![]() ![]() One uses the constant vector c T = (c 1, c 2) to determine the initial point x0 T = (x0 1, x0 2) of a particular solution trajectory, x T(t) = (x 1(t), x 2(t)). Where e f(t) is the exponential of the function f(t). The general solution for given constants c 1 and c 2 is: This matrix has two distinct eigenvalues, µ 1 = -4 and µ 2 = -2, with corresponding linearly independent eigenvectors, v 1 T = (1, 0) and v 2 T = (1, 1). The solution to a system of linear differential equations involves the eigenvalues and eigenvectors of the matrix A. For this linear differential equation system, the origin is a stable node because any trajectory proceeds to the origin over time. The origin is an equilibrium point for any matrix since A * o = o. From any initial point x0, the trajectory of x T(t) = (x 1(t), x 2(t)) converges over time to the origin o T = (0, 0) for this particular matrix A. The pattern that the arrows make in the diagram depends on the values of the coefficients of the matrix A. To obtain a rough trajectory of the solution vector x(t), start at an initial point x0, (t = 0), and follow the arrows as t increases in value. These vectors represented as arrows provide a picture of a vector field in the x 1-x 2 plane generated by the system of linear differential equations. The next diagram shows the normalized vectors of d x(t)/dt at points of x T = (x 1, x 2). If one starts at t = 0 at the initial point x0 T = (x0 1, x0 2), one can follow the trajectory of x(t) as time t proceeds as given by d x T(t)/dt = (dx 1(t)/dt, dx 2(t)/dt). The vector d x/dt begins at x with its direction and magnitude provided by the coefficients of the vector d x/dt. In the following diagram, the values of d x/dt and x are plotted for four sets of values in the x 1-x 2 plane. The values of d x T/dt = (dx 1/dt, dx 2/dt) depend on the matrix A, and on the values of x T = (x 1, x 2). In matrix form, this system of linear differential equations is: d x/dt Where D is a diagonal matrix formed from the eigenvalues of A, and S is matrix whose columns are their associated eigenvectors listed in the same order as the eigenvalues in D.Įxample Problem #1: A with two negative eigenvalues.įind equations x T(t) = (x 1(t), x 2(t)) for the system d x/dt = A * x, with x(0) = x0 = (x0 1, x0 2), where ![]() If the eigenvectors of the matrix A of dimension n form a basis of R n, then A can be diagonalized as: The Eigenvectors of A Form a Basis of R n. x n(t)) is the vector of unknown functions of time, and A = is the matrix of coefficients. ![]() The same problem expressed in matrix and vector form is: d x/dt = A * x The eigenvalues and eigenvectors of A can be real numbers and vectors in R n, or complex numbers and vectors in C n. If A has repeated eigenvalues, one might also need to compute their generalized eigenvectors. n, one must find the complete set of eigenvalues and eigenvectors of the matrix A. To obtain explicit formulae for the functions x i(t), i = 1. Objective: obtain formulae for the functions x i(t), i = 1. This system of linear differential equations is called autonomous because the coefficients of A are not explicit functions of time. n, and the matrix of constant coefficient is A =. Where x i(t) is a function of time, i = 1. Matrix exponential | eigenvector & generalized eigenvector basis | general solution - eigenvector & generalized eigenvector basisĮxample problems | general matrix phase diagram | equilibrium classification | summary stability diagram | referencesĪ System of Linear Differential Equations.Ī system of linear differential equations can be expressed as:ĭx 1/dt = a 1,1 x 1 + a 1,2 x 2 +. Linear differential equations | matrix representation | eigenvector basis | example problems | general solution - eigenvector basis Please show your support by joining Egwald Web Services as a Facebook Fan: Egwald Mathematics - Linear Algebra: Systems of Linear Differential EquationsĮgwald's popular web pages are provided without cost to users. ![]()
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