IB Maths AI HL
Coupled & Second Order Differential Equations
Paper 1 & 2
~6 min read
Equilibrium Points
An equilibrium point is where the system stops moving: both dxdt = 0 and dydt = 0. A trajectory starting there never leaves. Whether nearby trajectories settle towards it (stable) or escape (unstable) is decided entirely by the eigenvalues of M.
π What you need to know
- Definition: a point (x, y) where both dxdt = 0 and dydt = 0.
- Find it: solve the two rate equations simultaneously (set each to 0).
- Origin always works for the homogeneous system αΊ = Mx β (0, 0) is always an equilibrium.
- Stable: every nearby trajectory stays close to the point.
- Unstable: at least one nearby trajectory moves away.
- Saddle: an unstable point where some trajectories approach and others flee.
Finding an equilibrium point
Equilibrium condition
dxdt = 0 and dydt = 0 simultaneously
β standard definition for AI HL
π§ “Both rates zero β nothing moves”
At equilibrium neither variable is changing, so a particle placed there stays put forever. To locate it, set both derivatives to zero and solve the pair of equations.
Origin guaranteed: for αΊ = Mx (no constant terms), putting x = y = 0 makes both rates zero, so (0, 0) is always an equilibrium point. Constant terms shift it elsewhere.
Stable, unstable & the eigenvalue test
π€ Why do the eigenvalues decide stability?
Near the origin the solution is built from eΞ»t terms. A negative eigenvalue makes its term shrink towards zero (pulling trajectories in); a positive one makes it grow (pushing them out). For complex eigenvalues the real part plays the same role. So the signs of the eigenvalues’ real parts tell you directly whether the point attracts or repels.
| Eigenvalues of M | Nature of (0, 0) |
|---|
| Both positive, distinct | Unstable |
| Both negative, distinct | Stable |
| One positive, one negative | Unstable (saddle) |
| Complex, positive real parts | Unstable |
| Complex, negative real parts | Stable |
| Complex, zero real parts | Stable (a centre) |
π§ “Negative real parts pull in”
If every eigenvalue (or real part) is negative, the point is stable. If any is positive, it’s unstable. Purely imaginary eigenvalues give a stable centre of closed orbits.
π§ Recipe β locate and classify
- Set both rates to 0 and solve for (x, y).
- Find the eigenvalues of M.
- Check the signs (or real parts).
- Classify using the table: all-negative β stable; any positive β unstable; mixed real β saddle.
- Name it if asked (node, spiral, centre, saddle).
Worked examples
WE 1Show that (3, 4) is an equilibrium point of dxdt = 2x β 3y + 6, dydt = x + y β 7.
Substitute and check both rates are zero.
dxdt = 2(3) β 3(4) + 6 = 0
dydt = 3 + 4 β 7 = 0
both zero at (3, 4), so it is an equilibrium point
WE 2For dxdt = x + 3y, dydt = 2x + 2y (eigenvalues 4 and β1), find the coordinates and nature of the equilibrium point.
No constant terms, so the origin is the equilibrium; classify by the eigenvalues.
x = y = 0 makes both rates 0 β equilibrium at (0, 0)
eigenvalues 4 (> 0) and β1 (< 0): one each sign
(0, 0) is a saddle point β unstable
WE 3Find the equilibrium point of dxdt = 2x + 3y β 8, dydt = 3x β y β 1.
Constant terms shift the equilibrium off the origin β solve simultaneously.
2x + 3y β 8 = 0 and 3x β y β 1 = 0
from the second: y = 3x β 1
2x + 3(3x β 1) β 8 = 0 β 11x β 11 = 0 β x = 1
equilibrium point (1, 2)
WE 4A system has eigenvalues β2 and β5. Classify the equilibrium at the origin.
Both real and negative.
both negative and distinct
stable (a stable node)
WE 5A system has complex eigenvalues 2 Β± 3i. Classify the equilibrium at the origin.
Complex β look at the real part.
real part = 2 > 0
spirals away from the origin
unstable
π‘ Top tips
- Set both rates to zero and solve simultaneously to locate the point.
- Origin is free for αΊ = Mx with no constants.
- All real parts negative β stable; any positive β unstable.
- Mixed real signs β saddle (always unstable).
- Purely imaginary β centre, classed as stable.
- To “show” a point is equilibrium, just substitute and get 0 from both rates.
β Common mistakes
- Assuming the origin when constant terms move the equilibrium elsewhere.
- Calling a saddle stable β it’s always unstable.
- Ignoring the real part of complex eigenvalues.
- Only checking one rate β both must be zero.
- Mislabelling a centre as unstable β zero real part is stable.
Next up β Sketching Solution Trajectories. You can now locate an equilibrium point and judge whether it attracts or repels. The next topic draws the one trajectory a given start point follows: mark the initial point, find the initial direction from dxdt and dydt at t = 0, then steer it using the eigenvector lines and the stability you’ve just classified.
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