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Ordinary Differential Equation--System with Constant Coefficients

To solve the system of differential equations

(dx)/(dt)=Ax(t)+p(t),

(1)

where A is a matrix and x and p are vectors, first consider the homogeneous case with p=0. The solutions to

(dx)/(dt)=Ax(t)

(2)

are given by

x(t)=e^(At).

(3)

But, by the eigen decomposition theorem, the matrix exponential can be written as

e^(At)=uDu^(-1),

(4)

where the eigenvector matrix is

u=[u_1 ... u_n]

(5)

and the eigenvalue matrix is

D=[e^(lambda_1t) 0 ... 0; 0 e^(lambda_2t) ... 0; | | ... 0; 0 0 ... e^(lambda_nt)].

(6)

Now consider

The individual solutions are then

x_i=(e^(At)u)·e_i^^=u_ie^(lambda_it),

(10)

so the homogeneous solution is

x=sum_(i=1)^nc_iu_ie^(lambda_it),

(11)

where the c_is are arbitrary constants.

The general procedure is therefore

1. Find the eigenvalues of the matrix A (lambda_1, ..., lambda_n) by solving the characteristic equation.

2. Determine the corresponding eigenvectors u_1, ..., u_n.

3. Compute

x_i=e^(lambda_it)u_i

(12)

for i=1, ..., n. Then the vectors x_i which are real are solutions to the homogeneous equation. If A is a 2×2 matrix, the complex vectors x_j correspond to real solutions to the homogeneous equation given by R[x_j] and I[x_j].

4. If the equation is nonhomogeneous, find the particular solution given by

x^*(t)=X(t)intX^(-1)(t)p(t)dt,

(13)

where the matrix X is defined by

X(t)=[x_1 ... x_n].

(14)

If the equation is homogeneous so that p(t)=0, then look for a solution of the form

x=xie^(lambdat).

(15)

This leads to an equation

(A-lambdaI)xi=0,

(16)

so xi is an eigenvector and lambda an eigenvalue.

5. The general solution is

x(t)=x^*(t)+sum_(i=1)^nc_ix_i.

(17)

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Cite this as:

Weisstein, Eric W. "Ordinary Differential Equation--System with Constant Coefficients." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/OrdinaryDifferentialEquationSystemwithConstantCoefficients.html

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