Systems of Differential Equations
6.9 Nonhomogeneous Linear Systems
In this section, we study the nonhomogeneous linear system
[asciimath]bb(y)'=Abb(y)+bb(f)(t)[/asciimath](6.9.1)
where matrix [asciimath]A[/asciimath] is an [asciimath]nxxn[/asciimath] matrix function and [asciimath]bb(f)[/asciimath] is an n-vector forcing function. The associated homogenous system [asciimath]bb(y)'=Abb(y)[/asciimath] is called the complementary system.
The methods from Chapter 3, such as Undetermined Coefficients and Variation of Parameters, used for finding particular solutions to nonhomogeneous linear equations, can be extended to nonhomogeneous linear systems. We focus here on the method of Variation of Parameters.
Variation of Parameters
The method of variation of parameters, as discussed in Section 3.5 for linear equations, applies to linear systems. It requires a fundamental set of solutions to the complementary (homogeneous) equation.
Theorem. Suppose an [asciimath]nxxn[/asciimath] matrix [asciimath]A[/asciimath] and an n-th vector [asciimath]bb"f"[/asciimath] are continuous on an open interval [asciimath]I[/asciimath]. Let [asciimath]bb(y)_p[/asciimath] be a particular solution of system 6.9.1 on [asciimath]I[/asciimath], and let [asciimath]{bb(y)_1, bb(y)_2, ..., bb(y)_n}[/asciimath] be a fundamental set of solutions of the complementary system [asciimath]bb(y)'=A(t)bb(y)[/asciimath]. Then the general solution to 6.9.1 on [asciimath]I[/asciimath] is
[asciimath]bb(y)(t)=bb(y)_p+bb(y)_c[/asciimath]
where [asciimath]bb(y)_c=c_1bb(y)_1+c_2bb(y)_2+...+c_nbb(y)_n[/asciimath] is the complementary solution and where [asciimath]c_i[/asciimath] is an arbitrary constant. The general solution can be expressed as
[asciimath]bb(y)(t)=bb(y)_p+c_1bb(y)_1+c_2bb(y)_2+...+c_nbb(y)_n[/asciimath]
Method of Variation of Parameters for Nonhomogeneous Linear Systems
To find a particular solution to [asciimath]bb(y)'=Abb(y)+bb(f)(t)[/asciimath]
1. Find a fundamental set of solutions [asciimath]{bb(y)_1, bb(y)_2, ..., bb(y)_n}[/asciimath] to the corresponding complementary system [asciimath]bb(y)'=Abb(y)[/asciimath].
2. Form the fundamental matrix [asciimath]Y(t)[/asciimath] for the complementary system.
[asciimath]Y(t)= [bb(y)_1 \ bb(y)_2 \ ... \ bb(y)_n][/asciimath]
3. Find the inverse of the fundamental matrix, [asciimath]Y^-1(t)[/asciimath].
4. Determine [asciimath]bb(v)(t)=int\ Y^-1(t)bb(f)(t) dt[/asciimath]
5. A particular solution to the system is given by
[asciimath]bb(y)_p(t)=Y(t)*bb(v)(t)[/asciimath]
[asciimath]bb(y)_p(t)=Y(t)int \ Y^-1(t)bb(f)(t) dt[/asciimath]
6. A general solution to the system is then
[asciimath]bb(y)(t)=bb(y)_p+c_1bb(y)_1+c_2bb(y)_2+...+c_nbb(y)_n[/asciimath]
Find the general solution to the system
[asciimath]bb"y"'=[(-4,-3),(6,5)]bb"y"+[(2),(-2e^t)][/asciimath]
Show/Hide Solution
1. First we find a fundamental solution to the associated complementary (homogeneous) system.
The characteristic polynomial of the coefficient matrix [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda) =|(lambda+4,3),(-6,lambda-5) |[/asciimath]
[asciimath]=(lambda+4)(lambda-5)+18[/asciimath]
[asciimath]=lambda^2-lambda-2[/asciimath]
[asciimath]=(lambda-2)(lambda+1)[/asciimath]
The roots of [asciimath]c_A(lambda)[/asciimath], which are [asciimath]lambda_1=2[/asciimath] and [asciimath]lambda_2=-1[/asciimath], are the eigenvalues of [asciimath]A[/asciimath]. Then, we find the corresponding eigenvectors.
For [asciimath]lambda_1=2[/asciimath], we have
[asciimath](lambda_1 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(6,3),(-6,-3) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_1=2[/asciimath] are [asciimath]bb"u"_1=t[(-1),(2)][/asciimath] for [asciimath]t!=0[/asciimath].
For [asciimath]lambda_2=-1[/asciimath], we have
[asciimath](lambda_2 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(3,3),(-6,-6) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
The eigenvectors corresponding to [asciimath]lambda_2=-1[/asciimath] are [asciimath]bb"u"_2=t[(-1),(1)][/asciimath] for [asciimath]t!=0[/asciimath] .
Therefore, [asciimath]{e^(2t) [(-1),(2)], \ e^(-t) [(-1),(1)] }[/asciimath] is a fundamental solution set to the complementary system.
2. Thus the fundamental matrix [asciimath]Y(t)[/asciimath] for the complementary system is
[asciimath]Y(t)=[(-e^(2t),-e^-t),(2e^(2t),e^-t)][/asciimath]
3. We determine [asciimath]Y^-1(t)[/asciimath]
[asciimath]Y^-1(t)=1/(-e^(2t)e^-t+2e^(2t)e^-t)[(e^(-t),e^-t),(-2e^(2t),-e^(2t))][/asciimath]
[asciimath]=[(e^(-2t),e^(-2t)),(-2e^(t),-e^(t))][/asciimath]
4. Determine [asciimath]bb"v"(t)[/asciimath] letting the constant of integration be zero
[asciimath]bb"v"(t)=int\ Y^-1(t)bb"f"(t) dt[/asciimath]
[asciimath]=int \ [(e^(-2t),e^(-2t)),(-2e^(t),-e^(t))][(2),(-2e^t)] dt[/asciimath]
[asciimath]=int \ [(2e^(-2t)-2e^-t),(-4e^t+2e^(2t))] dt[/asciimath]
[asciimath]=[(-e^(-2t)+2e^-t),(-4e^t+e^(2t))][/asciimath]
5. Then, a particular solution to the system is
[asciimath]bb"y"_p(t)=Y(t).bb"v"(t)[/asciimath]
[asciimath]=[(-e^(2t),-e^-t),(2e^(2t),e^-t)][(-e^(-2t)+2e^-t),(-4e^t+e^(2t))][/asciimath]
[asciimath]=[(5-3e^t),(-6+5e^t)][/asciimath]
6. Thus, a general solution to the system is
[asciimath]bb"y"=bb"y"_p+c_1bb"y"_1+c_2bb"y"_2[/asciimath]
[asciimath]=[(5-3e^t),(-6+5e^t)][/asciimath] [asciimath]+c_1e^(2t) [(-1),(2)] +c_2e^(-t) [(-1),(1)][/asciimath]
Which can also be written as
[asciimath]bb"y"=[(5-3e^t),(-6+5e^t)][/asciimath] [asciimath]+[(-e^(2t),-e^-t),(2e^(2t),e^-t)] [(c_1),(c_2)][/asciimath]
Try an Example
Find the general solution to the system
[asciimath]bb(y)'=[(3,14,-13),(0,11,-7),(0,14,-10)]bb(y)+[(2e^t),(0),(-e^(2t))][/asciimath]
Show/Hide Solution
The complementary system is
[asciimath]bb(y)'=[(3,14,-13),(0,11,-7),(0,14,-10)]bb(y)[/asciimath]
The forcing vector is
[asciimath]bb(f)(t)=[(2e^t),(0),(-e^(2t))][/asciimath]
1. In Example 6.6.3 in Section 6.6, we found a fundamental solution set of the complementary system associated with the given system in this example.
[asciimath]{e^(3t) [(1),(0),(0)], \ e^(4t) [(1),(1),(1)] ,\ e^(-3t)[(2),(1),(2)] }[/asciimath]
2. The fundamental matrix [asciimath]Y(t)[/asciimath] for the complementary system is
[asciimath]Y"(t)=[(e^(3t),e^(4t),2e^(-3t)),(0,e^(4t),e^(-3t) ),(0,e^(4t),2e^(-3t) ) ][/asciimath]
3. We determine [asciimath]Y^-1(t)[/asciimath] using the row reduction method, involving augmenting the matrix [asciimath]Y(t)[/asciimath] with the identity matrix.
[asciimath][Y|I]~[I|Y^-1][/asciimath]
[asciimath]Y^-1=[(e^(-3t),0,-e^(-3t)),(0,2e^(-4t),-e^(-4t) ),(0,-e^(3t),e^(3t) ) ][/asciimath]
4. Determine [asciimath]bb(v)(t)[/asciimath] letting the constant of integration be zero.
[asciimath]bb"v"(t)=int\ Y^-1(t)bb"f"(t) dt[/asciimath]
[asciimath]=int \ [(e^(-3t),0,-e^(-3t)),(0,2e^(-4t),-e^(-4t) ),(0,-e^(3t),e^(3t) ) ] [(2e^t),(0),(-e^(2t))] dt[/asciimath]
[asciimath]=int \ [(2e^(-2t)+e^(-t)),(e^(-2t)),(-e^(5t))] dt[/asciimath]
[asciimath]=[(-e^(-2t)-e^(-t)),(-1/2e^(-2t)),(-1/5e^(5t))][/asciimath]
5. Then, a particular solution to the system is
[asciimath]bb(y)_p(t)=Y(t)*bb(v)(t)[/asciimath]
[asciimath]=[(e^(3t),e^(4t),2e^(-3t)),(0,e^(4t),e^(-3t) ),(0,e^(4t),2e^(-3t) ) ][/asciimath] [asciimath][(-e^(-2t)-e^(-t)),(-1/2e^(-2t)),(-1/5e^(5t))][/asciimath]
[asciimath]=1/10[(-10e^(t)-19e^(2t)),(-7e^(2t)),(-9e^(2t))][/asciimath]
6. Thus, a general solution to the system is
[asciimath]bb(y)(t)=bb(y)_p+c_1bb(y)_1+c_2bb(y)_2+...+c_nbb(y)_n[/asciimath]
[asciimath]bb(y)(t)=1/10[(-10e^(t)-19e^(2t)),(-7e^(2t)),(-9e^(2t))] +[/asciimath] [asciimath]c_1e^(3t) [(1),(0),(0)]+c_2 e^(4t) [(1),(1),(1)]+c_3e^(-3t)[(2),(1),(2)][/asciimath]
This can also be expressed as
[asciimath]bb"y"=1/10[(-10e^(t)-19e^(2t)),(-7e^(2t)),(-9e^(2t))][/asciimath] [asciimath]+[(e^(3t),e^(4t),2e^(-3t)),(0,e^(4t),e^(-3t) ),(0,e^(4t),2e^(-3t) ) ] [(c_1),(c_2),(c_3)][/asciimath]
Section 6.9 Exercises
- Find the general solution to the system of differential equations
[asciimath]bb"y'"=[(-25,36),(-18,26)] bb"y"[/asciimath] [asciimath]+[(-2),(5 e^t)][/asciimath]
Show/Hide Answer
[asciimath]bb(y)(t)=[(-26-90 e^t),(-18-65 e^t)]+[(3e^-t,4e^(2t)),(2e^-t,3e^(2t))][(c_1),(c_2)][/asciimath]
- Find the general solution to the system of differential equations
[asciimath]bb"y'"=[(5,3),(-6,-4)] bb"y"[/asciimath] [asciimath]+[(2),(5 e^t)][/asciimath]
Show/Hide Answer
[asciimath]bb(y)(t)=[(-4-7.5 e^t),(6+10 e^t)]+[(-e^-t,-e^(2t)),(2e^-t,e^(2t))][(c_1),(c_2)][/asciimath]
