknown: f(0+)f(0^+), f′(0+)f'(0^+), and f(∞)f(\infty). The values at 0+0^+ are reached through t=0−t = 0^- and the continuity rules; the value at ∞\infty is read from the final DC steady state.
  2. Write one differential equation in that unknown, valid for t>0t > 0. The element laws and one KCL or KVL pair are combined until a single variable survives.
  3. Read the characteristic equation off the differential equation. For f′′+Xf′+Yf=Zf'' + X f' + Y f = Z, the characteristic equation is s2+Xs+Y=0s^2 + X s + Y = 0; the forcing term ZZ takes no part in it.
  4. Classify the roots and select the matching solution template.
  5. Apply the two initial conditions to fix A1A_1 and A2A_2, and state the answer.
The two continuity rules

Neither state variable is permitted to jump. The capacitor voltage satisfies vc(0+)=vc(0−)v_c(0^+) = v_c(0^-) because a jump would demand infinite current, and the inductor current satisfies iL(0+)=iL(0−)i_L(0^+) = i_L(0^-) because a jump would demand infinite voltage. Every other quantity in the circuit, including vc′v_c' and iL′i_L', is free to change instantaneously.

Reading the steady states

In DC steady state no quantity varies, so iC=C dvc/dt=0i_C = C\,dv_c/dt = 0 and vL=L diL/dt=0v_L = L\,di_L/dt = 0. The capacitor behaves as an open circuit and the inductor as a short circuit. This single substitution supplies both t=0−t = 0^- and t=∞t = \infty.

The derivative condition f′(0+)f'(0^+) is the one that is easily forgotten. It is never read directly; it is computed from the element law of the storage element at t=0+t = 0^+, that is, from iC(0+)=C vc′(0+)i_C(0^+) = C\,v_c'(0^+) or from vL(0+)=L iL′(0+)v_L(0^+) = L\,i_L'(0^+).

§2 · Worked example A: from the equation to the answer

Solve vc′′+30vc′+200vc=600v_c'' + 30 v_c' + 200 v_c = 600

The differential equation and its boundary values are supplied here, so that steps 3 through 5 may be practised on their own before a circuit is reduced to an equation. Given: vc(0+)=12v_c(0^+) = 12 V and vc′(0+)=20v_c'(0^+) = 20 V/s.

The problem

The equation vc′′+30vc′+200vc=600v_c'' + 30 v_c' + 200 v_c = 600 is to be solved for t>0t > 0, subject to vc(0+)=12v_c(0^+) = 12 V and vc′(0+)=20v_c'(0^+) = 20 V/s. The two initial conditions are what the two arbitrary constants of a second-order equation require.

Step 1: the final value

In the steady state every derivative vanishes, so the equation collapses to 200vc=600200 v_c = 600:

vc(∞)=600200=3 V\tm{a.vinf}{v_c(\infty) = \frac{600}{200} = 3\ \text{V}}

The forcing term settles the final value and nothing else.

Step 2: the characteristic equation

The coefficients are transcribed directly, with s2s^2 standing for the second derivative, ss for the first, and 1 for the function itself:

vc′′+30vc′+200vc=600    ⟹    s2+30s+200=0v_c'' + 30 v_c' + 200 v_c = 600 \;\;\Longrightarrow\;\; s^2 + 30 s + 200 = 0

The right-hand side is dropped, because a constant forcing term shifts the final value without altering the natural behavior.

Step 3: the roots

s1,2=−30±302−4(1)(200)2(1)=−30±102=−20,  −10s_{1,2} = \frac{-30 \pm \sqrt{30^2 - 4(1)(200)}}{2(1)} = \frac{-30 \pm 10}{2} = -20,\; -10

The discriminant is positive, so the roots are real and distinct: the response is overdamped.

Step 4: the template

The overdamped template is written with the two roots in the exponents and the final value added:

vc(t)=A1es1t+A2es2t+vc(∞)=A1e−20t+A2e−10t+3v_c(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} + v_c(\infty) = A_1 e^{-20t} + A_2 e^{-10t} + 3

Both exponents are negative, so both transient terms decay and only the 3 V remains.

Step 5a: apply the value condition

Setting t=0t = 0 makes both exponentials equal to 1:

12=A1+A2+3    ⟹    A1+A2=912 = A_1 + A_2 + 3 \;\;\Longrightarrow\;\; A_1 + A_2 = 9

Step 5b: apply the derivative condition and solve

Differentiating the template brings each root down as a factor:

vc′(t)=−20A1e−20t−10A2e−10t    ⟹    20=−20A1−10A2v_c'(t) = -20 A_1 e^{-20t} - 10 A_2 e^{-10t} \;\;\Longrightarrow\;\; 20 = -20 A_1 - 10 A_2

Substituting A1=9−A2A_1 = 9 - A_2 gives 20=−180+20A2−10A220 = -180 + 20 A_2 - 10 A_2, hence A2=20A_2 = 20 and A1=−11A_1 = -11:

vc(t)=−11e−20t+20e−10t+3 V\boxed{v_c(t) = -11 e^{-20t} + 20 e^{-10t} + 3\ \text{V}}

A check confirms both conditions: vc(0)=−11+20+3=12v_c(0) = -11 + 20 + 3 = 12 V and vc′(0)=220−200=20v_c'(0) = 220 - 200 = 20 V/s. ✓

The plot shows why the overdamped label is deserved. The curve leaves 12 V with the prescribed positive slope, turns over almost immediately, and then decays to 3 V without ever crossing it.

§3 · The three damping cases

The discriminant selects the template

Everything about the shape of the response is decided by the sign of X2−4YX^2 - 4Y, the discriminant of s2+Xs+Y=0s^2 + X s + Y = 0. Three cases arise, and each carries its own template. The final value f(∞)f(\infty) is added to every template.

X2−4Y>0  ⟹  s1,s2 real and distinctX^2 - 4Y > 0 \;\Longrightarrow\; s_1, s_2 \ \text{real and distinct}
f(t)=A1es1t+A2es2t+f(∞)f(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} + f(\infty)

The steady state is reached without overshoot, and more slowly than in the critical case. Two separate decay rates are present, and the slower root dominates the tail.

X2−4Y=0  ⟹  s1=s2=s (repeated)X^2 - 4Y = 0 \;\Longrightarrow\; s_1 = s_2 = s \ \text{(repeated)}
f(t)=A1test+A2est+f(∞)f(t) = A_1 t e^{s t} + A_2 e^{s t} + f(\infty)

The factor tt on the first term is required, because A1est+A2estA_1 e^{st} + A_2 e^{st} would collapse into a single constant and could not satisfy two conditions. This case reaches the steady state fastest without overshoot.

X2−4Y<0  ⟹  s1,2=−α±jωdX^2 - 4Y < 0 \;\Longrightarrow\; s_{1,2} = -\alpha \pm j\omega_d
f(t)=A1e−αtcos⁡(ωdt)+A2e−αtsin⁡(ωdt)+f(∞)f(t) = A_1 e^{-\alpha t}\cos(\omega_d t) + A_2 e^{-\alpha t}\sin(\omega_d t) + f(\infty)

The real part −α-\alpha sets the decay envelope and the imaginary part ωd\omega_d sets the ringing frequency. The steady state is approached fastest of the three, but the response overshoots and oscillates on the way.

The three cases driven to the same final value. The underdamped response arrives first but rings; the critically damped response is the fastest arrival without overshoot; the overdamped response is the slowest.

underdamped critically damped overdamped

Watch the roots move

The constant term is held at Y=400Y = 400, so ω0=Y=20\omega_0 = \sqrt{Y} = 20 rad/s, and the coefficient XX is swept. While the roots are complex their distance from the origin is fixed at ω0\omega_0, so they slide along the dashed circle; at X=40X = 40 they collide at −20-20, and beyond that they separate along the real axis, one root running toward the origin and the other toward −∞-\infty. The response at the right is the step response with f(0)=0f(0) = 0, f′(0)=0f'(0) = 0, and f(∞)=1f(\infty) = 1.

The roots of s² + Xs + 400 = 0

The step response they produce

X² − 4Y = −1200     s1,2 = −10.00 ± j17.32     underdamped

The slowest root, not the average

An overdamped response is often described as slow, and the reason is visible in the root positions. As XX grows, one root races off toward −∞-\infty and contributes a transient that vanishes almost at once, while the other creeps toward the origin and holds the response back. The tail is governed by the root nearest the origin.

§4 · Where the characteristic equation comes from

One trial solution, one algebraic equation

The characteristic equation is not a rule to be memorized in isolation; it is what remains after an exponential trial solution is substituted into the differential equation. The source-free series loop makes the derivation short.

KVL around the loop

One loop carries one current, so a single KVL equation describes the circuit. The inductor voltage is L di/dtL\,di/dt, the resistor voltage is RiRi, and the capacitor voltage is the accumulated charge divided by CC:

Ldidt+Ri+1C∫i dt=0\tm{rl.l}{L \frac{di}{dt}} + \tm{rl.r}{R i} + \tm{rl.c}{\frac{1}{C}\int i\,dt} = 0

Differentiate, then normalize

Differentiating once removes the integral, and dividing by LL places the equation in standard form with a leading coefficient of 1:

Ld2idt2+Rdidt+iC=0    ⟹    d2idt2+RLdidt+1LCi=0L \frac{d^2 i}{dt^2} + R \frac{di}{dt} + \frac{i}{C} = 0 \;\;\Longrightarrow\;\; \frac{d^2 i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{1}{LC} i = 0

Try an exponential

Every term of the equation is a derivative of ii, and the exponential is the one function whose derivatives reproduce itself. Therefore i(t)=Aesti(t) = A e^{st} is proposed, with AA and ss still unknown:

As2est+ARLsest+A1LCest=0A s^2 e^{st} + A \frac{R}{L} s e^{st} + A \frac{1}{LC} e^{st} = 0

Factor and cancel

The common factor AestA e^{st} is removed. It is never zero for a nontrivial solution, because est≠0e^{st} \neq 0 for every finite tt and A=0A = 0 would leave no current at all:

Aest(s2+RLs+1LC)=0    ⟹    s2+RLs+1LC=0A e^{st}\left(s^2 + \frac{R}{L} s + \frac{1}{LC}\right) = 0 \;\;\Longrightarrow\;\; s^2 + \frac{R}{L} s + \frac{1}{LC} = 0

The differential equation has become an ordinary quadratic, whose two roots are the two natural frequencies of the circuit.

The rule in general form

Nothing in the argument depended on the particular circuit. For any second-order equation written with a leading coefficient of 1,

f′′+Xf′+Yf=Z    ⟹    s2+Xs+Y=0f'' + X f' + Y f = Z \;\;\Longrightarrow\;\; \boxed{s^2 + X s + Y = 0}

The forcing term ZZ is discarded, because it fixes f(∞)f(\infty) and leaves the natural response untouched. For the series loop, X=R/LX = R/L and Y=1/(LC)Y = 1/(LC).

§5 · Worked example B: an underdamped circuit

Find vc(t)v_c(t)

The source supplies 10u(t)10u(t) V, so it contributes nothing before t=0t = 0 and 10 V afterwards. The inductor and the resistor are in parallel between the source node and the capacitor node, and the capacitor returns to ground. The full procedure is applied.

The problem

Find vc(t)v_c(t) for t>0t > 0. Given: a 10u(t)10u(t) V source, a 15\tfrac{1}{5} H inductor, a 5 Ω resistor, and a 1100\tfrac{1}{100} F capacitor. The circuit has been undisturbed for all t<0t < 0.

t=0−t = 0^-: the DC steady state before the step

The source contributes 0 V, so the circuit is unenergized and has been at rest indefinitely. The inductor is replaced by a short circuit and the capacitor by an open circuit, and nothing drives either one:

iL(0−)=0 Avc(0−)=0 Vi_L(0^-) = 0\ \text{A} \qquad v_c(0^-) = 0\ \text{V}

t=0+t = 0^+: continuity, then the derivative

The two state variables carry across the switching instant unchanged, which gives vc(0+)=0v_c(0^+) = 0 V and iL(0+)=0i_L(0^+) = 0 A. The derivative is then found from KCL at the capacitor node, where the source now stands at 10 V:

iC(0+)=10−05−iL(0+)=2−0=2 Ai_C(0^+) = \tm{b.R}{\frac{10 - 0}{5}} - \tm{b.L}{i_L(0^+)} = 2 - 0 = 2\ \text{A}
vc′(0+)=iC(0+)C=21/100=200 V/sv_c'(0^+) = \frac{i_C(0^+)}{C} = \frac{2}{1/100} = 200\ \text{V/s}

The capacitor voltage starts at zero, but it is already climbing at 200 V/s.

t=∞t = \infty: the DC steady state after the step

The inductor becomes a short circuit again, which ties the capacitor node directly to the 10 V source node, and the capacitor becomes an open circuit and draws nothing:

vc(∞)=10 Vv_c(\infty) = 10\ \text{V}

t>0t > 0: the element laws

Two unknown functions are in play, vcv_c and iLi_L, so two equations are written and one variable is eliminated. The element laws supply the derivatives:

iC=Cdvcdt=1100vc′vL=LdiLdt=15iL′i_C = C \frac{dv_c}{dt} = \frac{1}{100} v_c' \qquad\qquad v_L = L \frac{di_L}{dt} = \frac{1}{5} i_L'

KCL at the capacitor node

Currents are summed out of the node that carries vcv_c. The resistor current flows out through 5 Ω toward the 10 V node, the inductor current arrives, and the capacitor current flows down:

vc−105−iL+1100vc′=0\tm{b.R}{\frac{v_c - 10}{5}} - \tm{b.iL}{i_L} + \tm{b.C}{\frac{1}{100} v_c'} = 0

KVL around the outer loop

The outer loop passes through the source, the inductor, and the capacitor:

−10+15iL′+vc=0\tm{b.vs}{-10} + \tm{b.L}{\frac{1}{5} i_L'} + \tm{b.C}{v_c} = 0

The loop through the resistor instead of the inductor gives the same equation, because the two elements are in parallel and share their voltage.

Eliminate iLi_L

The KCL equation is solved for iLi_L and differentiated once, which produces exactly the iL′i_L' that KVL requires:

iL=vc−105+1100vc′    ⟹    iL′=vc′5+vc′′100i_L = \frac{v_c - 10}{5} + \frac{1}{100} v_c' \;\;\Longrightarrow\;\; i_L' = \frac{v_c'}{5} + \frac{v_c''}{100}

Substituting into the KVL equation and multiplying through by 500 clears every fraction:

−10+15[vc′5+vc′′100]+vc=0    ⟹    vc′′+20vc′+500vc=5000-10 + \frac{1}{5}\left[\frac{v_c'}{5} + \frac{v_c''}{100}\right] + v_c = 0 \;\;\Longrightarrow\;\; \boxed{v_c'' + 20 v_c' + 500 v_c = 5000}

The final value may be checked against it at once: 500vc(∞)=5000500 v_c(\infty) = 5000 gives vc(∞)=10v_c(\infty) = 10 V, in agreement with the steady-state circuit. ✓

The characteristic equation and its roots

s2+20s+500=0    ⟹    s1,2=−20±400−20002=−20±j402=−10±j20s^2 + 20 s + 500 = 0 \;\;\Longrightarrow\;\; s_{1,2} = \frac{-20 \pm \sqrt{400 - 2000}}{2} = \frac{-20 \pm j40}{2} = -10 \pm j20

The discriminant is negative, so the roots are complex conjugates: the response is underdamped, with α=10\alpha = 10 and ωd=20\omega_d = 20 rad/s.

The template and the two constants

vc(t)=A1e−10tcos⁡20t+A2e−10tsin⁡20t+10v_c(t) = A_1 e^{-10t}\cos 20t + A_2 e^{-10t}\sin 20t + 10

At t=0t = 0 the cosine is 1 and the sine is 0, so 0=A1+100 = A_1 + 10 and A1=−10A_1 = -10. Differentiating by the product rule and evaluating at t=0t = 0 leaves only two surviving terms, −αA1-\alpha A_1 and ωdA2\omega_d A_2:

vc′(0+)=−10A1+20A2=200    ⟹    100+20A2=200    ⟹    A2=5v_c'(0^+) = -10 A_1 + 20 A_2 = 200 \;\;\Longrightarrow\;\; 100 + 20 A_2 = 200 \;\;\Longrightarrow\;\; A_2 = 5
vc(t)=−10e−10tcos⁡20t+5e−10tsin⁡20t+10 V\boxed{v_c(t) = -10 e^{-10t}\cos 20t + 5 e^{-10t}\sin 20t + 10\ \text{V}}

The capacitor voltage overshoots to 12.93 V at t=0.134t = 0.134 s and rings toward 10 V inside an envelope that decays as e−10te^{-10t}. The ringing period is 2π/ωd=0.3142\pi/\omega_d = 0.314 s.

Why the overshoot reaches 12.93 V

The two transient terms combine into a single sinusoid, −10cos⁡20t+5sin⁡20t=11.18cos⁡(20t−2.678)-10\cos 20t + 5\sin 20t = 11.18\cos(20t - 2.678), whose amplitude is 102+52=11.18\sqrt{10^2 + 5^2} = 11.18 V. The first maximum occurs when the cosine returns to 1, at t=2.678/20=0.134t = 2.678/20 = 0.134 s, where the envelope has decayed to 11.18e−1.339=2.9311.18 e^{-1.339} = 2.93 V. The peak is therefore 10+2.93=12.9310 + 2.93 = 12.93 V, an overshoot of 29.3 percent.

§6 · Worked example C: a critically damped circuit

Find iL(t)i_L(t)

Here the source steps from one nonzero value to another, so the circuit is already energized before t=0t = 0. The source supplies 2+2u(t)2 + 2u(t) V, that is, 2 V for t<0t < 0 and 4 V for t>0t > 0. The resistor and the capacitor are in parallel, and the inductor carries the current that is sought.

The problem

Find iL(t)i_L(t) for t>0t > 0. Given: a 2+2u(t)2 + 2u(t) V source, a 2 H inductor, a 5 Ω resistor, and a 150\tfrac{1}{50} F capacitor.

t=0−t = 0^-: the DC steady state at 2 V

The inductor is a short circuit and the capacitor is an open circuit, so the whole 2 V appears across the resistor and Ohm's law delivers the inductor current directly:

iL(0−)=25 Avc(0−)=2 Vi_L(0^-) = \frac{2}{5}\ \text{A} \qquad v_c(0^-) = 2\ \text{V}

t=0+t = 0^+: continuity, then the derivative

Continuity carries both state variables across: iL(0+)=25i_L(0^+) = \tfrac{2}{5} A and vc(0+)=2v_c(0^+) = 2 V. The source has jumped to 4 V while the capacitor voltage has not moved, so the difference appears across the inductor, and its element law gives the derivative:

vL(0+)=4−2=2 V    ⟹    iL′(0+)=vL(0+)L=22=1 A/sv_L(0^+) = 4 - 2 = 2\ \text{V} \;\;\Longrightarrow\;\; i_L'(0^+) = \frac{v_L(0^+)}{L} = \frac{2}{2} = 1\ \text{A/s}

t=∞t = \infty: the DC steady state at 4 V

The inductor is a short circuit and the capacitor is an open circuit once more, so the 4 V appears across the resistor:

iL(∞)=45 Ai_L(\infty) = \frac{4}{5}\ \text{A}

KCL at the node, KVL around the loop

The inductor current divides between the resistor and the capacitor, and the loop through the source, the inductor, and the capacitor closes the description:

iL=vc5+150vc′−4+2iL′+vc=0\tm{c.iL}{i_L} = \tm{c.R}{\frac{v_c}{5}} + \tm{c.C}{\frac{1}{50} v_c'} \qquad\qquad \tm{c.vs}{-4} + \tm{c.L}{2 i_L'} + \tm{c.C}{v_c} = 0

Eliminate vcv_c

This time the KVL equation is the one that is rearranged, because it isolates vcv_c in a single step, and its derivative follows:

vc=4−2iL′vc′=−2iL′′v_c = 4 - 2 i_L' \qquad\qquad v_c' = -2 i_L''

Substituting both into the KCL equation and multiplying through by 25 gives the differential equation:

iL=4−2iL′5−2iL′′50    ⟹    iL′′+10iL′+25iL=20i_L = \frac{4 - 2 i_L'}{5} - \frac{2 i_L''}{50} \;\;\Longrightarrow\;\; \boxed{i_L'' + 10 i_L' + 25 i_L = 20}

The final value confirms it: 25iL(∞)=2025 i_L(\infty) = 20 gives iL(∞)=45i_L(\infty) = \tfrac{4}{5} A, as the steady-state circuit already showed. ✓

The characteristic equation and its roots

s2+10s+25=0    ⟹    s1,2=−10±100−1002=−5,  −5s^2 + 10 s + 25 = 0 \;\;\Longrightarrow\;\; s_{1,2} = \frac{-10 \pm \sqrt{100 - 100}}{2} = -5,\; -5

The discriminant is exactly zero, so the roots are real and repeated: the response is critically damped.

The template and the two constants

iL(t)=A1te−5t+A2e−5t+45i_L(t) = A_1 t e^{-5t} + A_2 e^{-5t} + \frac{4}{5}

At t=0t = 0 the first term vanishes, so 25=A2+45\tfrac{2}{5} = A_2 + \tfrac{4}{5} and A2=−25A_2 = -\tfrac{2}{5}. Differentiating gives iL′(t)=A1e−5t−5A1te−5t−5A2e−5ti_L'(t) = A_1 e^{-5t} - 5 A_1 t e^{-5t} - 5 A_2 e^{-5t}, so at t=0t = 0:

1=A1−5A2=A1+2    ⟹    A1=−11 = A_1 - 5 A_2 = A_1 + 2 \;\;\Longrightarrow\;\; A_1 = -1
iL(t)=−te−5t−25e−5t+45 A\boxed{i_L(t) = -t e^{-5t} - \tfrac{2}{5} e^{-5t} + \tfrac{4}{5}\ \text{A}}

The inductor current rises from 25\tfrac{2}{5} A to 45\tfrac{4}{5} A with no overshoot and no ringing, which is the signature of the critical case.

The response is monotonic, as claimed

Differentiating the answer gives iL′(t)=−e−5t+5te−5t+2e−5t=(1+5t)e−5ti_L'(t) = -e^{-5t} + 5t e^{-5t} + 2 e^{-5t} = (1 + 5t)e^{-5t}, which is positive for every t≥0t \geq 0. The current therefore increases at every instant and approaches 45\tfrac{4}{5} A from below without ever exceeding it. The initial value iL′(0)=1i_L'(0) = 1 A/s also matches the condition that was imposed. ✓

§7 · Your turn

Practice: find vc(t)v_c(t) in a series loop

A series loop is driven by a 12u(t)12u(t) V source through a 6 Ω resistor and a 1 H inductor into a 18\tfrac{1}{8} F capacitor. The circuit is unenergized for t<0t < 0. The problem should be attempted on paper before the hints are opened, in order.

Solve for vc(t)v_c(t), the capacitor voltage, for t>0t > 0.

Hint 1: the three boundary values

Before the step the source contributes nothing and the circuit is unenergized, so vc(0−)=0v_c(0^-) = 0 V and i(0−)=0i(0^-) = 0 A; continuity carries both across. The derivative follows from the capacitor law, vc′(0+)=i(0+)/Cv_c'(0^+) = i(0^+)/C, and the inductor current is still zero at that instant. After the step the capacitor is an open circuit, so no current flows and no voltage is dropped across the resistor or the inductor.

Hint 2: one loop, one equation

A single loop carries a single current, so one KVL equation suffices: −12+Ri+L di/dt+vc=0-12 + Ri + L\,di/dt + v_c = 0. Every term is then written in vcv_c alone through i=C dvc/dti = C\,dv_c/dt, which turns L di/dtL\,di/dt into LC vc′′LC\,v_c''.

Hint 3: the differential equation
LC vc′′+RC vc′+vc=12    ⟹    vc′′+RLvc′+1LCvc=12LCLC\,v_c'' + RC\,v_c' + v_c = 12 \;\;\Longrightarrow\;\; v_c'' + \frac{R}{L} v_c' + \frac{1}{LC} v_c = \frac{12}{LC}

With R=6R = 6 Ω, L=1L = 1 H, and C=18C = \tfrac{1}{8} F this is vc′′+6vc′+8vc=96v_c'' + 6 v_c' + 8 v_c = 96. The discriminant 36−3236 - 32 is positive, so the case is overdamped.

Solution

The boundary values are vc(0+)=0v_c(0^+) = 0 V, vc′(0+)=i(0+)/C=0v_c'(0^+) = i(0^+)/C = 0 V/s, and vc(∞)=96/8=12v_c(\infty) = 96/8 = 12 V. The roots of s2+6s+8=0s^2 + 6s + 8 = 0 are −2-2 and −4-4, so the overdamped template applies:

vc(t)=A1e−2t+A2e−4t+12v_c(t) = A_1 e^{-2t} + A_2 e^{-4t} + 12

The two conditions give A1+A2=−12A_1 + A_2 = -12 and −2A1−4A2=0-2A_1 - 4A_2 = 0, hence A1=−2A2A_1 = -2A_2, so −A2=−12-A_2 = -12, A2=12A_2 = 12, and A1=−24A_1 = -24:

vc(t)=−24e−2t+12e−4t+12 V\boxed{v_c(t) = -24 e^{-2t} + 12 e^{-4t} + 12\ \text{V}}

An independent check substitutes the answer back into the differential equation. With vc′=48e−2t−48e−4tv_c' = 48 e^{-2t} - 48 e^{-4t} and vc′′=−96e−2t+192e−4tv_c'' = -96 e^{-2t} + 192 e^{-4t}, the coefficients of e−2te^{-2t} collect to −96+288−192=0-96 + 288 - 192 = 0 and those of e−4te^{-4t} to 192−288+96=0192 - 288 + 96 = 0, leaving 8⋅12=968 \cdot 12 = 96 on the left, which matches the forcing term exactly. ✓

The loop current follows as i=Cvc′=6(e−2t−e−4t)i = C v_c' = 6(e^{-2t} - e^{-4t}) A, which peaks at t=12ln⁡2=0.347t = \tfrac{1}{2}\ln 2 = 0.347 s with a value of 1.5 A, and which correctly starts and ends at zero.