Non-Homogeneous Systems MATH 374 Lecture 25 Non-Homogeneous Systems

8.7: Non-Homogeneous Systems Consider the problem X’ = AX + B(t) (1) where A is a constant n x n, real-valued matrix and B(t) is a known n x 1 vector function, say

Non-Homogeneous Systems X’ = AX + B(t) (1) Non-Homogeneous Systems From Theorem 8.6, we know that the general solution to (1) is: X(t) = Xc + Xp = c1X1 + … + cnXn + Xp, where X1, … , Xn are linearly independent solutions of X’ = AX (3) and Xp is a particular solution of (1).

Non-Homogeneous Systems X’ = AX + B(t) (1) Non-Homogeneous Systems As we did in the section on Variation of Parameters (see our class notes, Section 7.2), let’s look for a particular solution of (1) of the form: Xp(t) = b1(t)X1 + … + bn(t)Xn (4) where b1, … , bn are functions of t to be determined. Putting (4) into (1), we find:

Non-Homogeneous Systems X’ = AX + B(t) (1) Non-Homogeneous Systems X’ = AX (3) (b1(t)X1 + … + bn(t)Xn)’ = A(b1(t)X1 + … + bn(t)Xn) + B(t) which implies (b1’(t)X1 + … + bn’(t)Xn) + (b1(t)X1’ + … + bn(t)Xn’) = b1(t)AX1 + … + bn(t)AXn + B(t). (5) Since each Xi solves (3), (5) reduces to b1’(t)X1 + … + bn’(t)Xn = B(t), which is equivalent to

Non-Homogeneous Systems X’ = AX + B(t) (1) Non-Homogeneous Systems X’ = AX (3) n x n matrix, called a fundamental matrix for (3)

Non-Homogeneous Systems X’ = AX + B(t) (1) Non-Homogeneous Systems

Example 1 Solution: First, we first need to solve the homogeneous problem We find that the general solution to (9) is Thus, two linearly independent solutions to (9) are

Example 1 To find a particular solution to (8) of the form (4), we need to solve (6), which, for this problem is The solution to (10) can be found via Cramer’s Rule.

Example 1

Example 1 Use Mathematica to find b1(t) and b2(t).