When the matrix has two distinct real eigenvalues, every solution is C₁·e^(λ₁t)·v₁ + C₂·e^(λ₂t)·v₂. The constants come from a two-by-two system at t = 0, and the exponents decide which term the curve follows as t grows. A start that is on neither eigenline picks up both.
Core ideas
Three ideas. A sum of solutions is a solution, the constants are fixed by where you start, and the exponents say where you end up.
If x₁ and x₂ solve x′ = A·x, so does C₁·x₁ + C₂·x₂ for any numbers C₁ and C₂, because both sides of the equation are linear in x. With the two straight-line solutions this gives a two-parameter family, and when the eigenvectors point in different directions the family contains a solution through every point.
At t = 0 every exponential is 1, so the general solution is C₁·v₁ + C₂·v₂. Setting that equal to the start (x₀, y₀) is a pair of linear equations in C₁ and C₂. The eigenvectors are the columns of its matrix, and a pair of different directions makes it solvable exactly.
As t grows, e^(λ₁t) and e^(λ₂t) separate, and the term with the larger eigenvalue dwarfs the other. The curve bends to run parallel to that eigenline. Running time backwards swaps the roles: the term with the smaller eigenvalue wins as t → −∞.
The material
Why the sum is a solution, how to fit it in fractions, what each term does, and the place the lab stops.
If x₁ and x₂ solve x′ = A·x, then so does C₁·x₁ + C₂·x₂ for every pair of constants.
The derivative of the sum is C₁·x₁′ + C₂·x₂′, because constants come out of derivatives.
Each xᵢ′ equals A·xᵢ, so the derivative is C₁·A·x₁ + C₂·A·x₂ = A·(C₁·x₁ + C₂·x₂). That is A times the sum, which is the equation again.
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For distinct real eigenvalues λ₁, λ₂ with eigenvectors v₁, v₂, the general solution of x′ = A·x is x(t) = C₁·e^(λ₁t)·v₁ + C₂·e^(λ₂t)·v₂.
The constants C₁ and C₂ are fixed by a start, and the solution through a given start is unique. The lab prints the formula with the eigenvectors substituted in.
For rows 2 1 and 1 2 the eigenpairs are (1, (1, −1)) and (3, (1, 1)). To start at (1, 0) solve C₁·(1, −1) + C₂·(1, 1) = (1, 0).
That is C₁ + C₂ = 1 and −C₁ + C₂ = 0. Adding gives 2C₂ = 1, so C₂ = 1/2 and C₁ = 1/2. The solution is (1/2)·e^(t)·(1, −1) + (1/2)·e^(3t)·(1, 1).
The check is exact, and it is the one to keep doing. The start matches at t = 0, and the first equation holds at every t, because the two sides simplify to the same expression. Doing only the first check would miss a wrong exponent; doing only the second would miss a wrong constant.
In the solution above the exponent 3 beats the exponent 1. For large t the first term is a thin correction on the second, so the curve points almost exactly along (1, 1); its distance from the diagonal, which the first term sets, still grows with e^(t), only far more slowly than the curve itself runs out. Going the other way, toward negative t, both terms shrink and the first is the bigger, so the curve comes out of the origin tangent to the line through (1, −1). That is the shape of every node: orbits hug the slow eigenline at the origin and run parallel to the fast one far away. A saddle is different, because its exponents have opposite signs: there the orbit closes in on one eigenline as t → ∞ and on the other as t → −∞, which the second preset shows.
A start on an eigenline has the other constant zero and never bends. A start on neither has both. For the second preset, with rows 1 2 and 3 0 and the eigenpairs (−2, (2, −3)) and (3, (1, 1)), the start (3, 0) lies on neither line, and 2C₁ + C₂ = 3, −3C₁ + C₂ = 0 gives C₁ = 3/5 and C₂ = 9/5. The nearer eigenline to (3, 0) is the diagonal, and the solution is still not the diagonal solution: the constant C₁ is not zero.
For rows −1 0 and 0 −2 the eigenvalues are −1 and −2 with the axes as eigenvectors. From (2, 2) the constants are 2 and 2, and the solution is (2e^(−t), 2e^(−2t)).
Both terms die away. The one with exponent −1 dies more slowly, so it wins at the end, and the orbit comes into the origin tangent to the x-axis.
The lab fits the constants only when its arithmetic stays rational. For eigenvalues that are surds the constants would carry square roots inside the formula, and the lab prints a dash rather than a rounded guess. When the eigenvalues are complex or repeated the general solution has a different shape, and the next lesson shows what the picture does in each case.
Interactive
Trace, determinant, eigenvalues and exact Euler steps
Method
Eigenpairs first, the system for the constants second, and the long run last.
Solve λ² − τ·λ + Δ = 0 and one equation (A − λ·I)·v = 0 for each root. Keep the order, and keep each vector with its own eigenvalue.
C₁·e^(λ₁t)·v₁ + C₂·e^(λ₂t)·v₂. Do not put in the start yet.
Every exponential is 1, so C₁·v₁ + C₂·v₂ = (x₀, y₀). That is two equations, one for each coordinate.
Eliminate one constant and solve for the other, in fractions. Substitute back into both equations to check the start is matched.
Compare the exponents. The term with the larger eigenvalue wins as t → ∞ and the smaller as t → −∞. A constant that is zero removes its term altogether.
The first preset. The eigenpairs from the last lesson are λ = 1 with (1, −1) and λ = 3 with (1, 1).
The lab prints the constants as C₁ = 1/2, C₂ = 1/2, and the formula in the general-solution tile has the same two terms. The start (1, 0) is on neither eigenline, which is why both constants are non-zero.
For large t the diagonal term is far the larger, and the orbit runs out almost along the line through (1, 1). It reaches that direction only in the limit. At every finite t the small term is still bending the orbit away from the line.
You should be able to write C₁·e^(λ₁t)·v₁ + C₂·e^(λ₂t)·v₂, set t = 0, solve for the constants in exact fractions, check both the start and the equation, and say which term dominates at each end of time.
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