Naming the rate v = x′ turns one equation of second order into two of first order, x′ = v and v′ = −(c/a)·x − (b/a)·v. The grid of four numbers on the right has a trace −b/a and a determinant c/a, and the quadratic built from them is the characteristic equation divided by a. The same roots appear, so the same motion does.
Core ideas
Three ideas: the new unknown, what its equation looks like as a matrix, and why nothing about the motion is new.
Write v for x′. Then x′ = v is true by the choice of name, and v′ = x″ can be replaced using the original equation. The result is a pair of first-order equations in the two unknowns x and v, which a later course studies in the plane of (x, v).
The right-hand sides are 0·x + 1·v and −(c/a)·x − (b/a)·v, so the four coefficients form a grid, the matrix A. Its trace is the sum of the diagonal entries, 0 + (−b/a) = −b/a, and its determinant is 0·(−b/a) − 1·(−c/a) = c/a.
Given the start x(0) and the start v(0) = x′(0) the system and the equation have the same solution: the first component of the system is the function the equation asks for, and the second is its rate. The two initial values of the equation are the two components of the system's starting point.
The material
The substitution, the matrix and its trace and determinant, a check that the system and the equation agree, and what the lab's last tile names.
From here the unknown is called x, because the system treats position and velocity as a pair, and the equation under study is a·x″ + b·x′ + c·x = 0 with a not zero. Divide by a, and solve for the second rate.
The first line is the original equation, solved for its highest rate. The last two lines are a system of two first-order equations; the second is the first line with v written in place of x′. Nothing has been lost, since v is the rate of x by definition. The system is written as a matrix with one row for each equation; the words trace and determinant are those of Algebra's Systems and Matrices, and the two formulas in the claim below are all this lesson needs of them.
The matrix A has trace τ = −b/a and determinant Δ = c/a, and the equation λ² − τ·λ + Δ = 0 is the characteristic equation a·r² + b·r + c = 0 divided by a.
The trace is 0 + (−b/a) and the determinant is 0·(−b/a) − 1·(−c/a), which are −b/a and c/a.
Then λ² − τ·λ + Δ = λ² + (b/a)·λ + c/a. Multiplying by a gives a·λ² + b·λ + c, which is the characteristic quadratic with λ in place of r. The eigenvalues of A, the numbers a later course defines as the roots of this quadratic, are the roots r of the equation.
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The first preset is x″ + 3x′ + 2x = 0, and the solution of the equation from the earlier lesson with x(0) = 1 and x′(0) = 0 is x = 2·e^(−t) − e^(−2t). Its rate is v = −2·e^(−t) + 2·e^(−2t). Both equations of the system hold exactly.
The pair (x, v) is a solution of the system, and its first component is the solution of the equation. The pair also starts at (x(0), v(0)) = (1, 0), which is the two initial values of the equation written as a point.
The lab takes the matrix as its input and prints the trace, the determinant, the eigenvalues and a name for the picture the system makes in the plane. The matrix for x″ + 3x′ + 2x = 0 has τ = −3 and Δ = 2, and its eigenvalues are −2 and −1. The matrix for x″ + 4x = 0 has τ = 0 and Δ = 4, eigenvalues ±2i. The matrix for x″ + 2x′ + 5x = 0 has τ = −2 and Δ = 5, eigenvalues −1 ± 2i. These are the roots from the earlier lessons, one for one.
The name the lab gives the picture follows the roots: negative real roots, imaginary roots and complex roots with a negative real part are named a stable node, a centre and a stable spiral. What the names mean is the subject of Systems and the Phase Plane; here they are a label on roots already understood.
The lab prints six more tiles that belong to that course: eigenvectors, a general solution written with them, the constants fitted to the start, and two Euler tiles that the next lesson reads. One of them is worth a look now. For the first preset the constants tile says C₁ = −1, C₂ = 2, while “Real Distinct Roots” fitted C₁ = 2, C₂ = −1 to the same start. The two labs number the roots in opposite orders, this one from the smaller eigenvalue up, so here C₁ = −1 goes with e^(−2t), and the first component is 2·e^(−t) − e^(−2t) either way.
Interactive
Trace, determinant, eigenvalues and exact Euler steps
Method
Five moves. The one that is easy to skip is the division by a.
Solve for the highest rate: x″ = −(c/a)·x − (b/a)·x′. The rate x″ must stand alone, with coefficient 1.
Write v = x′, and x′ = v is the first equation of the system.
The second equation is v′ = −(c/a)·x − (b/a)·v. Its coefficients of x and v are the bottom row of the matrix.
τ = −b/a and Δ = c/a. They are the sum of the diagonal entries and the diagonal product minus the cross product.
They are the same quadratic up to the factor a. Compute the roots once and check them against the lab's eigenvalue tile.
From the equation to the matrix, and from the matrix back to the quadratic.
Here a = 1, so dividing by a changes nothing. With 2x″ + 8x′ + 6x = 0, the bottom row would be −3 and −4, not −6 and −8.
The roots are those of the earlier lessons, and so are the solutions: the first component of every solution of the system is a solution of the equation.
You should be able to divide by the leading coefficient, name v = x′, write the matrix, compute its trace and determinant, and show that λ² − τ·λ + Δ = 0 is the characteristic equation divided by a.
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