Maths › Differential equations › Modelling with differential equations
Modelling with differential equations
Springs, dampers and linked populations are described by the same kind of equation. The roots of the auxiliary equation determine whether the behaviour is oscillation, damping or decay.
Builds on Second order equations and Forces and Newton's laws.
IN THIS TOPIC
- Recognise simple harmonic motion from its differential equation and state period and amplitude.
- Model a damped oscillator and interpret each term physically.
- Classify light, critical and heavy damping via the discriminant.
- Reduce a coupled pair of first order equations to one second order equation.
COMMON MISCONCEPTION
Adding any amount of damping to an oscillator removes the oscillation entirely.
Simple harmonic motion
Strip out the resistance and a mass on a spring obeys ẍ = −ω²x, acceleration proportional to displacement and pointing back towards the centre. The auxiliary equation m² + ω² = 0 has roots ±ωi, so x = A cos ωt + B sin ωt. Written in harmonic form that is R cos(ωt − α), which shows the amplitude R = √(A² + B²) and the period 2π/ω at a glance. The initial displacement and velocity fix A and B, and with them the amplitude; only the period 2π/ω is independent of where the mass started, which is the property that makes a pendulum keep time, provided its swings stay small enough for the small-angle model behind ẍ = −ω²x.
Damping in three strengths
Add resistance and Newton's second law gives y'' + by' + cy = 0, where the y term pulls back and the y' term drains energy. The discriminant of the auxiliary equation sorts the outcomes. b² − 4c < 0 is light damping, oscillation inside a shrinking envelope: resistance reshapes the motion before it stops it. b² − 4c = 0 is critical damping, the fastest non-oscillatory return this standard model allows, though for some starting conditions the solution still crosses equilibrium once on its way home. Larger b gives heavy damping, a slow creep home.
WORKED EXAMPLE
A lightly damped oscillator
Solve y'' + 2y' + 5y = 0 with y(0) = 1, y'(0) = −1.
The auxiliary equation m² + 2m + 5 = 0 has roots m = −1 ± 2i.
So y = e−t(A cos 2t + B sin 2t).
y(0) = 1 gives A = 1. Differentiating and setting y'(0) = −1 gives B = 0.
y = e−tcos 2t. Frequency 2 from the imaginary part, decay rate 1 from the real part.
Push the system with an external force and the equation gains a right-hand side, y'' + by' + cy = f(t). The complementary function is the transient, which damping removes, and the particular integral is the steady state that survives. Questions phrased as 'describe the long-term behaviour' are asking you to say exactly that.
Coupled systems
Predator and prey, or two connected tanks, arrive as a pair. dx/dt involves y and dy/dt involves x. Differentiate one equation, substitute the other, and the pair collapses to a single second order equation in one variable. Solve it, then recover the second variable from the first equation.
WORKED EXAMPLE
Collapsing a coupled pair
Solve dx/dt = y and dy/dt = −x with x(0) = 1, y(0) = 0.
Differentiating the first gives x'' = dy/dt = −x, so x'' + x = 0.
Auxiliary m² + 1 = 0 gives x = A cos t + B sin t.
x(0) = 1 gives A = 1. Since y = x' = −A sin t + B cos t, the condition y(0) = 0 gives B = 0.
x = cos t, y = −sin t. The pair orbit the unit circle clockwise, for ever.
INDEPENDENT PRACTICE
Reading the damping
A shock absorber obeys y'' + 6y' + 9y = 0. Classify the damping and give the general solution.
Show the working
The discriminant is 36 − 36 = 0, so the damping is critical.
Repeated root m = −3, so y = (A + Bt)e−3t.
Critical damping is the design target for car suspension, the quickest settle without oscillation.
ASSESSMENT FOCUS
- Interpret constants in context. The y' coefficient is resistance and the y coefficient is stiffness.
- Quote the discriminant when classifying damping. The word on its own is not a justification.
- For SHM, read ω from the equation and give the period as 2π/ω without solving anything.
- In coupled systems, state which equation you differentiate and where you substitute.
- Recover the second variable from a first order equation, never by integrating from scratch.
CHECK YOURSELF
Classify the damping in y'' + 6y' + 9y = 0 and state the long-term behaviour of any solution.
Show a hint
Compute b² − 4c and look for a repeated root.
Show the answer
36 − 36 = 0, so the damping is critical. The repeated root m = −3 gives y = (A + Bt)e−3t, and every solution returns to zero without oscillating, as fast as the system allows.
SHM is ẍ = −ω²x, with period 2π/ω and amplitude fixed by the starting conditions.
Damping is read from b² − 4c: negative oscillates in an envelope, zero settles fastest, positive creeps.
Collapse coupled pairs by differentiating one equation and substituting the other.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the modelling with differential equations questions page.
CHECK YOUR PROGRESS
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- Recognise simple harmonic motion from its differential equation and state period and amplitude.
- Model a damped oscillator and interpret each term physically.
- Classify light, critical and heavy damping via the discriminant.
- Reduce a coupled pair of first order equations to one second order equation.
Open the full revision checklist to see every objective in the course in one place.