Systems of Differential Equations
6.7 Constant-Coefficient Homogeneous Systems: Complex Eigenvalues
In this section, we examine solutions to the homogeneous system with constant coefficients [asciimath]bb"y"'=Abb"y"[/asciimath] for the case where the eigenvalues of the coefficient matrix are complex. Typically, these eigenvalues are conjugates of each other, denoted as [asciimath]lambda=alpha+-ibeta[/asciimath], where [asciimath]i[/asciimath] is the imaginary unit, and [asciimath]alpha[/asciimath] and [asciimath]beta[/asciimath] are real numbers. As in the complex case of second-order differential equations, we utilize Euler’s formula to convert complex exponentials into real trigonometric functions, starting from the guessed solution form [asciimath]y=e^(rt)bb"u"[/asciimath].
Theorem. If an [asciimath]nxxn[/asciimath] matrix [asciimath]A[/asciimath] has complex conjugate eigenvalues [asciimath]lambda=alpha+-ibeta[/asciimath] with the corresponding eigenvector [asciimath]bb"u"=bb"a"+-ibb"b"[/asciimath], then two linearly independent solutions to the homogeneous system [asciimath]bb"y"'=Abb"y"[/asciimath] are
[asciimath]bb"y"_1=e^(alphat)(bb"a"cos(betat)-bb"b"sin(beta t))[/asciimath]
[asciimath]bb"y"_2=e^(alphat)(bb"a"sin(betat)+bb"b"cos(beta t))[/asciimath]
The general solution to the system is then given by
[asciimath]bb"y"(t)=c_1bb"y"_1+c_2 bb"y"_2[/asciimath]
[asciimath]bb"y"(t)=[/asciimath] [asciimath]c_1e^(alphat)(bb"a"cos(betat)-bb"b"sin(beta t))+[/asciimath] [asciimath]c_2e^(alphat)(bb"a"sin(betat)+bb"b"cos(beta t))[/asciimath] (6.7.1)
where [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath] are arbitrary constants.
Find a general solution of
[asciimath]bb"y"'=[(5,6),(-3,-1)]bb"y"[/asciimath]
Show/Hide Solution
1. First we need to 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]=|(5-lambda,6),(-3,-1-lambda) |[/asciimath]
[asciimath]=(5-lambda)(-1-lambda)+18[/asciimath]
[asciimath]=lambda^2-4lambda+13[/asciimath]
[asciimath]=(lambda-2)^2+9[/asciimath]
Therefore, the roots of [asciimath]c_A(lambda)[/asciimath], which are [asciimath]lambda=2+-i3[/asciimath], are the eigenvalues of [asciimath]A[/asciimath].
2. Next we find the corresponding eigenvectors by finding the solution to the equation [asciimath](lambda I-A)bb"u"=bb"0"[/asciimath]. However, we only need to find the eigenvector associated with one of the eigenvalues, e.g., [asciimath]lambda_1=2+i3[/asciimath].
[asciimath](A-lambda_1 I)bb"u"=bb"0"[/asciimath]
[asciimath][(5-(2+i3),6),(-3,-1-(2+i3)) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(3-i3,6),(-3,-3-i3) ][/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-i3,6,|,0),(-3,-3-i3,|,0) ]~[/asciimath][asciimath][(1,1+i,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to eigenvalue [asciimath]lambda_1=2+i3[/asciimath] are [asciimath]bb"u"_1=t[(-1-i),(1)][/asciimath] for [asciimath]t!=0[/asciimath]. Letting [asciimath]t=1[/asciimath], we have a basic eigenvector
[asciimath]bb"u"_1=[(-1-i),(1)][/asciimath] [asciimath]=[(-1),(1)]+i[(-1),(0)][/asciimath]
The real part of [asciimath]bb"u"_1[/asciimath] is [asciimath]bb"a"=[(-1),(1)][/asciimath] and the imaginary part of [asciimath]bb"u"_1[/asciimath] is [asciimath]bb"b"=[(-1),(0)][/asciimath].
The eigenvector corresponding to the conjugate eigenvalue is the conjugate of eigenvector [asciimath]bb"u"_1[/asciimath]. Thus, the eigenvector associated with the eigenvalue [asciimath]lambda_2=2-i3[/asciimath] is
[asciimath]bb"u"_2=[(-1+i),(1)][/asciimath] [asciimath]=[(-1),(1)]-i[(-1),(0)][/asciimath]
3. Therefore, a general solution to the system is given by Equation 6.7.1.
[asciimath]bb"y"(t)=c_1e^(2t)([(-1),(1)] cos(3t)-[(-1),(0)] sin(3 t))[/asciimath] [asciimath]+c_2e^(2t)([(-1),(1)] sin(3t)+[(-1),(0)] cos(3 t))[/asciimath]
[asciimath]=c_1e^(2t)([(-cos(3t)+sin(3t)),(cos(3t))] )[/asciimath] [asciimath]+c_2e^(2t)([(-sin(3t)-cos(3t)),(sin(3t))] )[/asciimath]
Try an Example
Solve the system of differential equations with initial conditions
[asciimath]bb"y"' = [[1,3],[-15,-11]] \ bb"y", \quad \ bb"y"(0) = [(-3), (12)][/asciimath].
Show/Hide Solution
1. First we need to 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]=|(1-lambda,3),(-15,-11-lambda) |[/asciimath]
[asciimath]=(1-lambda)(-11-lambda)+45[/asciimath]
[asciimath]=lambda^2+10lambda+34[/asciimath]
[asciimath]=(lambda+5)^2+9[/asciimath]
Therefore, the roots of [asciimath]c_A(lambda)[/asciimath], which are [asciimath]lambda=-5+-i3[/asciimath], are the eigenvalues of [asciimath]A[/asciimath].
2. Next we find the corresponding eigenvectors by finding the solution to the equation [asciimath](A-lambda I)bb"u"=bb"0"[/asciimath]. However, we only need to find the eigenvector associated with one of the eigenvalues, e.g., [asciimath]lambda_1=-5-i3[/asciimath].
[asciimath](lambda_1 I-A)bb"u"=bb"0"[/asciimath]
[asciimath][(1-lambda,3),(-15,-11-lambda) ][/asciimath] [asciimath][(u_1),(u_2)]=[(0),(0)][/asciimath]
[asciimath][(6+i3,3),(-15,-6+i3) ][/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+i3,3,|,0),(-15,-6+i3,|,0) ]~[/asciimath] [asciimath][(1,2/5-i1/5,|,0),(0,0,|,0) ][/asciimath]
Therefore, the eigenvectors corresponding to eigenvalue [asciimath]lambda_1=-5-i3[/asciimath] are [asciimath]bb"u"_1=t[(-2/5+i1/5),(1)][/asciimath] for [asciimath]t!=0[/asciimath] . Letting [asciimath]t=5[/asciimath], we have a basic eignevector
[asciimath]bb"u"_1=[(-2+i),(5)][/asciimath] [asciimath]=[(-2),(5)]+i[(1),(0)][/asciimath]
The real part of [asciimath]bb"u"_1[/asciimath] is [asciimath]bb"a"=[(-2),(5)][/asciimath] and the imaginary part of [asciimath]bb"u"_1[/asciimath] is [asciimath]bb"b"=[(1),(0)][/asciimath].
The eigenvector corresponding to the conjugate eigenvalue is the conjugate of eigenvector [asciimath]bb"u"_1[/asciimath]. Thus, the eigenvector associated with eigenvalue [asciimath]lambda_2=-5+i3[/asciimath] is
[asciimath]bb"u"_2=[(-2-i),(5)][/asciimath] [asciimath]=[(-2),(5)]-i[(1),(0)][/asciimath]
3. Therefore, a general solution to the system is given by Equation 6.7.1.
[asciimath]bb"y"(t)=c_1e^(-5t)([(-2),(5)] cos(3t)-[(1),(0)] sin(3 t))[/asciimath] [asciimath]+c_2e^(-5t)([(-2),(5)] sin(3t)+[(1),(0)] cos(3 t))[/asciimath]
4. We apply the initial conditions to find constants [asciimath]c_1[/asciimath] and [asciimath]c_2[/asciimath].
[asciimath]bb"y"(0)=[(-3),(12)][/asciimath]
[asciimath]c_1e^(0)([(-2),(5)] cos(0)-[(1),(0)] sin(0))[/asciimath] [asciimath]+c_2e^(0)([(-2),(5)] sin(0)+[(1),(0)] cos(0)) =[/asciimath] [asciimath][(-3),(12)][/asciimath]
[asciimath]c_1[(-2),(5)]+c_2[(1),(0)]=[/asciimath] [asciimath][(-3),(12)][/asciimath]
This gives a system of two equations and two unknowns.
[asciimath]{(-2c_1+c_2=-3),(5c_1=12 ):}[/asciimath]
Solving the system yields
[asciimath]c_1=12/5,[/asciimath] [asciimath]c_2=9/5[/asciimath]
Therefore the solution to the initial value problem is
[asciimath]bb"y"(t)=12/5e^(-5t)([(-2),(5)] cos(3t)-[(1),(0)] sin(3 t))[/asciimath] [asciimath]+9/5e^(-5t)([(-2),(5)] sin(3t)+[(1),(0)] cos(3 t))[/asciimath]
Try an Example
Section 6.7 Exercises
- Find a solution to the system of differential equations
[asciimath]bb"y"'=[(-7,-12),(6,5)] bb"y"[/asciimath]
Show/Hide Answer
[asciimath]bb"y"(t)=c_1e^(-t)([(-1),(1)]cos(6t)-[(1),(0)]sin(6t))+[/asciimath] [asciimath]c_2e^(-t)([(-1),(1)]sin(6t)+[(1),(0)]cos(6t))[/asciimath]
- Solve the system of differential equations with initial conditions
[asciimath]bb"y"' = [[4,2],[-29,-10]] \ bb"y", \quad \ bb"y"(0) = [(-2), (19)][/asciimath].
Show/Hide Answer
[asciimath]y_1(t)=-2 e^(-3t)cos(3t)+8 e^(-3t)sin(3t)[/asciimath]
[asciimath]y_2(t)=19 e^(-3t)cos(3t)-25 e^(-3t)sin(3t)[/asciimath]
- Solve the system of differential equations with initial conditions
[asciimath]bb"y"' = [[2,1],[-17,-6]] \ bb"y", \quad \ bb"y"(0) = [(-2), (11)][/asciimath].
Show/Hide Answer
[asciimath]y_1(t)=-2 e^(-2t)cos(t)+3 e^(-2t)sin(t)[/asciimath]
[asciimath]y_2(t)=11 e^(-2t)cos(t)-10 e^(-2t)sin(t)[/asciimath]
