Systems of Differential Equations
6.6 Constant-Coefficient Homogeneous Systems: Real Eigenvalues
In Section 6.5, we explored how solutions to homogeneous systems with constant coefficients
[asciimath]\mathbf{y}' = A\mathbf{y}[/asciimath](6.6.1)
are in the form
[asciimath]\mathbf{y} = e^{rt}\mathbf{u}[/asciimath]
where [asciimath]r[/asciimath] is an eigenvalue, and [asciimath]\mathbf{u}[/asciimath] is the corresponding eigenvector of the coefficient matrix [asciimath]A[/asciimath].
In this section, we focus on the case where the eigenvalues of matrix [asciimath]A[/asciimath] are distinct and real.
Theorem: If an [asciimath]n xx n[/asciimath] matrix [asciimath]A[/asciimath] has [asciimath]n[/asciimath] real, distinct eigenvalues [asciimath]\lambda_1, \lambda_2, \ldots, \lambda_n[/asciimath], and [asciimath]\mathbf{u}_i[/asciimath] is an eigenvector associated with the eigenvalue [asciimath]\lambda_i[/asciimath], then the vectors [asciimath]\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_n[/asciimath] are linearly independent.
In this context, the solutions for each eigenvalue take the form [asciimath]y = e^{\lambda_i t}\mathbf{u}_i[/asciimath]. Collectively, the set [asciimath]{e^{\lambda_1 t} \mathbf{u}_1, e^{\lambda_2 t} \mathbf{u}_2, \ldots, e^{\lambda_n t} \mathbf{u}_n}[/asciimath] forms a fundamental solution set for the homogeneous system 6.6.1.
Consequently, the general solution to the system can be expressed as a linear combination of these individual solutions.
[asciimath]\mathbf{y}(t) = c_1e^{\lambda_1 t} \mathbf{u}_1 + c_2 e^{\lambda_2 t} \mathbf{u}_2 + \ldots + c_n e^{\lambda_n t} \mathbf{u}_n[/asciimath](6.6.2)
where [asciimath]c_i[/asciimath] is an arbitrary constant.
Find a general solution of
[asciimath]bb"y"'=[(-6,-3),(1,-2)]bb"y"[/asciimath]
Show/Hide Solution
1. First we need to find the eigenvalues of the coefficient matrix [asciimath]A[/asciimath].
The characteristic polynomial of [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(-6-lambda,-3),(1,-2-lambda) |[/asciimath]
[asciimath]=(lambda+6)(lambda+2)+3[/asciimath]
[asciimath]=lambda^2+8lambda+15[/asciimath]
[asciimath]=(lambda+5)(lambda+3)[/asciimath]
The roots of [asciimath]c_A(lambda)[/asciimath] , which are [asciimath]lambda_1=-5[/asciimath] and [asciimath]lambda_2=-3[/asciimath] , are the eigenvalues of [asciimath]A[/asciimath].
2. Next to find the corresponding eigenvectors, we need to find the solution to the equation [asciimath](lambda I-A)bb"u"=bb"0"[/asciimath] for each eigenvalue.
For [asciimath]lambda_1=-5[/asciimath] , we have
[asciimath](lambda_1 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(-5+6,3),(-1,-5+2) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(1,3),(-1,-3) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_1=-5[/asciimath] are [asciimath]bb"u"_1=t[(-3),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Using [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_1=-5[/asciimath] is [asciimath]bb"u"_1[/asciimath] [asciimath]=[(-3),(1)][/asciimath].
For [asciimath]lambda_2=-3[/asciimath] , we have
[asciimath](lambda_2 I-A)bb(u)=bb(0)[/asciimath]
[asciimath][(-3+6,3),(-1,-3+2) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(3,3),(-1,-1) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(3,3,|,0),(-1,-1,|,0) ]~[/asciimath] [asciimath][(1,1,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_2=-3[/asciimath] are [asciimath]bb"u"_2=t[(-1),(1)][/asciimath] for [asciimath]t!=0[/asciimath] . Using [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_2=-3[/asciimath] is [asciimath]bb"u"_2[/asciimath] [asciimath]=[(-1),(1)][/asciimath].
3. A general solution to the system is given by Equation 6.6.2.
[asciimath]bb"y"(t)=c_1e^(-5t) [(-3),(1)]+c_2 e^(-3t) [(-1),(1)][/asciimath]
Try an Example
Solve the system of differential equations with the given initial values.
[asciimath]{(y_1' = 4 y_1 -15 y_2),(y_2' = 2 y_1 -7 y_2):},[/asciimath] [asciimath]y_1(0) = 7, \quad y_2(0) = 3[/asciimath]
Show/Hide Solution
1. We first express the system in the matrix notation.
[asciimath]bb"y"'=[(4,-15),(2,-7) ]bb"y",[/asciimath] [asciimath]bb"y"(0)=[(7),(3)][/asciimath]
2. Next, we find the eigenvalues of [asciimath]A[/asciimath].
The characteristic polynomial of the coefficient matrix [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(4-lambda,-15),(2,-7-lambda) |[/asciimath]
[asciimath]=(4-lambda)(-7-lambda)+30[/asciimath]
[asciimath]=lambda^2+3lambda+2[/asciimath]
[asciimath]=(lambda+1)(lambda+2)[/asciimath]
The roots of [asciimath]c_A(lambda)[/asciimath], which are [asciimath]lambda_1=-1[/asciimath] and [asciimath]lambda_2=-2[/asciimath] , are the eigenvalues of [asciimath]A[/asciimath].
3. We then find the corresponding eigenvectors.
For [asciimath]lambda_1=-1[/asciimath], we have
[asciimath](lambda_1 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(4+1,-15),(2,-7+1) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(5,-15),(2,-6) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(5,-15,|,0),(2,-6,|,0) ]~[/asciimath] [asciimath][(1,-3,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_1=-1[/asciimath] are [asciimath]bb"u"_1=t[(3),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Using [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_1=-1[/asciimath] is [asciimath]bb"u"_1[/asciimath] [asciimath]=[(3),(1)][/asciimath].
For [asciimath]lambda_2=-2[/asciimath], we have
[asciimath](lambda_2 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(4+2,-15),(2,-7+2) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(6,-15),(2,-5) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(6,-15,|,0),(2,-5,|,0) ]~[/asciimath] [asciimath][(1,-5/2,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_2=-2[/asciimath] are [asciimath]bb"u"_2=t[(5/2),(1)][/asciimath] for [asciimath]t!=0[/asciimath] . Using [asciimath]t=2[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_2=-2[/asciimath] is [asciimath]bb"u"_2[/asciimath] [asciimath]=[(5),(2)][/asciimath].
4. A general solution to the system is given by Equation 6.6.2.
[asciimath]bb"y"(t)=c_1e^(-t) [(3),(1)]+c_2 e^(-2t) [(5),(2)][/asciimath]
5. Finally, we apply the initial conditions to find constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
[asciimath]bb"y"(0)=[(7),(3)][/asciimath]
[asciimath]c_1e^(0) [(3),(1)]+c_2 e^(0) [(5),(2)]=[(7),(3)][/asciimath]
[asciimath][(3c_1+5c_2),(c_1+2c_2)]=[/asciimath] [asciimath][(7),(3)][/asciimath]
This gives a system of two equations and two unknowns.
[asciimath]{(3c_1+5c_2=7),(c_1+2c_2=3 ):}[/asciimath]
Solving the system yields
[asciimath]c_1=-1,[/asciimath] [asciimath]c_2=2[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]bb"y"(t)=-e^(-t) [(3),(1)]+2 e^(-2t) [(5),(2)][/asciimath]
Try an Example
Solve the system of differential equations with the given initial values.
[asciimath]{(y_1' = 3 y_1 +14 y_2-13y_3),(y_2' = 11 y_2 -7 y_3),(y_3' = 14 y_2 -10 y_3) :},[/asciimath] [asciimath]y_1(0) = -3, \ y_2(0) = -1, \ y_3(0)=1[/asciimath]
Show/Hide Solution
1. We first express the system in the matrix notation.
[asciimath]bb"y"'=[(3,14,-13),(0,11,-7),(0,14,-10) ]bb"y",[/asciimath] [asciimath]bb"y"(0)=[(-3),(-1),(1)][/asciimath]
2. Next, we find the eigenvalues of the coefficient matrix [asciimath]A[/asciimath].
The characteristic polynomial of [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(3-lambda,14,-13),(0,11-lambda,-7),(0,14,-10-lambda)|[/asciimath]
[asciimath]=(3-lambda)|(11-lambda,-7),(14,-10-lambda)|[/asciimath]
[asciimath]=(3-lambda)(lambda^2-lambda-12)[/asciimath]
[asciimath]=(3-lambda)(lambda+3)(lambda-4)[/asciimath]
The roots of [asciimath]c_A(lambda)[/asciimath], which are [asciimath]lambda_1=3[/asciimath], [asciimath]lambda_2=4[/asciimath], and [asciimath]lambda_3=-3[/asciimath], are the eigenvalues of [asciimath]A[/asciimath].
3. We then find the corresponding eigenvectors.
For [asciimath]lambda_1=3[/asciimath], we have
[asciimath](lambda_1 I-A)bb"u"=bb"0"[/asciimath]
[asciimath]=[(0,14,-13),(0,8,-7),(0,14,-13)][/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(0,14,-13,|,0),(0,8,-7,|,0),(0,14,-13,|,0)]~[/asciimath] [asciimath][(0,1,0,|,0),(0,0,1,|,0),(0,0,0,|,0)][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_1=3[/asciimath] are [asciimath]bb"u"_1=t[(1),(0),(0)][/asciimath] for [asciimath]t!=0[/asciimath]. Taking [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_1=3[/asciimath] is [asciimath]bb"u"_1[/asciimath] [asciimath]=[(1),(0),(0)][/asciimath].
For [asciimath]lambda_2=4[/asciimath], we have
[asciimath](lambda_2 I-A)bb"u"=bb"0"[/asciimath]
[asciimath]=[(-1,14,-13),(0,7,-7),(0,14,-14)][/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(-1,14,-13),(0,7,-7),(0,14,-14)]~[/asciimath] [asciimath][(1,0,-1),(0,1,-1),(0,0,0)][/asciimath]
Thus, the eigenvectors corresponding to [asciimath]lambda_2=4[/asciimath] are [asciimath]bb"u"_2=t[(1),(1),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Taking [asciimath]t=1[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_2=4[/asciimath] is [asciimath]bb"u"_2[/asciimath] [asciimath]=[(1),(1),(1)][/asciimath].
For [asciimath]lambda_3=-3[/asciimath], we have
[asciimath](lambda_3 I-A)bb"u"=bb"0"[/asciimath]
[asciimath]=[(6,14,-13),(0,14,-7),(0,14,-7)][/asciimath] [asciimath][(u_1),(u_2),(u_3)]=[(0),(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(6,14,-13),(0,14,-7),(0,14,-7)]~[/asciimath] [asciimath][(1,0,-1),(0,1,-1/2),(0,0,0)][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_3=-3[/asciimath] are [asciimath]bb"u"_3=t[(1),(1/2),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Taking [asciimath]t=2[/asciimath], a basic eigenvector corresponding to [asciimath]lambda_3=-3[/asciimath] is [asciimath]bb"u"_3[/asciimath] [asciimath]=[(2),(1),(2)][/asciimath].
4. A general solution to the system is given by Equation 6.6.2.
[asciimath]bb"y"(t)=c_1e^(3t) [(1),(0),(0)]+c_2 e^(4t) [(1),(1),(1)]+c_3e^(-3t)[(2),(1),(2)][/asciimath]
5. Finally, we apply the initial conditions to find constants [asciimath]c_1[/asciimath], [asciimath]c_2[/asciimath], and [asciimath]c_3[/asciimath].
[asciimath]bb"y"(0)=[(-3),(-1),(1)][/asciimath]
[asciimath]c_1e^0 [(1),(0),(0)]+c_2 e^0 [(1),(1),(1)]+c_3e^0[(2),(1),(2)] =[(-3),(-1),(1)][/asciimath]
This gives a system of three equations and three unknowns.
[asciimath]{(c_1+c_2+2c_3=-3),(c_2+c_3=-1 ),(c_2+2c_3=1):}[/asciimath]
Solving the system yields
[asciimath]c_1=-4,[/asciimath] [asciimath]c_2=-3,[/asciimath] [asciimath]c_3=2[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]bb"y"(t)=-4e^(3t) [(1),(0),(0)]-3 e^(4t) [(1),(1),(1)]+2e^(-3t)[(2),(1),(2)][/asciimath]
Try an Example
In Section 6.4, we explored how higher-order linear differential equations can be converted into systems of first-order linear equations. This transformation, coupled with the matrix method offers several advantages, like better organization of the problem and ease of computation. While this approach might not always be shorter than the characteristic polynomial method discussed in Section 3.2, especially for solving homogeneous second-order differential equations with constant coefficients, it is beneficial to understand this process. To illustrate how it is applied, let’s work through an example.
Convert the given differential equation to a linear first-order system and find the solution.
[asciimath]y''+3y'-10 y = 0,[/asciimath] [asciimath]y(0) = -3, \quad y'(0) = -13[/asciimath]
Show/Hide Solution
a. Converting the equation to a system:
1a. Introduce a new variable [asciimath]x_i[/asciimath] :
[asciimath]x_1=y[/asciimath]
[asciimath]x_2=y'[/asciimath]
2a. Express the derivatives by differentiating the above equations and rearrange the original differential equation to isolate [asciimath]y''[/asciimath]:
[asciimath]x_1'=y'=x_2[/asciimath]
[asciimath]x_2'=y''[/asciimath] [asciimath]=-3y'+10y[/asciimath] [asciimath]->[/asciimath] [asciimath]x_2'=-3x_2+10x_1[/asciimath]
We also express the initial conditions in terms of the new variables:
[asciimath]x_1(0)=y(0)=-3[/asciimath]
[asciimath]x_2(0)=y'(0)=-13[/asciimath]
3a. The system of first-order equations is then
[asciimath]x_1'=x_2[/asciimath]
[asciimath]x_2'=10x_1-3x_2[/asciimath]
[asciimath]x_1(0)=-3, \ x_2(0)=-13[/asciimath]
1b. We express the system in the matrix form.
[asciimath]bb"x"'=[(0,1),(10,-3) ]bb"x",[/asciimath] [asciimath]bb"x"(0)=[(-3),(-13)][/asciimath]
2b. Next, we find the eigenvalues of the coefficient matrix [asciimath]A[/asciimath].
The characteristic polynomial of [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(0-lambda,1),(10,-3-lambda) |[/asciimath]
[asciimath]=(-lambda)(-3-lambda)-10[/asciimath]
[asciimath]=lambda^2+3lambda-10[/asciimath]
[asciimath]=(lambda-2)(lambda+5)[/asciimath]
The roots of [asciimath]c_A(lambda)[/asciimath], which are [asciimath]lambda_1=2[/asciimath] and [asciimath]lambda_2=-5[/asciimath], are the eigenvalues of [asciimath]A[/asciimath].
3b. We then find the corresponding eigenvectors.
For [asciimath]lambda_1=2[/asciimath], we have
[asciimath](lambda_1 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(-2,1),(10,-5) ][/asciimath][asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(-2,1,|,0),(10,-5,|,0) ]~[/asciimath][asciimath][(1,-1/2,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_1=2[/asciimath] are [asciimath]bb"u"_1=t[(1/2),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Using [asciimath]t=2[/asciimath], a basic eigenvector is [asciimath]bb"u"_1[/asciimath] [asciimath]=[(1),(2)][/asciimath].
For [asciimath]lambda_2=-5[/asciimath], we have
[asciimath](lambda_2 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(5,1),(10,2) ][/asciimath][asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
To solve the system, we form the augmented matrix and bring it to RREF using row operations.
[asciimath][(5,1,|,0),(10,2,|,0) ]~[/asciimath][asciimath][(1,1/5,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to [asciimath]lambda_2=-5[/asciimath] are [asciimath]bb"u"_2=t[(-1/5),(1)][/asciimath] for [asciimath]t!=0[/asciimath] . Using [asciimath]t=5[/asciimath], a basic eigenvector is [asciimath]bb"u"_2[/asciimath] [asciimath]=[(-1),(5)][/asciimath].
4b. A general solution to the system is given by Equation 6.6.2.
[asciimath]bb"x"(t)=c_1e^(2t) [(1),(2)]+c_2 e^(-5t) [(-1),(5)][/asciimath]
5b. We apply the initial conditions to find constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
[asciimath]bb"x"(0)=[(-3),(-13)][/asciimath]
[asciimath]c_1e^(0) [(1),(2)]+c_2 e^(0) [(-1),(5)]=[(-3),(-13)][/asciimath]
[asciimath][(c_1-c_2),(2c_1+5c_2)]=[/asciimath] [asciimath][(-3),(-13)][/asciimath]
This gives a system of two equations and two unknowns.
[asciimath]{(c_1-c_2=-3),(2c_1+5c_2=-13 ):}[/asciimath]
Solving the system yields
[asciimath]c_1=-4,[/asciimath] [asciimath]c_2=-1[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]bb"x"(t)=-4e^(2t) [(1),(2)]- e^(-5t) [(-1),(5)][/asciimath]
c. Determining the solution to the original equation
Given [asciimath]bb"x"(t)=[(x_1),(x_2)]=[(y),(y')][/asciimath] , we see that the solution to the original 2nd-order differential equation [asciimath]y[/asciimath] is the top row of the system’s solution. Therefore, the solution to the original equation is
[asciimath]y=-4e^(2t)+e^(-5t)[/asciimath]
Try an Example
Section 6.6 Exercises
- Solve the system of differential equations with the given initial values.
[asciimath]{(y_1' = 7 y_1 -12 y_2),(y_2' = 2 y_1 -3 y_2):},[/asciimath] [asciimath]y_1(0) = 4, \quad y_2(0) = 1[/asciimath]
Show/Hide Answer
[asciimath]y_1(t)=6e^(3t)-2e^t[/asciimath]
[asciimath]y_2(t)=2e^(3t)-e^t[/asciimath]
- Solve the system of differential equations
[asciimath]bb"y"'=[(-17,-30),(4,5)] bb"y"[/asciimath]
Show/Hide Answer
[asciimath]bb"y"(t)=c_1e^(-5t)[(-5),(2)]+c_2e^(-7t)[(-3),(1)][/asciimath]
- Solve the system of differential equations
[asciimath]bb"y"'=[(1,-12,7),(0,-10,6),(0,-12,8)] bb"y",[/asciimath] [asciimath]mathbfy'(0)=[(3),(4),(7)][/asciimath]
Show/Hide Answer
[asciimath]bb"y"(t)=-4e^t[(1),(0),(0)]+3e^(2t)[(2),(1),(2)] +e^(-4t)[(1),(1),(1)][/asciimath]
