Chapter 10 Linear Systems

8/18/2019 Chapter 10 Linear Systems

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10.1. Systems of Linear Differential Equations. Theory of Linear Systems.

Linear system of DEs:

x

1 = a11(t)x1(t) + a12(t)x2(t) + ... + a1n(t)xn(t) + f 1(t)

x

2 = a21(t)x1)(t) + a22(t)x2(t) + ... + a2n(t)xn(t) + f 2(t)

......

......

x

n = an1(t)x1(t) + an2(t)x2(t) + ... + ann(t)xn(t) + f n(t)

ai j(t), f i(t)-continuous on an interval I . If all free functions f i(t) = 0 , i = 1,...,n are equal to0 then the system is called homogeneous; otherwise non-homogeneous, i.e., if at least one of thefunctions f i(t) = 0, then the system is called non-homogeneous.

Matrix Form of a Linear system

X =

x1x2...xn

X

= d

dt

x1x2...xn

=

x

1

x

2

...x

n

F (t) =

f 1(t)f 2(t)...f n(t)

An×n(t) =

a11(t) a12(t) ... a1n(t)a21(t) a22(t) ... a2n(t)...

......

an1(t) an2(t) ... ann(t)

Remark. We shall consider mainly systems with a coefficients matrix An×n(t) having only constantterms, in other words all akj(t) will be constant functions, i.e., numbers.

X

= A X + F(t)

x

1

x

2...

x

n

=

a11 a12 ... a1na21 a22 ... a2n...

......

an1 an2 ... ann

x1x2...xn

+

f 1(t)f 2(t)...f n(t)

n × 1 n × n n × 1 n × 1Solution of a system on an interval I is a vector

X =

x1(t)x2(t)...xn(t)

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satisfying the system. It is also n × 1 one column matrix. The set of functions (or the vector)

(x1(t), x2(t), . . . , xn(t))

can be interpreted geometrically as one-parametric (t)-representation of a space curve in Rn.

Theorem. If the entries of the coefficient matrix A(t) and the free column vector F (t) are continuous functions on an interval I , containing t0, then the IVP:

X

= AX + F

X (t0) = X 0,

has a unique solution(x1(t), x2(t), . . . , xn(t))

on the interval I .

The initial value conditions

X (t0) = X 0in a non-matrix form are:

x1(t0) = γ 1; x2(t0) = γ 2; . . . ; xn(t0) = γ n,

where

X 0 =

γ 1γ 2...γ n

,

and the numbers γ 1, γ 2, . . . , γ n are n arbitrary chosen numbers.

Remark. In the case n = 2 the unique solution X = [x1(t), x2(t)] can be interpreted as a planecurve passing trough the point with coordinates (γ 1, γ 2).

In the case n = 3 the unique solution X = [x1(t), x2(t), x3(t)] can be interpreted as a space curvepassing trough the point with coordinates (γ 1, γ 2, γ 3).

The unique solutionX = (x1(t), x2(t), . . . , xn(t))

in the n-dimensional case can be interpreted as a curve in the n-dimensional space Rn passingthrough the point

[γ 1, γ 2, . . . , γ n] ∈ Rn.

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Superposition PrincipleLet X 1, X 2, .., X k be a set of solution to the homogeneous system

X

= An×n X.

Then any linear combination X = c1x1+ ...+ckX k, where c1, c2,...,ck are constants is also a solutionof the system.

Criterion for Linearly Independent Solutions Wronskian determinant:

Consider n solutions of the homogeneous linear system with n × n coefficient matrix:X

= An×n X :

X 1 =

x11x21...xn1

, X 2 =

x12x22...xn2

, · · · , X n =

x1nx2n...xnn

.

The solutions X 1, X 2, . . . , X n are linearly independent on an interval I if their Wronskian deter-minant:

W (X 1, X 2, . . . , X n) =

x11 x12 ... x1nx21 x22 ... x2n...

......

xn1 xn2 ... xnn

= 0.

distinct from zero for any t on the interval I .

Fundamental set of solutions. General Solution for a homogeneous system with n × ncoefficient matrix.(1) If X 1, X 2, ..., X n are n linearly independent solutions on an interval I then they form a funda-mental set of solutions.

Note that for a homogeneous system

X

= An×n X

with n × n coefficient matrix An×n;each fundamental set of solutions contains exactly n solutions.

(2) If X 1, X 2, ..., X n is a fundamental set of solutions for

X

= An×n X ,

thenX = c1X 1 + c2X 2 + · · · + cnX n

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is the set of all solutions, i.e., the general solution of the system. Here, c1, c2, . . . , cn arearbitrary constants. So, we have n-parameter family of solutions.

Problem 1. Write the linear system in matrix form:(a)

dx

dt = 3x − 5y

dy

dt = 4x + 8y.

Solution. x

y

=

3 −5

4 8

x

y

(b)

dx

dt = x − y

dydt

= x + 2z

dz

dt = −x + z .

Solution. x

y

z

=

1 −1 01 0 2

−1 0 1

xy

z

(c)

dx

dt = −3x + 4y + e−t

sin(2t)dy

dt = 5x + 9z + 4e−t cos(2t)

dz

dt = y + 6z − e−t .

Solution. x

y

z

=

−3 4 05 0 9

0 1 6

xy

z

+

e

−t sin(2t)4e−t cos(2t)−e−t

Problem 2. Verify that

X =

xy

z

=

sin(t)−1

2 sin(t) − 12 cos(t)− sin(t) + cos(t)

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is a solution of the given homogeneous linear system of DEs:

X

=

1 0 11 1 0

−2 0

−1

X .

Solution. cos(t)−1

2 cos(t) + 12 sin(t)− cos(t) − sin(t)

= ?

1 0 11 1 0

−2 0 −1

sin(t)−1

2 sin(t) − 12 cos(t)− sin(t) + cos(t)

=

1 · sin(t) + 0 ·

−12 sin(t) − 12 cos(t)

+ 1 · (− sin(t) + cos(t))

1 · sin(t) + 1 · −12 sin(t) − 12 cos(t) + 0 · (− sin(t) + cos(t))−2 · sin(t) + 0 · −12 sin(t) − 12 cos(t)− 1 · (− sin(t) + cos(t))

=

cos(t)−12 cos(t) + 12 sin(t)

−cos(t)

−sin(t)

.

General Solution of a Non-Homogeneous SystemConsider a non-homogeneous linear system of the form

X

= An×n X + F,

where An×n is the coefficient matrix having n rows and n columns and

F = F n×1 =

f 1(t)f 2(t)

...f n(t)

is the column of the free functions. Now suppose that

X c = c1X 1 + c2X 2 + · · · + cnX nis the general solution of the associated homogeneous system

X

= An×n X

and suppose that X p is a particular solution of the given non-homogeneous system. Then

X = Xc + Xp

is the general solution (the set of all solution) of the given non-homogeneous linearsystem of first-order DEs.Problem 3. Verify that

X p =

2−1

t +

5

1

=

2t + 5−t + 1

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is a particular solution of the non-homogeneous system

dx

dt = x + 4y + 2t + 7

dy

dt = 3x + 2y − 4t − 18Solution. Representing the system in matrix form

x

y

=

1 4

3 2

x

y

+

2t − 7−4t − 18

we verify the solution:

2−1

= ?

1 43 2

2t + 5−t + 1

+

2t − 7−4t − 18

=

2t + 5 − 4t + 46t + 15

−2t + 2

+

2t − 7

−4t

−18

=

2

−1

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10.2 Matrix Method to Solve Homogeneous Linear Systems. The D-Method (theMethod of Systematic Elimination) to Solve Linear Systems.

Consider a homogenous systemX

= An×n X .

The general solution of a homogeneous linear system is based on the eigenvalues and the eigenvectorsof the coefficient matrix An×n.(A) Distinct Real Eigenvalues. Suppose that the matrix An×n has n distinct real eigenvalues

λ1, λ2, . . . , λn.

Then, there exists a set of n linearly independent eigenvectors K 1 corresponding to λ1; K 2corresponding to λ2,...,K n corresponding to λn. Then

X 1 = K 1eλ1t, X 2 = K 2e

λ2t, . . . , X n = K 1eλnt

are n solutions of the system that form a fundamental set of solutions. Hence,

X = c1X 1 + c2X 2 + · · · + cnX nthat is

X = c1K 1eλ1t + c2K 2e

λ2t + · · · + cnK neλnt

is the general solution (the set of all solution) of the given homogeneous system.Important remark. Eigenvalue can be equal to 0. However, an eigenvector is always anon-zero vector, i.e, at least one component of an eigenvector must be distinct from 0.Many students have mistakes on this point when solving problems on linear systems of DEs.Problem 1. Find the general solution of the given system

X

=

10 −5

8 −12

X .

Solution. More detailed matrix form of the system

x

y

=

10 −5

8 −12

xy

.

First we compute the eigenvalues of the coefficient matrix.

det

10 − λ −5

8 −12 − λ

= λ2 + 2λ − 80 = 0 ⇒ λ1,2 = −1 ±√

81

⇒ λ1 = −10, λ2 = 8.Hence, we have the case of distinct real eigenvalues. Next we determine the eigenvectors:

λ1 = −10,

20 −58 −2

k1

k2

=

0

0

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4k1 − k2 = 0 ⇒ k1 = 1, k2 = 4 ⇒ K 1 =

k1k2

=

14

.

λ2 = 8,

2 −5

8

−20

k1

k2

=

0

0

2k1 − 5k2 = 0 ⇒ k1 = 5, k2 = 2 ⇒ K 1 =

k1k2

=

52

.

Hence, the general solution of the system is


λ2t

X = c1

14

e−10t + c2

52

e8t

X =

x

y

=

c1e−10t + 5c2e8t

4c1e−10t + 2c2e8t

or in a classical form the general solution is:

x(t) = c1e−10t + 5c2e8t

y(t) = 4c1e−10t + 2c2e

8t .

(B) Repeated Real Eigenvalues.Problem 2. Solve the following homogeneous system of first-order linear DEs:

x

y

z

=

1 −2 2−2 1 −2

2 −2 1

xy

z

.

Solution. First we compute the eigenvalues of the coefficient matrix:

det(A − λI ) = det 1

−λ

−2 2

−2 1 − λ −22 −2 1 − λ

= −(λ + 1)2(λ − 5) = 0.From here λ1 = λ2 = −1 (repeated eigenvalues) and λ3 = 5.First we find two linearly independent eigenvectors corresponding to the repeated eigenvalue λ1 =−1. Note that this it not always possible but in our concrete problem it is. That isbecause later we shall consider another method in the case of repeated eigenvalues.

2 −2 2−2 2 −2

2 −2 2

k1k2

k3

=

00

0

⇒ k1 − k2 + k3 = 0

and we compute two linearly independent eigenvectors:

K 1 =

11

0

, K 2 =

10

−1

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that give two linearly independent solutions:

X 1 = K 1eλ1t = K 1e

−t =

110

e−t, X 2 = K 2e

λ1t = K 2e−t =

10

−1

e−t

Next, we compute an eigenvector corresponding to the eigenvalue λ3 = 5: −4 −2 2−2 −4 −2

2 −2 −4

k1k2

k3

=

00

0

⇒

2k1 + k2 − k3 = 0k1 + 2k2 + k3 = 0

k1 − k2 − 2k3 = 0and from here

K 3 =

1

−11

and the general solution is:

X =

xy

z

= c1K 1e−t + c2K 2e−t + c3K 3e5t

= c1

11

0

e−t + c2

10

−1

e−t + c3

1−1

1

e5t

=

c1e

−t + c2e−t + c3e

5t

c1e−t − c3e5t

−c2e−t + c3e

5t

Second Method of finding General Solution in case of Repeated Eigenvalues. Supposethat we solve 2 × 2 homogeneous system and λ1 = λ2 is repeated eigenvalue. Then it is impossibleto find two linearly independent eigenvectors associated with the eigenvalue. There is only oneeigenvector K 1 (up to multiplication with a non-zero constant) associated with the eigenvalue λ1.Then, we solve the next linear algebraic system for the eigenvector K 1 and the vector P 1:

K 1 =

k1k2

and P 1 =

p1 p2

:

(A − λ1I ) K 1 = O(A − λ1I ) P 1 = K 1

The first equation in fact determines the eigenvector K 1 but the second equation determines anothervector P 1. Then,

X 1 = K 1eλ1t, X 2 = K 1te

λ1t + P 1eλ1t

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are two linearly independent solution and the general solution is given by

X = c1X 1 + c2X 2 = c1K 1eλ1t + c2

K 1te

λ1t + P 1eλ1t

.

Problem 3. Find the general solution of the given system of DEs. (Solve the given system.)

dx

dt = 3x − y

dy

dt = 9x − 3y

Solution. Matrix form: x

y

=

3 −19 −3

x

y

.

Coefficient matrix:

A2×2 =

3 −19 −3

.

Eigenvectors:det|A − λI | =

3 − λ −1

9 −3 − λ

= λ2 − 9 + 9 = λ2 = 0

and from here λ1 = λ2 = 0 is repeated eigenvalue. Find an eigenvector K 1 associated with theeigenvalue λ1 = 0:

3 − 0 −19 −3 − 0

k1

k2

=

00

⇒ 3k1 − k2 = 0 ⇒ k1 = 1, k2 = 3

K 1 =

13

3 − 0 −19 −3 − 0

p1

p2 = k1

k2

3 − 0 −19 −3 − 0

p1 p2

=

13

⇒

3 p1 − p2 = 19 p1 − 3 p2 = 3 ⇒ p1 = 1, p2 = 2 ⇒ P 1 =

12

From here we proceed to determine the general solution:

X =

x

y

= c1X 1 + c1X 2

X = c1K 1eλ1t

+ c2

K 1teλ1t

+ P 1eλ1t

X = c1

13

e0t + c2

1

3

te0t +

12

e0t

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X = c1

13

+ c2

13

t +

12

X =

c1 + c2(t + 1)

3c1 + c2(3t + 2)

or in classical form the general solution is:

x(t) = c1 + c2(t + 1)

y(t) = 3c1 + c2(3t + 2)

Problem 4. Find the general solution of the linear system

X

=

−1 3−3 5

X .

Solution 1 by using a matrix method based on eigenvalues and eigenvectors. First wecompute the eigenvalues:

det(A

−λI ) = det

−1 − λ 3

−3 5 − λ = λ2

−4λ

−5 + 9 = (λ

−2)2 = 0.

From here λ1 = λ2 = 2 and we have 2 repeated eigenvalues.Next, we determine eigenvectors associated with the repeated eigenvalue: −1 − 2 3

−3 5 − 2

k1k2

=

0

0

⇒

−3 3−3 3

k1

k2

=

0

0

⇒ −k1 + k2 = 0 ⇒ k1 = k2 = 1 ⇒ K 1 =

k1k2

=

1

1

Note that in this problem it is not possible to obtain another eigenvector that formsa linearly independent set of vectors together with K 1. That is because we have toproceed with computing the vector P1.

−3 3−3 3 p1 p2

=

11 ⇒ −3 p1 + 3 p2 = 1 ⇒ p1 = −1/3, p2 = 0

⇒ P 1 =

p1 p2

=

−130

In view of this

X 1 = K 1eλ1t =

1

1

e2t

X 2 = K 1teλ1t + P 1e

λ1t =

1

1

te2t +

−1/30

e2t

and the general solution of the given system is:

X = xy = c1X 1 + c2X 2 = c1K 1eλ1t + c2 K 1teλ1t + P 1eλ1t

= c1

11

e2t + c2

11

te2t +

−1/30

e2t

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or in classical form the general solution is

x(t) = c1e2t + c2

te2t − 1

3e2t

y(t) = c1e2t + c2te

2t,

where c1 and c2 are arbitrary constants.Solution 2 by using th D-Method. Consider the given linear system:

x

= −x + 3yy

= −3x + 5y,By using the D symbol the system is:

Dx = −x + 3yDy = −3x + 5y, ⇒

or equivalently,

(D + 1)x − 3y = 0 × (−3)3x + (D − 5)y = 0, ×(D + 1)

and after multiplying from the left-side the first equation by -3 and the second equation by ( D +1),we obtain:

−3(D + 1)x + 9y = 03(D + 1)x + (D + 1)(D − 5)y = 0 .

We sum both equations to eliminate the function x and obtain the following equation:

(D + 1)(D − 5)y + 9y = 0 ⇒ (D2 − 4D − 5)y + 9y = 0and going back to prime notation of a derivative we obtain a second-order DE in terms of y andsolve it:

y − 4y + 4y = 0

m2 − 4m + 4 = 0 ⇒ m1 = m2 = 2 repeated solutiony = c1e

2t + c2te2t .

We have determined y and here we proceed in a much more simple way than what is givenin the textbook. We express x from the second equation in terms of y and y

andreplacing them by the explicit expressions using that y = c1e

2t + c2te2t we obtain x:

x = −13

y

− 5y

= −1

3

2c1e

2t

+ c2e

2t

+ 2c2te

2t

− 5c1e2t

− 5c2te2t

= c1e2t + c2

te2t − 1

3e2t

.

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Finally, the general solution (the set of all solutions) of the given system is:

x(t) = c1e2t + c2

te2t − 1

3e2t

y(t) = c1e

2t

+ c2te

2t

.(C) Complex Eigenvalues. Suppose that we solve the homogeneous system

X

= A X,

where A is 2 × 2 coefficient matrix so, the system can written as:

x

y

=

a11 a12a21 a22

x

y

Note that the matrix A has real entries but it can have complex eigenvalues. Suppose that λ1 = α+i β is a complex eigenvalue (i2 = −1) and let K 1 be an eigenvector associated with λ1. Note that K 1has complex entries. Then, because A has only real entries the complex conjugate value of λ1:

λ2 = λ̄1 = α − iβ

will be the second eigenvalue of A and also the complex conjugate vector of K 1:

K 2 = K̄ 1

will be the eigenvector associated with λ2. Then,

X̃ 1 = K 1eλ1t, X̃ 2 = K 2e

λ2t = K̄ 1eλ̄1t

are two linearly independent but complex solutions of the system and the general solu-tion in complex form is given by

X = c1 X̃ 1 + c2 X̃ 2

⇒ X = c1K 1eλ1t + c2K 2eλ2t⇒ X = c1K 1eλ1t + c2 K̄ 1eλ̄1t.

From a complex representation of the general solution to a real representation of thegeneral solution:Method 1 from a complex to a real solution described in the textbook [5], also. Formthe vectors

B1 = 1

2

K 1 + K̄ 1

, B2 =

i

2

−K 1 + K̄ 1 ,where K 1 is the eigenvector associated with the eigenvalue λ1 = α + i β . Then

X 1

= [B1

cos(βt)−

B2

sin(βt)] eαt

X 2 = [B2 cos(βt) + B1 sin(βt)] eαt

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are two linearly independent real solutions of the system and the general solution inreal form will be:

X = c1 X 1 + c2 X 2.

Note that in fact

B1 = Re (K 1) , B2 = I m (K 1) .Method 2 from a complex to a real solution. It is similar to Method 1 but does notneed any memorizing. Here it is. Consider one of the two complex solutions K 1e

λ1t or K 2eλ2t.

For exampleK 1e

λ1t.

Then,X 1 = Re

K 1e

λ1t

, X 2 = I m

K 1eλ1t

are two real solution of the system that are linearly independent and the general solution in realform will be:

X = c1 X 1 + c2 X 2.

Hence, we have to separate the real part and the imaginary part of one of the complex solutions

K 1eλ1t or K 2eλ2t in order to obtain a real fundamental set of solutions.Problem 5. Find the general solution of the system

x

= 6x − yy

= 5x + 2y .

Solution 1 by using a matrix mathod based on eigenvalues and eigenvectors. Matrixform of the system:

x

y

=

6 −15 2

x

y

.

The eigenvalues:

det(A

−λI ) = det

6 − λ −15 2 − λ

= λ2

−8λ + 17 = 0.

λ1,2 = 4 ±√

16 − 17 = 4 ± i ⇒λ1 = 4 + i, λ2 = λ̄1 = 4 − i

and we have two complex eigenvalues. Next, we determine eigenvectors associated with the eigen-values:

6 − (4 + i) −15 2 − (4 + i)

k1k2

=

00

⇒

2 − i −1

5 −2 − i

k1k2

=

00

⇒ (2 − i)k1 − k2 = 0 ⇒ k1 = 1, k2 = 2 − i ⇒ K 1 =

k1k2

=

12 − i

Then, the eigenvector associated with λ2 = λ̄1 = 4−

i is

K 2 =

12 + i

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The general solution in complex form is:


λ2t

that is

X = c1

12 − i

e(4+i)t + c2

12 + i

e(4−i)t .

From complex to real general solution by using Method 1.

B1 = Re(K 1) =

1

2

, B2 = I m(K 1) =

0−1

.

Also, λ1 = 4 + i ⇒ α = 4, β = 1. Then,

X 1 = [B1 cos(βt) − B2 sin(βt)] eαt

=

12

cos(t) −

0−1

sin(t)

e4t

andX 2 = [B2 cos(βt) + B1 sin(βt)] e

αt

=

0−1

cos(t) +

12

sin(t)

e4t

and the general solution in real form is:

X =

x

y

= c1X 1 + c2X 2

c1

cos(t)

2 cos(t) + sin(t)

e4t + c2

sin(t)

2sin(t) − cos(t)

e4t .

and in classical form

x(t) = [c1 cos(t) + c2 sin(t)] e4t

y(t) = [c1(2 cos(t) + sin(t)) + c2(2 sin(t) − cos(t))] e4t

From complex to real general solution by using Method 2. We have

K 1eλ1t =

12 − i

e(4+i)t =

eit

(2 − i)eit

e4t .

=

cos(t) + i sin(t)

(2 − i)(cos(t) + i sin(t))

e4t =

cos(t) + i sin(t)

2cos(t) + sin(t) + i(2 sin(t) − cos(t))

e4t

and separating the real and the imaginary parts gives 2 real linearly independent solutions:

X 1 =

cos(t)

2cos(t) + sin(t)

e4t

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and

X 2 =

sin(t)

2sin(t) − cos(t)

e4t

The general solution in real form is

X = c1X 1 + c2X 2and we complete the solution the same way as when applying Method 1. As you can see the secondmethod does not need any memorizing.In the process of applying Method 2 we have used the following Euler’s formula (see http://mathworld.wolframSuppose that g(t) is a real valued function. Then,

ei g(t) = cos(g(t)) + i sin(g(t)).

Solution 2 of the above problem by using the D-method. Consider the linear system

x

= 6x − yy

= 5x + 2y .

We use the symbol D to denote derivatives, for example: Dy = y

, DDy = D2y = D(Dy) =D(y

) = y

, (D − 6)y = y − 6y. From here the linear system

x

= 6x − yy

= 5x + 2y .

can be written in a D-form as it follows:

(D − 6)x + y = 0−5x + (D − 2)y = 0 .

Our goal is to obtain a DE either only in terms of x or in terms of y and to solve it (in other wordsto exclude one of the dependent variables). Multiplying the first equation by

−(D

−2) we obtain:

−(D − 2)(D − 6)x − (D − 2)y = 0−5x + (D − 2)y = 0

and summing both equations:

−(D − 2)(D − 6)x − (D − 2)y + (D2)y − 5x = 0−(D − 2)(D − 6)x − 5x = 0−(D2 − 8D + 12)x − 5x = 0(D2 − 8D + 12)x + 5x = 0D2x − 8Dx + 12x + 5x = 0x

−8x

+ 17x = 0

m2 − 8m + 17 = 0 ⇒ m1,2 = 4 ± √ 16 − 17 = 4 ± ix(t) = c1e

4t cos(t) + c2e4t sin(t)

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Hence,

x(t) = c1e4t cos(t) + c2e

4t sin(t)

x(t) = e4t [c1 cos(t) + c2 sin(t)] .

From the first equation:

y = 6x − x

y = 6c1e4t cos(t) + 6c2e

4t sin(t)

−[4c1e4t cos(t) − c1e4t sin(t) + 4c2e4t sin(t) + c2e4t cos(t)]y = 6c1e

4t cos(t) + 6c2e4t sin(t)

−4c1e4t cos(t) + c1e4t sin(t) − 4c2e4t sin(t) − c2e4t cos(t)y = c1(2 cos(t) + sin(t))e

4t + c2(2 sin(t) − cos(t))e4t

Problem 6. Find the general solution of the system

dx

dt

= 2x + 8y

dy

dt = −x − 2y .

Solution 1 by using a matrix method based on eigenvalues and eigenvectors. Matrix formof the system:

x

y

=

2 8−1 −2

x

y

.

The eigenvalues:

det(A − λI ) = det 2 − λ 8−1 −2 − λ

= λ2 + 4 = 0.λ1,2 = ± 2 i ⇒

λ1

=−

2 i, λ2

= λ̄1

= 2 i

and we have two complex eigenvalues. Next, we determine eigenvectors associated with the eigen-values. Consider λ1 = −2 i.

2 + 2i 8−1 −2 + 2i

k1

k2

=

00

⇒ −k1 + (−2 + 2i)k2 = 0 ⇒ k1 = −2 + 2i, k2 = 1 ⇒ K 1 =

k1k2

=

−2 + 2i1

Then, the eigenvector associated with λ2 = λ̄1 = 2i is

K 2 =

−2 − 2i1

The general solution in complex form is:


λ2t

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that is

X = c1

−2 + 2i1

e−2it + c2

−2 − 2i1

e2it .

From complex to real general solution by using Method 1.

B1 = Re(K 1) = −2

1

, B2 = I m(K 1) =

20

.

Also, λ1 = −2i ⇒ α = 0, β = −2. Then,X 1 = [B1 cos(βt) − B2 sin(βt)] eαt

=

−21

cos(−2t) −

20

sin(−2t)

e0t

andX 2 = [B2 cos(βt) + B1 sin(βt)] e

αt

= 2

0 cos(−

2t) + −2

1 sin(−

2t) e0t

and the general solution in real form is (note that cos(◦) is an even function and sin(◦) is an oddfunction, i.e., cos(−2t) = cos(2t), sin(−2t) = − sin(2t)):

X =

x

y

= c1X 1 + c2X 2

c1

−2 cos(2t) + 2 sin(2t)cos(2t)

+ c2

2 cos(2t) + 2 sin(2t)

− sin(2t)

.


x(t) = 2c1(sin(2t) − cos(2t)) + 2c2(cos(2t) + sin(2t))y(t) = c1 cos(2t)

−c2 sin(2t)

From complex to real general solution by using Method 2. We have

K 1eλ1t =

−2 + 2i1

e−2it =

(−2 + 2i)e−2it

e−2it

.

=

(−2 + 2i)(cos(2t) − i sin(2t))

cos(−2t) + i sin(−2t))

=

2 sin(2t) − 2 cos(2t) + i(2 cos(2t) + 2 sin(2t)

cos(2t) − i sin(2t)

We used in the above computations the fact that cos(◦) is an even function and sin(◦) is an oddfunction. Separating the real part and the imaginary part gives 2 real linearly independent solutions:

X 1 =

2 sin(2t) − 2 cos(2t)

cos(2t)

and

X 2 =

2 cos(2t) + 2 sin(2t)

− sin(2t)

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The general solution in real form is

X =

x

y

= c1X 1 + c2X 2

c1 2 sin(2t) −

2 cos(2t)

cos(2t)

+ c2 2 cos(2t) + 2 sin(2t)− sin(2t) .


x(t) = 2c1(sin(2t) − cos(2t)) + 2c2(cos(2t) + sin(2t))y(t) = c1 cos(2t) − c2 sin(2t)

Method 2 to obtain a real general solution does not need any memorizing of readyformulas.Remark. In the process of applying Method 2 we have used the following Euler’s formula (seehttp://mathworld.wolfram.com/EulerFormula). Suppose that g(t) is a real valued function.Then,

ei g(t) = cos(g(t)) + i sin(g(t)).

Solution 2 of the above problem by using the D-method. Consider the linear system

x

= 2x + 8y

y

= −x − 2y .We use the symbol D to denote derivatives, for example: Dx = x

, DDx = D(Dx) = D2x =D(x

) = x

, (D − 2)x = x − 2x. From here the linear systemx

= 2x + 8y

y

= −x − 2y .can be written in a D-form as it follows:

(D − 2)x − 8y = 0x + (D + 2)y = 0 .

Our goal is to obtain a DE either in terms of x or in terms of y and to solve it. Multiplying thesecond equation by −(D − 2) we obtain:

(D − 2)x − 8y = 0−(D − 2)x − (D − 2)(D + 2)y = 0

and summing both equations yields:

−(D − 2)(D + 2)y − 8y = 0−(D2 − 4)y − 8y = 0−D2y + 4y − 8y = 0D2y + 4y = 0

y

+ 4y = 0m2 + 4 = 0 ⇒ m2 = −4 ⇒ m = ±√ −4 ⇒ m1,2 = ± 2iy(t) = c1 cos(2t) + c2 sin(2t)

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Hence,

y(t) = c1 cos(2t) + c2 sin(2t)

From the second equation of the system we obtain x(t):

x = −2y − y

x = −2c1 cos(2t) − 2c2 sin(2t) + 2c1 sin(2t) − 2c2cos(2t)x(t) = 2c1(sin(2t) − cos(2t)) − 2c2(cos(2t) + sin(2t)).

and the general solution of the system is:

x(t) = 2c1(sin(2t) − cos(2t)) − 2c2(cos(2t) + sin(2t))y(t) = c1 cos(2t) + c2 sin(2t) .

The form of the general solution is a bit different from that obtained in the Solution 1. However, c1and c2 are arbitrary constants and replacing c2 by −c2 will give the same as in the Solution 1 formof the general solution.

Here we solve a non-homogeneous system by using the D- method.

Problem 7. Find the general solution of the system

dx

dt = −y + 2

dy

dt = x + 4 .

Solution by the D-Method. By using the symbol Dy = y

the system is

Dx + y = 2

−x + Dy = 4 .Now our goal is to eliminate x in order to obtain a DE only in terms of y . Multiplying the secondequation from left by D we obtain:

Dx + y = 2

−Dx + D2y = D4 ⇒ −Dx + D2y = 0,

observing that D4 = [4]

= 0. Summing both equation eliminates x and we obtain a second-orderDE in terms of y :

D2y + y = 2 ⇒ y + y = 2.We solve the above non-homogeneous DE. First we find the general solution yc(t) of the associated

homogeneous DE: y

+ y = 0 ⇒ m2 + 1 = 0 ⇒ m1,2 = ±i⇒ yc(t) = c1 cos(t) + c2 sin(t).

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By using the method of undetermined coefficients we are looking for y p in the form y p = A:

[A]

+ A = 2 ⇒ A = 2.

Hence,

y(t) = yc(t) + y p(t) = c1 cos(t) + c2 sin(t) + 2.

Expressing x(t) in terms of y(t) by using the second equation we obtain:

x(t) = y

(t) − 4 = −c1 sin(t) + c2 cos(t) − 4

or in matrix form:

X = X c + X p

=

x

y

= c1

− sin(t)cos(t)

+ c2

cos(t)

sin(t)

+

−42

.

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10.4. Non-Homogeneous Linear Systems of Differential Equations

Given a linear system of DEs:

x

1 = a11x1(t) + a12x2(t) + ... + a1nxn(t) + f 1(t)

x2 = a21x1)(t) + a22x2(t) + ... + a2nxn(t) + f 2(t)

......

......

x

n = an1x1(t) + an2x2(t) + ... + annxn(t) + f n(t) .

If all free functions f i(t) = 0 , i = 1,...,n are equal to 0 then the system is called homogeneous;otherwise non-homogeneous, i.e., if at least one of the functions f i(t) = 0, then the system iscalled non-homogeneous.

Matrix Form of Linear System

X

= A X + F (t),

where

X =

x1x2

...xn

X

= d

dt

x1x2

...xn

=

x

1x

2...

x

n

F (t) =

f 1(t)f 2(t)...f n(t)

A =

a11 a12 ... a1na21 a22 ... a2n...

......

an1 an2 ... ann

.

General solution of a non-homogenous linear system. Given a non-homogeneous linearsystem

X

= A X + F (t).

If X c is the general solution of the associated homogenous system (see topics 10.2 and 10.3):

X

= A X

and X p is a particular solution of the given non-homogeneous system, then

X = X c + X p

is the general solution (the set of all solutions) of the given non-homogenous system. Inthe previous section we study how to find the general solution of a homogeneous system (see topics10.2 and 10.3). Here we shall exercise these method together with methods to find a particularsolution.Problem 1. Find the general solution of the system

dx

dt = 2x + 3y − e2t

dy

dt = −x − 2y + e2t

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Here is the system in matrix form:

x

y

=

−2 3−1 −2

x

y

+

−e2te2t

.

Solution 1 of Problem 1 by using a matrix method to solve the associated homogeneoussystem and the method of undetermined coefficients to find a particular solution of thegiven system.(1) First, we find the general solution of the associated homogeneous system by using a matrixmethod based on eigenvalues and eigenvectors.We solve the associated homogeneous system:

x

y

=

−2 3−1 −2

x

y

.

We apply a matrix method based on eigenvalues and eigenvectors.First we compute the eigenvalues of the coefficient matrix.

det 2 − λ 3−1 −2 − λ = λ2 − 1 = 0 ⇒ λ1,2 = ± 1

⇒ λ1 = −1, λ2 = 1.Hence, we have the case of distinct real eigenvalues. Next we determine the eigenvectors:

λ1 = −1,

3 3−1 −1

k1k2

=

0

0

k1 + k2 = 0 ⇒ k1 = 1, k2 = −1 ⇒ K 1 =

k1k2

=

1−1

.

Then,

X 1 = K 1eλ1 t =

1

−1

e−t

is a solution of the associated homogeneous system.

λ2 = 1,

1 3−1 −3

k1

k2

=

00

k1 + 3k2 = 0 ⇒ k1 = 3, k2 = −1 ⇒ K 2 =

k1k2

=

3−1

.

Then,

X 2 = K 2eλ2 t =

3−1

et

is a solution of the associated homogeneous system.The general solution X c of the associated homogeneous system is

X c = c1X 1 + c2X 2

X c = c1K 1eλ1t + c2K 2e

λ2t

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X c =

xc

yc

= c1

1−1

e−t + c2

3−1

et .

or written in a classical form:

xc(t) = c1

e−t + 3c2

et

yc(t) = −c1e−t − c2et

(2). Next, we find a particular solution X p of the given non-homogeneous system byusing the Method of Undetermined Coefficients.The main step here is to determine the preliminary form of the particular solution.Here is a method that does not need any memorizing. Consider the free functions and itsderivatives:

f 1(t) = −e2t, f 1(t) = −2e2t, f

1 (t) = −4e2t, . . .f 2(t) = e

2t, f

1(t) = 2e2t, f

1 (t) = 4e2t, . . .

We observe that f 1 and f 2, and all their derivatives have the form Ce2t, where C is a

number (constant). Then, we look for particular solutions x p, y p in the same form:

x p = Ae2t, y p = Be

2t (A, B numbers).

Note that there is no duplication of x p and y p with the general solution of the associated homogenoussystem.Then, looking for x p and y p we compute:

x

p = 2x p + 3y p − e2t

y

p = −x p − 2y p + e2t

and the above system with x p = Ae2t, y p = Be

2t takes the form:

2Ae2t = 2Ae2t + 3Be2t − e2t ⇒ (3B − 1)e2t = 0 ⇒ 3B − 1 = 02Be

2t

= −Ae2t

− 2Be2t

+ e2t

⇒ (A + 4B − 1)e2t

= 0 ⇒ A + 4B − 1 = 0and from here B = 1/3 and A = −4B + 1 = −4/3 + 1 = −1/3. Then

x p(t) = −13

e2t, y p(t) = 1

3e2t

and a particular solution is:

X p =

x p

y p

=

−13

e2t13e

2t

.

(3) The general solution of the system in matrix form:

X =

x

y

= X c + X p = c1K 1e

λ1 t + c2K 2eλ2 t + X p

= c1

1−1

e−t + c2

3−1

et +

−13e2t13

e2t

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⇒ λ1 = −1, λ2 = 1.Hence, we have the case of distinct real eigenvalues. (2) Next we determine the correspondingeigenvectors:

λ1 = −1, 3 3−1 −1 k1k2 = 00

k1 + k2 = 0 ⇒ k1 = 1, k2 = −1 ⇒ K 1 =

k1k2

=

1−1

.

λ2 = 1,

1 3−1 −3

k1

k2

=

00

k1 + 3k2 = 0 ⇒ k1 = 3, k2 = −1 ⇒ K 2 =

k1k2

=

3−1

.

(3)Form the matrix P = (K 1 K 2) and find the inverse matrix P −1.

P = (K 1 K 2) = 1 3

−1 −1 , P −1 =

1

2

−1 −31 1

= −1/2 −3/2

1/2 1/2

(4) With X = P Y that is Y = P −1X , compute P −1F and solve the new uncoupled system:

Y

= DY + P −1F.

P −1F =

−1/2 −3/21/2 1/2

−e2te2t

=

−e2t0

The new uncoupled system:

y

1

y

2

=

−1 00 1

y1y2

+

−e2t0

The new system in classical form:

y

1 = −y1 − e2ty

2 = y2

and these are two independent linear first-order DEs. We solve them by using the Method of Integrating factor.Consider:

y

1 = −y1 − e2t ⇒ y

1 + y1 = −e2t, p(t) = 1, e p(t)dt = e

1dt = et

[ety1]

= −e3t ⇒ ety1 = −

e3tdt ⇒ ety1 = −13

e3t + c1

y1 = −13

e2t + c1e−t ⇒ y1 = c1e−t − 1

3e2t.

The second equation is separable, also:dy2y2

= dt ⇒

dy2y2

=

dt ⇒ ln |y2| = t + c2

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|y2| = et+c2 = c2et (c2 > 0) ⇒ y2 = c2et c2 an arbitary constant(5) Find the general solution of the given system by using the formula:

X = x

y = P Y = P

y1

y2 .

X =

x

y

=

1 3−1 −1

c1e

−t − 13e2tc2e

t

=

c1e

−t + 3c2et − 13e2t−c1e−t − c2et + 13e2t

.

= c1

1−1

e−t + c2

3−1

et +

−13e2t13

e2t

The general solution of the system in classical form:

x(t) = c1e−t + 3c2e

t − 13

e2t

y(t) = −c1e−t − c2et + 13

e2t .

Solution 3 of Problem 1 by using the D-Method:

(D − 2)x − 3y = −e2tx + (D + 2)y = e2t .

Multiply the second equation by (D − 2) on the left:(D − 2)x − 3y = −e2t(D

−2)x + (D

−2)(D + 2)y = (D

−2)e2t = 2e2t

−2e2t = 0 .

Subtract the first equation from the second in order to eliminate x:

(D − 2)(D + 2)y + 3y = e2t ⇒ (D2 − 4)y + 3y = e2t ⇒ y − y = e2t.

Solve the DE y − y = e2t to find y(t):

m2 − 1 = 0, m1,2 = ± 1, yc = c1e−t + c2et

Find y p by using undetermined coefficients.

f (t) = e2t, f

(t) = 2e2t, . . . ⇒ y p = Ae2t not duplication with yc

y

p − y p = 4Ae2t

− Ae2t

= 3Ae2t

= e2t

⇒ A = 1

3

y p = 1

3e2t.

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Then

y = yc + y p = c1e−t + c2e

t + 1

3e2t.

From the second equation:

x = e2t − y − 2y = e2t + c1e−t − c2et − 23

e2t − 2c1e−t − 2c2et − 23

e2t

x = −c1e−t − 3c2et − 13

e2t.

Hence, the general solution of the given non-homogeneous system is:

x(t) = −c1e−t − 3c2et − 13

e2t

y(t) = c1e−t + c2e

t + 1

3e2t .

(Note that substituting c1 → −c1 and c2 → −c2 being arbitrary constants we obtain the same as inthe previous Solutions form of the general solution.)Problem 2. Solve the linear system by using a matrix method

x

= 4y + sin(t)

y

= −x + cos(t) .

Solution. (A) First we find the general solution X c of the associated homogeneous system by usinga matrix method based on eigenvalues and eigenvectors:

x

= 4y

y

= −x .

Matrix form of the associated homogeneous system:

x

y

=

0 4−1 0

x

y

We compute the eigenvalues of the coefficient matrix.

det

−λ 4−1 −λ

= λ2 + 4 = 0 ⇒ λ1,2 = ± 2 i

⇒ λ1 = −2i, λ2 = 2i.Hence, we have the case of distinct real eigenvalues. Next we determine the eigenvectors:

λ1 = −2i,

2i 4

−1 2i

k1k2

=

00

2ik1 + 4k2 = 0, −k1 + 2ik2 = 0 ⇒ k1 = 2i, k2 = 1 ⇒ K 1 =

k1k2

=

2i

1

.

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Then,

X̃ 1 = K 1eλ1 t =

2i

1

e−2it

is a complex solution of the associated homogeneous system.

λ2 = λ̄1 = 2i

K 2 = K̄ 1 =

−2i1

.

Then,

X̃ 2 = K 2eλ2 t =

−2i1

e2it

is a complex solution of the associated homogeneous system. The general solution X c of theassociated homogeneous system in complex form is

X c = c1 X̃ 1 + c2 X̃ 2

X c = c1K 1eλ1t + c2K 2e

λ2t

X c = c1

2i

1

e−2it + c2

−2i1

e2it

We find X c in a real form. In order to do this we propose two methods. The two meth-ods are similar but the second one is more straight and does not need any memorizing.

Method 1 to find a real form of X c. It can be found also in the textbook: Considerλ1 = −2i = α + iβ , where α = 0 and β = −2 and the associated eigenvector vector K 1. Then

B1 = Re(K 1) =

0

1

, B2 = I m(K 1) =

20

Then,

X 1 = [B1 cos(−2t) − B2 sin(−2t)] e0t =

01

cos(2t) +

20

sin(2t) =

2 sin(2t)

cos(2t)

is a real solution of the homogeneous system and

X 2 = [B2 cos(−2t) + B1 sin(−2t)] e0t =

20

cos(2t) −

01

sin(2t) =

2 cos(2t)− sin(2t)

is another real solution. X 1 and X 2 are two linearly independent solutions of the associated homo-geneous linear system. Then the general solution of the associated homogeneous linear system inreal form:

X c = c1X 2 + c2X 2 = c1

2 sin(2t)

cos(2t)

+ c2


.

In the above considerations we have used the fact the sin(−x) = − sin(x), i.e., sin(x) is an oddfunction; and cos(−x) = cos(x), i.e., cos(x) is an even function.

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Method 2 to find a real form of X c. Consider one of the complex solutions of the associatedhomogeneous system:

X̃ 1 = K 1eλ1 t =

2i

1

e−2it

Then, X 1 = Re

K 1eλ1 t

, X 2 = I m

K 1e

λ1 t

are two real linearly independent solutions of the associated homogeneous system. We compute byusing the Euler’s formula: e−2it = cos(2t) − i sin(2t).

2i

1

e−2it =

2i[cos(2t) − i sin(2t)]

1[cos(2t) − i sin(2t)]

=

2 sin(2t) + 2i cos(2t)

cos(2t) − i sin(2t)]

=

2 sin(2t)

cos(2t)

+ i


From here:

X 1 = Re K 1eλ1 t = 2 sin(2t)

cos(2t) , X 2 = I m K 1eλ1 t =

2 cos(2t)

− sin(2t) are two real linearly independent solutions of the associated homogeneous system. Hence,

X c = c1X 2 + c2X 2 = c1

2 sin(2t)

cos(2t)

+ c2


.

(B) We find a particular solution X p of the given non-homogeneous system by using the Method of Undetermined Coefficients. Consider the free functions and their derivatives:

f 1(t) = sin(t), f

1(t) = cos(t), f

1 (t) = sin(t), . . .

f 2(t) = cos(t), f

1(t) = − sin(t), f

1 (t) = − cos(t), . . .Then, the free functions and their derivatives have the form: C 1 sin(t) + C 2 cos(t) and this will be

the form to look for a particular solutions:

X p =

x p

y p

=

A sin(t) + B cos(t)C sin(t) + D cos(t)

Computing with x p = A sin(t) + B cos(t) and y p = C sin(t) + D cos(t) in the given system gives:

x

p = 4y p + sin(t)

y

p = −x p + cos(t) .that is

A cos(t) − B sin(t) = 4C sin(t) + 4D cos(t) + sin(t)

⇒ (A − 4D) cos(t) − (4C + B + 1) sin(t) = 0C cos(t) − D sin(t) = −A sin(t) − B cos(t) + cos(t)⇒ (A − D)sin(t) + (C + B − 1)cos(t) = 0 .

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and from here A −4D = 0, A−D = 0 ⇒ A = D = 0, 4C + B = −1, C + B = 1 ⇒ C = −2/3, B =5/3. Hence,

x p = A sin(t) + B cos(t) = 5

3 cos(t), y p = C sin(t) + D cos(t) = −2

3 sin(t).

(C) Construct the general solution. Matrix form of the general solution:

X =

x

y

= X c + X p = c1

2 sin(2t)

cos(2t)

+ c2


+

5

3 cos(t)−23 sin(t)

Classical form of the general solution:

x = 2c1 sin(2t) + 2c2 cos(2t) + 5

3 cos(t)

y = c1 cos(2t) − c2 sin(2t) − 23

sin(t) .

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Chapter 10 Linear Systems

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