First-order differential equations
Derive the mathematical skeleton shared by one-storage-element linear circuits before specializing to RC or RL. Reduce Kirchhoff plus one constitutive law to a first-order linear ODE, derive natural and forced response by integrating factor, show the equivalent final-value form x(t)=x∞+[x(0+)−x∞]e^(−t/τ), interpret τ as the time for the error to fall by e^(−1), and separate state continuity, initial conditions, final conditions and switching topology into a reusable solution workflow.
- Recognize a one-state linear circuit equation in the standard form a dx/dt+b x=c u(t).
- Normalize the equation to dx/dt+x/τ=K u(t) and identify τ=a/b when the coefficients are positive and constant.
- Derive the homogeneous natural response x_n=Ae^(−t/τ).
- Derive a particular response and the complete response using an integrating factor rather than memorizing a waveform.
- For a constant post-switch input, derive x(t)=x∞+[x(0+)−x∞]e^(−t/τ).
- Interpret x∞ from the post-switch DC equilibrium and x(0+) from the storage-element continuity constraint.
- Explain the physical meaning of τ, including the remaining-error fractions after 1τ, 2τ, 3τ and 5τ.
- Distinguish stable τ>0 decay from an unstable sign structure that produces exponential growth.
- Construct a topology-first switching workflow: solve t<0 state, apply continuity, derive t>0 network equation, find τ and final value, then verify limits.
RC and RL formulas look different only because their state variables and physical constants differ. Mathematically, both are examples of the same one-state linear dynamical system. Deriving that common structure once prevents a large collection of disconnected charging/discharging formulas.
02.1From Kirchhoff plus one storage law to one first-order ODE
After choosing a state variable x(t), many linear one-storage-element circuits reduce to
where a, b and c are constants for a fixed post-switch topology and u(t) is the independent excitation.
If a and b are nonzero, divide by a:
For a passive stable first-order network with positive effective coefficients, define
so the equation becomes
This normalized equation is the mathematical skeleton we will later obtain from both RC and RL networks.
02.2Natural response from the homogeneous equation
Set the independent forcing to zero:
Separate variables:
Integrating gives
The constant A is set by the state immediately after switching. The exponential is not a capacitor formula or an inductor formula: it is the natural mode of a stable first-order linear differential equation.
02.3Forced response by integrating factor
For a general input u(t), write
The integrating factor is
Multiplying the ODE gives
Integrate from 0+ to t:
Therefore
This expression separates memory of the initial state from the accumulated effect of the input.
02.4Constant post-switch input and the final-value form
If u(t)=U is constant for t≥0, a constant particular solution satisfies
For a stable circuit this is also the long-time equilibrium x∞. The complete response is
This is the most reusable first-order switching formula. It says:
No sign guessing is required if x(0+) and x∞ are computed with the same reference.
02.5The state cannot be chosen arbitrarily at t=0+
The switching operation can change topology instantly, but the energy-storage state follows the continuity constraints from chapter 01.
For a capacitor state x=vC under finite non-impulsive current:
For an inductor state x=iL under finite non-impulsive voltage:
Other circuit variables may jump. For example, resistor current can change instantly because an ideal resistor stores no state.
The correct workflow therefore solves the pre-switch circuit first, carries the state across the switching instant, and only then solves the post-switch dynamics.
02.6The final value comes from the post-switch topology
For a stable circuit with constant sources after t=0, x∞ is found from the t→∞ equilibrium of the new topology. In ordinary DC steady state:
The initial state may be determined by one topology, while the final state belongs to another. Mixing the pre-switch and post-switch networks is a common source of wrong transient formulas.
02.7What τ means physically
Define the error from the final value:
For a stable first-order response,
Therefore after one time constant, the remaining error is e⁻¹≈0.3679, meaning 63.21% of the initial distance to the final value has been traversed.
Figure 2.1 — What does a time constant really mean?
The one question is: what fraction of the distance to the final value remains after n time constants?
For a stable first-order response, normalized error is e^(−t/τ). After 1τ, 36.8% remains; after 3τ, about 5%; after 5τ, about 0.67%. “Settled” therefore needs a tolerance, not a magical boundary.
Thus “five time constants means settled” is only a shorthand for a sub-1% error criterion in a single first-order exponential.
02.8Initial slope is an independent check
Differentiate the final-value form:
At the switching instant,
This gives a powerful sanity check. If the final value is above the initial value, the initial slope must be positive for τ>0. If a derived waveform rises toward a lower final value, a sign or reference error is present.
For RC and RL circuits, the same slope can also be computed directly from i=C dv/dt or v=L di/dt at 0+, providing an independent circuit-law check.
02.9Stable decay versus exponential growth
The equation
decays only if τ>0. More generally, for
Re(a)>0 produces decay; Re(a)<0 produces growth. Passive RC/RL networks built from positive R, C and L give stable positive time constants, but active circuits can create effective negative damping or unstable poles.
Do not use the phrase “time constant” to hide a sign. Inspect the actual exponential exponent.
02.10Reusable switching workflow & mastery exercises
- Choose the state: capacitor voltage or inductor current.
- Solve t<0: find x(0−), often from settled DC.
- Apply continuity: obtain x(0+).
- Freeze the t>0 topology: derive the ODE with Kirchhoff and the element law.
- Find x∞: solve the post-switch equilibrium.
- Find τ: from the normalized ODE or equivalent resistance seen by the storage element when that method is justified.
- Assemble: x=x∞+[x(0+)−x∞]e^(−t/τ).
- Verify: initial value, final value, initial slope, units and energy direction.
- Solve τ dx/dt+x=4 for x(0+)=1.
- For the solution in exercise 1, find x at t=τ and t=3τ.
- A first-order state starts at 8 and tends to 2 with τ=5 ms. Write x(t) and its initial slope.
- How long until the error is below 1% of its initial value?
- How long until the error is below 0.1%?
- Why does x(0+) come from continuity but x∞ comes from the post-switch equilibrium?
- For dx/dt−200x=0, is the response stable? What is the exponential time scale?
- Why is a separate initial-slope check valuable even after x(0+) and x∞ are correct?
Solutions
1. x∞=4, so x(t)=4−3e^(−t/τ).
2. x(τ)=4−3/e≈2.896; x(3τ)=4−3e⁻³≈3.851.
3. x=2+6e^(−t/5ms); dx/dt|0+=−6/(5 ms)=−1200 units/s.
4. e^(−t/τ)<0.01 ⇒ t>τ ln100≈4.605τ.
5. t>τ ln1000≈6.908τ.
6. The storage law constrains the instantaneous state across switching; the final value belongs to the equilibrium of the new topology.
7. x=Ae^(+200t), so it grows and is unstable. The magnitude time scale is 1/200=5 ms but the exponent has the wrong sign for decay.
8. The slope tests whether the waveform initially moves in the physically correct direction and checks the differential law independently of endpoint matching.
Chapter 03 will now specialize this framework to RC circuits and show that τ emerges from the resistance seen by the capacitor times C, while retaining the topology/continuity/final-value discipline developed here.