Differential Equations in Engineering

Almost every law of engineering says how something changes: current changes the charge on a capacitor, temperature difference drives the rate of cooling, and force sets the rate of change of momentum. Writing those laws down produces differential equations, and solving them lets engineers predict how a system behaves over time. This article shows how the most common types arise and how to solve them, with realistic problems worked in full.

Why Engineers Love Differential Equations

A differential equation relates an unknown function to its derivatives. In engineering the independent variable is usually time \( t \), and the solution tells us the transient (how the system settles) and the steady state (where it ends up).

Common examples include:

• Electrical: voltage across RC and RL circuits, and RLC oscillations.
• Thermal: heating and cooling of components.
• Mechanical: vibrations of beams and suspensions.
• Chemical: concentration in a stirred tank reactor.
• Structural: beam deflection, \( EI\,\dfrac{d^4 y}{dx^4} = w(x) \).

First-Order Linear Equations

Many systems follow the form

$$ \tau \frac{dy}{dt} + y = y_\infty $$

where \( \tau \) is the time constant and \( y_\infty \) the final value. Separating variables gives the universal solution

$$ y(t) = y_\infty + \left(y_0 - y_\infty\right)e^{-t/\tau} $$

After one time constant the system has completed about 63.2% of its change, and after five it is within 1% of steady state. Engineers use this "five tau" rule constantly.

Solved Problem 1: Charging a Capacitor

A 100 μF capacitor, initially uncharged, is connected through a 10 kΩ resistor to a 12 V battery. Find the time constant, the capacitor voltage after 2 s and the time to reach 11 V.

Step 1: Derive the equation. Kirchhoff's voltage law around the loop gives \( V_s = iR + v_C \), and the capacitor current is \( i = C\,dv_C/dt \):

$$ RC\frac{dv_C}{dt} + v_C = V_s $$

This is exactly the standard first-order form with \( \tau = RC \) and \( y_\infty = V_s \).

Step 2: Time constant.

$$ \tau = RC = (10\,000\ \Omega)(100 \times 10^{-6}\ \text{F}) = 1.0\ \text{s} $$

(Ohms multiplied by farads give seconds.)

Step 3: Solution. With \( v_C(0) = 0 \):

$$ v_C(t) = 12\left(1 - e^{-t/1.0}\right)\ \text{V} $$

Step 4: Voltage at 2 s.

$$ v_C(2) = 12\left(1 - e^{-2}\right) = 12(1 - 0.1353) = 10.38\ \text{V} $$

Step 5: Time to reach 11 V.

$$ 11 = 12\left(1 - e^{-t}\right) \Rightarrow e^{-t} = \frac{1}{12} \Rightarrow t = \ln 12 = 2.48\ \text{s} $$

Sanity check. After two time constants we expect about 86.5% of 12 V, which is 10.38 V. The voltage approaches but never exceeds 12 V, and the 11 V point (91.7%) sensibly lies between two and three time constants.

Solved Problem 2: Cooling a Steel Forging

A steel forging at 900 °C is left to cool in a workshop at 30 °C. After 10 minutes it is at 500 °C. Using Newton's law of cooling, find how long it takes to reach 100 °C, safe for handling with gloves.

Step 1: Set up the model. Newton's law states that the cooling rate is proportional to the temperature excess:

$$ \frac{dT}{dt} = -k\left(T - T_a\right) $$

Step 2: Solve by separating variables.

$$ \int \frac{dT}{T - T_a} = -k\int dt \quad \Rightarrow \quad T(t) = T_a + (T_0 - T_a)e^{-kt} $$

So \( T(t) = 30 + 870e^{-kt} \) with \( t \) in minutes.

Step 3: Find \( k \) from the data. At \( t = 10 \) min, \( T = 500 \) °C:

$$ 470 = 870e^{-10k} \Rightarrow k = \frac{1}{10}\ln\frac{870}{470} = \frac{0.6158}{10} = 0.0616\ \text{min}^{-1} $$

Step 4: Time to reach 100 °C.

$$ 70 = 870e^{-kt} \Rightarrow t = \frac{1}{k}\ln\frac{870}{70} = \frac{2.520}{0.06158} = 40.9\ \text{min} $$

Sanity check. The time constant is \( 1/k \approx 16 \) minutes, and the excess must fall from 870 to 70, a factor of about 12.4, which takes \( \ln 12.4 \approx 2.5 \) time constants, or about 41 minutes. In practice, radiation makes the hot early stage faster than this simple model predicts, so 41 minutes is a reasonable, slightly conservative estimate.

Second-Order Equations: Oscillation

Systems with two energy stores, such as mass-spring or inductor-capacitor, give second-order equations:

$$ a\ddot{x} + b\dot{x} + cx = 0 $$

Try \( x = e^{\lambda t} \) to get the characteristic equation \( a\lambda^2 + b\lambda + c = 0 \). Its roots decide the behaviour:

• Two distinct real roots: overdamped, no oscillation.
• Repeated real root: critically damped.
• Complex roots \( \alpha \pm i\beta \): decaying oscillation \( e^{\alpha t}(A\cos\beta t + B\sin\beta t) \).

Solved Problem 3: A Damped Oscillation

A suspension model obeys \( \ddot{x} + 4\dot{x} + 13x = 0 \), with \( x \) in centimetres, \( x(0) = 1 \) cm and \( \dot{x}(0) = 0 \). Find \( x(t) \).

Step 1: Characteristic equation.

$$ \lambda^2 + 4\lambda + 13 = 0 \Rightarrow \lambda = \frac{-4 \pm \sqrt{16 - 52}}{2} = -2 \pm 3i $$

Step 2: General solution.

$$ x(t) = e^{-2t}\left(A\cos 3t + B\sin 3t\right) $$

Step 3: Apply \( x(0) = 1 \). This gives \( A = 1 \).

Step 4: Apply \( \dot{x}(0) = 0 \). Differentiating, \( \dot{x}(0) = -2A + 3B = 0 \), so \( B = 2/3 \).

$$ x(t) = e^{-2t}\left(\cos 3t + \tfrac{2}{3}\sin 3t\right)\ \text{cm} $$

Sanity check. At \( t = 0 \) this gives 1 cm, and the \( e^{-2t} \) envelope means the motion dies out within about 2.5 s (five time constants of 0.5 s), which is how a healthy car suspension should behave after a bump.

Did You Know?

• Isaac Newton and Gottfried Leibniz developed calculus independently in the late 17th century, and Newton's cooling law appeared in a paper of 1701.
• The same equation that describes an RLC circuit also describes a mass-spring-damper, which is why early analogue computers solved mechanical problems with electrical circuits.
• Engineers often solve these equations with Laplace transforms, which turn calculus into algebra.

Key Takeaways

• First-order systems follow \( y = y_\infty + (y_0 - y_\infty)e^{-t/\tau} \) and settle in about five time constants.
• Find unknown constants such as \( k \) from one measured data point.
• Second-order systems are solved through the characteristic equation; complex roots mean decaying oscillation.
• Always check initial values and long-term behaviour against physical intuition.

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