20. Variation of Parameters for Systems - Michigan State University

November 26, 2012

20-1

20. Variation of Parameters for Systems

Now, we consider non-homogeneous linear systems. Thus, we consider the system

x = Ax + g(t)

(1)

where g(t) is a continuous vector valued function, and A is an n ? n matrix.

Let (t) be a fundamental matrix for the associated homogeneous system

x = Ax

(2)

We try to find a particular solution of the form

x(t) = (t)v(t)

where v(t) is a non-constant vector valued function of t. We have

x = v + v = Av + v

and, since x = v is a solution of (1),

x = Ax(t) + g(t) = Av + g(t)

which leads to Av + v = Av + g(t),

or v = g(t).

Since, = (t) is a fundamental matrix, it is invertible. So, we get v = (t)-1g(t).

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20-2

That is, we get a system of linear equations to solve for v . We then get v(t) by integrating.

This is similar to what happened for the case of variation of parameters in second order scalar differential equations.

Examples. Example 1: p. 411, 1 Find the general solution to the system

x=

Write x =

x y

.

We have

2 -1 3 -2

x+

et t

x = 2x - y + et y = 3x - 2y + t

Step 1. Two independent homogeneous solutions. The characteristic polynomial z(r) = r2 - 1 with roots r = ?1.

For r, we have the eigenvector (r - 2)1 + 2 = 0, or

2 = (2 - r)1.

Let us call an eigenvalue pair a pair (r, v) where r is an eigenvalue and v is an associated eigenvector.

So, the eigenvalue pairs are

(1,

1 1

), (-1,

1 3

)

So, two independent solutions are

x1(t) = et

1 1

, e-t

1 3

The general homogeneous solution is

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20-3

xh(t) = c1et

1 1

+ c2e-t

1 3

A fundamental matrix is

(t) =

et e-t et 3e-t

.

To get a particular solution xp(t) by the variation of parameters method, we proceed as follows.

where We get v as follows.

xp(t) = (t)v(t)

(t)v =

et t

etv1 + e-tv2 = et etv1 + 3e-tv2 = t

or, in matrix form,

et e-t et 3e-t

v1 v2

=

et t

.

Hence, using the formula for the inverse of a 2 ? 2 matrix (formula(1) in section 18), we get

This gives

v1 v2

=

et e-t -1 et

et 3e-t

t

1 =

2

3e-t -e-t -et et

et t

.

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20-4

Integrating, we get

v1

=

1 (3 - te-t) 2

v2

=

1 (-e2t + tet). 2

Finally, we get

v1

=

1 (3t + e-t(1 + t)) 2

v2

=

1 (- 1 e2t + tet - et). 22

xp(t) =

etv1 + e-tv2 etv1 + 3e-tv2

.

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20-5

Example 2:

x = 2x + 2y + e2t y = 2x - y - et

Characteristic polynomial: r2 - r - 6 = (r - 3)(r + 2).

eigenpairs: (r - 2)1 - 22 = 0.

(r,

1

r-2

).

2

General homogeneous solution:

xh(t) = c1e3t

1

1

2

+ c2e-2t

1 -2

.

Fundamental Matrix:

e3t e-2t

1 2

e3t

-2e-2t

.

Particular solution:

xp(t) =

e3t e-2t

1 2

e3t

-2e-2t

v1 v2

,

where

e3t e-2t

1 2

e3t

-2e-2t

v1 v2

=

e2t -et .

Finding v1, v2: We have

e3tv1 + e-2tv2 = e2t

1 2

e3tv1

-

2e-2tv2

=

-et

e3tv1 - 4e-2tv2 = -2et

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