Unit 11: Systems of First-Order Linear Equations
Where differential equations and linear algebra become one subject: eigenvalues and eigenvectors as the straight-line solutions of a whole system.
10 topics · 34 questions
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7.1 Converting a Higher-Order Equation into a System
- How it worksHow do you introduce new variables for each derivative to turn y'' + 3y' + 2y = 0 into a first-order system?
- Worked exampleBuild a system of two first-order equations equivalent to y'' - 5y' + 6y = 0.
- Going furtherWhat happens to the size of the resulting system when the original equation has a fourth derivative instead of a second?
- Common mistakeWhy doesn't converting a higher-order equation into a system change how many initial conditions the problem needs?
Your turn1 of 2
Convert y'' + 4y' + 4y = 0 into the system x1'=x2, x2' = -4x1 - 4x2 (with x1=y, x2=y'). If x1(0) = 4 and x2(0) = 2, find x2'(0).7.1 Writing a System in Matrix Form
- How it worksHow do the coefficients of a linear system of first-order equations become the entries of the matrix A?
- Worked exampleFind the matrix form x' = Ax for the system x1' = 3x1 - 2x2, x2' = 4x1 - x2.
- Common mistakeWhy doesn't writing a system in matrix form change anything about what the actual solutions are?
Your turn1 of 2
For the system x1' = 4x1 + 4x2, x2' = 5x1 + 2x2, find the trace of the matrix A (the sum of its diagonal entries, a+d).7.3 Eigenvalues, Eigenvectors, and Straight-Line Solutions
- How it worksHow does substituting x = elambda t v into x' = Ax reduce to the equation Av = lambda v?
- Worked exampleFind the eigenvalues and eigenvectors of the matrix A = [[4, 2], [3, -1]].
- Going furtherWhat happens to the straight-line solution x = elambda t v when the initial condition doesn't lie along any eigenvector?
- Worked exampleFind the straight-line solution of x' = Ax that starts at the eigenvector (1, 3) for an eigenvalue of -2.
Your turn1 of 2
Find the larger eigenvalue of A = [[5, 4], [2, 0]], where lambda = (trace plus or minus √(trace2-4·det))/2, trace=a+d, det=ad-bc.7.5 Solving a System with Real, Distinct Eigenvalues
- Worked exampleSolve x' = Ax for A = [[1, 1], [4, 1]] with initial condition x(0) = (2, -1).
- Worked exampleFind the general solution of the system x1' = 2x1, x2' = -3x1 + x2.
- Common mistakeWhy doesn't the general solution of x' = Ax equal just c1 v1 plus c2 v2 with no exponential factors attached?
Your turn1 of 2
Solve the triangular system x1' = 3x1, x2' = 3x1 + -1x2, with x1(0) = 2 and x2(0) = -2 (p is not equal to r). Find x2(t).7.6 Solving a System with Complex Eigenvalues
- How it worksHow do you extract two real, independent solutions from one complex eigenvalue and its complex eigenvector?
- Worked exampleSolve x' = Ax for A = [[1, -5], [1, -3]], which has complex eigenvalues.
- Going furtherWhat happens to a spiraling solution's direction when the real part of the complex eigenvalues changes sign?
Your turn1 of 2
For A = [[0, 6], [-5, 0]] with complex eigenvalues, find the real part of the eigenvalues: trace/2 = (a+d)/2.7.8 Solving a System with a Repeated Eigenvalue
- How it worksHow does a generalized eigenvector fill in the missing second solution when a repeated eigenvalue has just one eigenvector?
- Worked exampleSolve x' = Ax for A = [[3, -4], [1, -1]], which has a repeated eigenvalue with only one independent eigenvector.
- Common mistakeWhy isn't t times the eigensolution automatically a valid second solution the way it was for a single equation?
Your turn1 of 2
The general solution of x'=Ax for a repeated eigenvalue r = 2 with one eigenvector v=(4,5) is x(t) = c1 ertv + c2 ert(tv+w). Find the coefficient of the t·ert term in the second component, i.e., c2·v2, given c2 = 2.7.7 The Matrix Exponential
- How it worksHow does the matrix exponential eAt, defined by its own power series, package every solution of x' = Ax into one operator?
- Worked exampleFind eAt for the diagonal matrix A = [[2, 0], [0, -3]].
- Going furtherWhat happens to eAt when the matrix A itself is not diagonalizable?
- Common mistakeWhy doesn't eAt just mean raising e to the power of each entry of the matrix A separately?
Your turn1 of 2
For the diagonal matrix A = [[3, 0], [0, -2]], find the (1,1) entry of eAt at t = 1.42.7.7 Fundamental Matrices
- How it worksHow does a fundamental matrix Psi(t) let you write the solution to any initial condition as Psi(t) Psi(0) inverse x0?
- Worked exampleBuild a fundamental matrix for the system with eigenvalues 1 and -2 and eigenvectors (1,1) and (1,-2).
- Going furtherWhat happens to a fundamental matrix's determinant, the system's Wronskian, as t changes?
- Ties togetherHow does a fundamental matrix for a system connect to the Wronskian you'd compute for a single higher-order equation?
Your turn1 of 2
Build a fundamental matrix Psi(t) for the system with eigenvalues 2 and -2 and eigenvectors (1,1) and (1,-1) — i.e., Psi(t) = [[elam1 t, elam2 t], [elam1 t, -elam2 t]]. Find det(Psi(t)) at t = 1.139.7.9 Nonhomogeneous Linear Systems
- How it worksHow does the formula x = Psi(t) times the integral of Psi(t) inverse g(t) dt extend variation of parameters to systems?
- Worked exampleSolve the nonhomogeneous system x1' = x2, x2' = -x1 + sec(t) using variation of parameters for systems.
- Going furtherWhat happens to the variation of parameters formula for a system when the forcing vector g(t) is constant?
Your turn1 of 2
For the nonhomogeneous system x' = Ax + g with A = [[0, 4], [4, 0]] and constant g = (5, 2), find the first component of the equilibrium (particular constant) solution x* = -A−1g.7.5 Why Eigenvalues Alone Don't Determine the Full Solution
- Common mistakeWhy isn't a system's solution fully pinned down until you also fix the eigenvectors and an initial condition?
- Worked exampleShow how two different initial conditions produce two different trajectories for a 2x2 system with eigenvalues 2 and -1.
- Going furtherWhat happens to a trajectory's shape when the initial condition lines up exactly with one eigenvector instead of a mix of both?
Your turn1 of 2
A 2x2 system has eigenvalues 2 and -2 with eigenvectors (1,1) and (1,-1). For the initial condition x(0)=(5,2), find c1 in x(0)=c1(1,1)+c2(1,-1), i.e., c1=(x10+x20)/2.