Mr. Integral
Math notes/Differential Equations/First-Order Methods

Cooling, Mixing and RC Circuits

2 MIN READ · PRACTISE AT YOUR OWN PACE   ·   FREE

LESSON 6 OF 6See the unit map ↗

Build first-order models from a balance law and interpret their time constant.

Builds on Equilibria, Stability and Logistic Growth

The bigger question: Which structure makes a first-order equation solvable?

On this page

The idea

Many engineering models follow accumulation = input − output. State what is well mixed, constant or proportional before writing the equation. The assumptions determine whether the resulting model is linear and whether its coefficients are constant.

Visual guide

VISUAL GUIDEOne time constant completes about 63% of a step response
A normalized charging or filling response is 1 − e⁻ᵗ⁄ᵗᵃᵘ. At t/τ = 1 it reaches 1 − e⁻¹ ≈ 0.632. The remaining distance to equilibrium, rather than the whole response, decays exponentially.001.250.32.50.63.750.951.2t / τfraction
  • Normalized response
  • Equilibrium
A normalized charging or filling response is 1 − e⁻ᵗ⁄ᵗᵃᵘ. At t/τ = 1 it reaches 1 − e⁻¹ ≈ 0.632. The remaining distance to equilibrium, rather than the whole response, decays exponentially.

Method and assumptions

A stable constant-coefficient model x′+(1/τ)x=bx'+(1/\tau)x=b has equilibrium x∗=bτx_*=b\tau and solution x=x∗+(x0−x∗)e−t/τx=x_*+(x_0-x_*)e^{-t/\tau}. After one time constant, the deviation retains the fraction e−1e^{-1}, about 37%37\%. This is a statement about deviation from equilibrium, not always the measured value.

Worked example: a mixing tank

A well-mixed 100100 L tank receives 22 L/min of salt solution at 33 g/L and drains at the same rate. If mm is salt mass in grams, m′=6−(2/100)mm'=6-(2/100)m. Starting fresh, m=300(1−e−t/50)m=300(1-e^{-t/50}). Volume stays constant because inflow equals outflow.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

In y′=−(y−y∞)/τ with τ>0, what fraction of the initial deviation remains after time τ?

Hint 1 · Find a starting point

The deviation follows exponential decay.

Hint 2 · Take the next step

It is multiplied by e⁻ᵗ/τ after time t.

Show the reasoning

Answer: e⁻¹

At t=τ the remaining fraction is e⁻¹≈0.368; one time constant does not mean the transient is gone.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: charging a capacitor

Kirchhoff’s voltage law gives RCv′+v=VRCv'+v=V for a series resistor and capacitor under a constant source. With v(0)=0v(0)=0, v=V(1−e−t/(RC))v=V(1-e^{-t/(RC)}). For RC=2RC=2 s, v(2)=V(1−e−1)v(2)=V(1-e^{-1}), about 0.632V0.632V.

Interpreting the result

Newton cooling uses the same shape with temperature difference from a fixed ambient value. If volume, ambient temperature or input varies, revise the coefficients or forcing instead of reusing a constant-parameter formula.

Practice

  1. What is the time constant of x′+5x=10x'+5x=10?
  2. What is its equilibrium?
  3. How much initial deviation remains after three time constants?
Show worked solutions
  1. τ=1/5\tau=1/5 in the time unit used.
  2. x∗=10/5=2x_*=10/5=2.
  3. The fraction is e−3≈0.050e^{-3}\approx0.050, about 5%5\%.

Further study

MIT OpenCourseWare: Differential Equations provides a full university course with additional lectures and exercises.

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

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