IB Maths AI HLCoupled & Second Order Differential EquationsPaper 1 & 2~6 min read
Sketching Solution Trajectories
A phase portrait shows typical trajectories; given an initial condition there’s just one. To sketch it, mark the start point, work out which way it sets off from the rates at t = 0, then steer using the eigenvector lines and the system’s overall shape.
š What you need to know
One start, one trajectory: the initial condition picks a single curve from the family.
Mark the start: it’s given, or substitute t = 0 into the exact solution.
Initial direction: find dxdt and dydt at t = 0 ā sign of dxdt gives left/right, sign of dydt gives up/down.
Use the eigenvectors: draw the eigenvector lines (real eigenvalues) and shape the curve around them.
Larger eigenvalue dominates as t ā ā ā the trajectory ends up parallel to its line.
Sketch only: capture the start, the direction, and the limiting behaviour.
Getting the initial direction
š§ “Start point, first step, then the flow”
Plot the start point, take the first step in the direction (dxdt, dydt) at t = 0, then follow the flow set by the eigenvalues.
Reading the first step: evaluate both rates at the start. dxdt > 0 means moving right, < 0 left; dydt > 0 means up, < 0 down. Together they give the initial direction vector.
š§ Recipe ā sketch a solution trajectory
Eigen-data: find (or read off) the eigenvalues and eigenvectors of M.
Start point: given, or substitute t = 0 into the exact solution.
Initial direction: compute dxdt and dydt at the start.
Draw the eigenvector lines (real eigenvalues) for reference.
Sketch the curve from the start, in that direction, becoming parallel to the larger eigenvalue’s line as it moves away.
Worked example
For dxdt = x ā 5y, dydt = ā3x + 3y, the exact solution is x = e6t(ā1, 1) ā 2eā2t(5, 3). Sketch the trajectory as t increases from 0.
Solution trajectory from (ā11, ā5)
Eigenvalues 6 and ā2 (a saddle). The curve starts at (ā11, ā5) heading up and to the right (direction (14, 18)), then bends to run parallel to the y = āx line ā the eigenvector for the positive eigenvalue ā as it moves away.
WE 1
Find the start point of the trajectory.
Substitute t = 0 into the exact solution.
x = (ā1, 1) ā 2(5, 3) = (ā1 ā 10, 1 ā 6)start point (ā11, ā5)
WE 2
Find the initial direction of the trajectory.
Evaluate both rates at the start point.
dxdt = (ā11) ā 5(ā5) = 14dydt = ā3(ā11) + 3(ā5) = 18direction (14, 18) ā up and to the right
WE 3
Which eigenvector line does the trajectory approach as t ā ā?
The larger eigenvalue dominates as t grows.
eigenvalues 6 and ā2; larger is 6eigenvector (ā1, 1) ā line y = āxparallel to y = āx as t ā ā
WE 4
A trajectory starts at (2, ā3) in a system with dxdt = y, dydt = āx. Find the initial direction.
Substitute the start point into both rates.
dxdt = y = ā3 (left)dydt = āx = ā2 (down)direction (ā3, ā2) ā down and to the left
WE 5
For the worked-example system, write the equations of both eigenvector lines.
Each line is through the origin parallel to its eigenvector.
(ā1, 1) ā gradient ā1 ā y = āx(5, 3) ā gradient ā ā y = ā xy = āx and y = ā x
š” Top tips
Start point from t = 0 ā substitute into the exact solution if not given.
Initial direction = (dxdt, dydt) at the start.
Draw the eigenvector lines for reference when eigenvalues are real.
End parallel to the larger eigenvalue’s line as t ā ā.
Don’t cross an eigenvector line.
Sketch, not plot ā show start, direction, and limiting behaviour.
ā Common mistakes
Wrong start point ā forgetting e0 = 1 when substituting t = 0.
Direction sign slips in dxdt or dydt.
Approaching the wrong line ā it’s the larger eigenvalue’s as t ā ā.
Crossing an eigenvector line.
Over-drawing ā an accurate plot isn’t expected, just the shape.
Next up ā Second Order Differential Equations. You’ve now handled coupled systems from every angle ā exact solutions, phase portraits, equilibrium points, and trajectories. The final topic connects back to them: a second order equation like dĀ²xdtĀ² + adxdt + bx = 0 becomes a coupled first-order system through the substitution y = dxdt, so everything you’ve built here applies straight away.
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