Use eigenvectors to organize coupled rates

Try

A second-order equation can become two first-order equations by tracking position y and velocity v=y′. For y″+3y′+2y=0, write y′=v and v′=−2y−3v. The state is the ordered pair (y,v). Two initial measurements now specify a single point in a two-dimensional state plane.

This change of representation does not add a physical assumption. It organizes the same information so matrix methods and phase portraits can describe coupled evolution.

Predict

For a constant matrix A, try a vector solution x(t)=e^(λt)v, where v is a fixed nonzero vector. Predict the condition on A,λ,v that makes this a solution of x′=Ax. Then ask why knowing only λ is insufficient to draw the direction of the corresponding trajectory in the state plane.

Experiment

Consider x′=2x+y and y′=x+2y. The matrix A has rows (2,1) and (1,2). Multiplication gives A(1,1)=3(1,1) and A(1,−1)=(1,−1). Thus the eigenpairs are λ₁=3,v₁=(1,1) and λ₂=1,v₂=(1,−1).

Initial-state decomposition	First coordinate	Second coordinate
c₁v₁+c₂v₂	c₁+c₂	c₁−c₂
Required state	4	2
Solved coefficients	c₁=3	c₂=1

The solution is 3e^(3t)(1,1)+e^t(1,−1). Check both coordinates at zero, then differentiate and multiply by A independently. As time grows, the first mode dominates because its exponential grows faster and its coefficient is nonzero.

Understand

Substitution of x=e^(λt)v gives λe^(λt)v=e^(λt)Av. Canceling the nonzero exponential yields Av=λv, the eigenvector equation. An eigenvector supplies a state-space direction preserved by the matrix; its eigenvalue supplies the growth or decay rate along that direction.

For a two-dimensional matrix with two independent real eigenvectors, the general solution is c₁e^(λ₁t)v₁+c₂e^(λ₂t)v₂. Distinct real eigenvalues guarantee independence. Initial data determine c₁,c₂ by a linear system. A repeated eigenvalue does not automatically supply two independent eigenvectors, so this formula cannot be assumed without checking the basis.

For a diagonal system x′=2x,y′=−y with initial state (3,4), the solution is (3e^(2t),4e^(−t)). The initial x rate is six, while the initial y rate is negative four. One coordinate grows and the other decays. The origin is unstable despite decay in one direction, because arbitrarily small components in the growing direction eventually move away.

Phase-plane axes label state variables, not time. A curve in that plane records how the pair changes; a separate time plot records each coordinate against t. Mixing these representations can make a correct exponential solution look like an incorrect straight-line time graph.

Apply

Draw both eigenvector lines for the coupled example and decompose the initial point into them. Then sketch the evolving contributions at two later times using the same coordinate scale. Explain why an initial state exactly on the slower eigenvector stays on it, whereas a nearby state with a faster-mode component eventually turns toward the faster direction.

For the position-velocity system at the start, recover the characteristic polynomial of the original second-order equation from its matrix. This is an independent consistency check between scalar and vector descriptions, not a separate model.

Continue

The next lesson extends phase portraits to spirals, saddles, centers, and matrices missing an eigenvector. Qualitative classification must account for both eigenvalue signs and the available eigenvector structure.

Use fundamental matrices and coordinate changes to normalize a full solution operator and preserve the physical state across a change of basis.