Maths › Differential equations › Modelling with differential equations
Modelling with differential equations
Springs, dampers and linked populations all speak the same language: read the roots of an auxiliary equation and the physics falls out as oscillation, damping or decay.
Builds on Second order equations and Forces and Newton's laws.
IN THIS TOPIC
- Model a damped oscillator with y'' + by' + cy = 0 and interpret each term.
- Classify light, critical and heavy damping via the discriminant b² − 4c.
- Reduce a coupled pair of first order equations to one second order equation.
WHAT YOU PROBABLY THINK
Adding any amount of damping to an oscillator kills the oscillation outright.
Damping in three strengths
For a mass on a spring with resistance, Newton's second law gives y'' + by' + cy = 0: the y term pulls back, 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, so the opening claim overstates its case. b² − 4c = 0 is critical damping, the fastest return with no overshoot; larger b is heavy damping, a slow creep home.
WORKED EXAMPLE
A lightly damped oscillator
Solve y'' + 2y' + 5y = 0 with y(0) = 1, y'(0) = −1.
Auxiliary: m² + 2m + 5 = 0, so m = −1 ± 2i.
General solution: 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.
Coupled systems
Predator and prey, or two connected tanks, arrive as a pair: dx/dt involving y, dy/dt involving x. Differentiate one equation and 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.
Differentiate the first: x'' = dy/dt = −x, so x'' + x = 0.
Auxiliary m² + 1 = 0: x = A cos t + B sin t.
x(0) = 1 gives A = 1; y = x' = −sin t at t = 0 gives B = 0.
x = cos t, y = −sin t: the pair orbit the unit circle clockwise, for ever.
TRY IT UNSEEN
Reading the damping
A shock absorber obeys y'' + 6y' + 9y = 0. Classify the damping and give the general solution.
Show the working
Discriminant: 36 − 36 = 0, so the damping is critical.
Repeated root m = −3: y = (A + Bt)e−3t.
Critical damping is the design target for car suspension: the quickest settle with no bounce.
THE EXAM BIT
- Interpret constants in context: the y' coefficient is resistance, the y coefficient stiffness.
- Quote the discriminant when classifying damping; the word alone does not earn the mark.
- 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
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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.
CHECK YOUR PROGRESS
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- Model a damped oscillator with y'' + by' + cy = 0 and interpret each term.
- Classify light, critical and heavy damping via the discriminant b² − 4c.
- 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.
No animated video for this topic yet; these notes stand alone.