Lecture Notes 3

1. Systems of first order linear ODE’s

A system of linear ODE’sx′1(t) = p11(t)x1(t) + p12(t)x2(t) + p1n(t)xn(t) + g1(t),

x′1(t) = p21(t)x1(t) + p22(t)x2(t) + p2n(t)xn(t) + g2(t)),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

x′1(t) = pn1(t)x1(t) + pn2(t)x2(t) + pnn(t)xn(t) + gn(t),

(1)

where x1(t), ..., xn(t) are unknown functions, pij(t), gj(t), i, j = 1, ...n aregiven functions, is called a nonhomogeneous linear system of ODE’s.If P (t),x(t),g(t) denote respectively

P (t) =

p11(t) p12(t) . . . p1n(t)p21(t) p22(t) . . . p2n(t). . . . . . . . . . . .pn1(t) pn2(t) . . . pnn(t)

,

x(t) =

x1(t)x2(t)

...xn(t)

, g(t) =

g1(t)g2(t)

...gn(t)

,then the system (1) can be written in the following form

x′(t) = P (t)x(t) + g(t). (2)

The system of equations

x′(t) = P (t)x(t) (3)

is called the homogeneous system of linear ODE’s. In what follows we as-sume that all functions pij(t), gj(t), i, j = 1, · · · , n are continuous functionson their domains of definition.

Theorem 1.1. If the the vector functions

x(1)(t) =

x11(t)x21(t)

...xn1(t)

, and x(2)(t) =

x12(t)x22(t)

...xn2(t)

1

2

are solutions of the homogeneous system (3) then each linear combinationof this vector functions C1x

(1)(t) +C2x(2)(t) is also a solution of the homo-

geneous system (3).

Theorem 1.2. If the functions pij(t), gj(t), i, j = 1, · · · , n are continuouson some interval (a, b) and t0 ∈ (a, b), then the initial value problem{

x′(t) = P (t)x(t) + g(t),

x(t0) = x0

(4)

has a unique solution on the interval (a, b).

Theorem 1.3. If the functions pij(t), gj(t), i, j = 1, · · · , n are continuouson some interval (a, b), t0 ∈ (a, b) and the vector functions

x(1)(t) =

x11(t)x21(t)

...xn1(t)

, · · · , x(n)(t) =

x1n(t)x2n(t)

...xnn(t)

are solutions of the system (3) such that∣∣∣∣∣∣∣∣

x11(t0) x12(t0) . . . x1n(t0)x21(t0) x22(t0) . . . x2n(t0). . . . . . . . . . . .

xn1(t0) xn2(t0) . . . xnn(t0)

∣∣∣∣∣∣∣∣ 6= 0.

Then the general solution of the system (3) has the form

x(t) = C1x(1)(t) + C2x

(2)(t) + · · ·+ Cnx(n)(t).

The determinant

W [x(1),x(2), · · · ,x(n)](t) :=

∣∣∣∣∣∣∣∣x11(t) x12(t) . . . x1n(t)x21(t) x22(t) . . . x2n(t). . . . . . . . . . . .

xn1(t) xn2(t) . . . xnn(t)

∣∣∣∣∣∣∣∣is called the Wronskian of the vector functions

x(1)(t),x(2)(t), · · · ,x(n)(t).

The following theorem is the analog of Abel’s theorem for solutions ofsecond order linear homogeneous equations.

3

Theorem 1.4. (Abel’s theorem) If the vector functions

x(1)(t),x(2)(t), · · · ,x(n)(t)

are solutions of the homogeneous system (3) on some interval (a, b) andt0 ∈ (a, b), then

W [x(1), · · · ,x(n)](t) = W [x(1), · · · ,x(n)](t0)e∫ t

0(p11(τ)+···+pnn(τ))dτ . (5)

The proof of Abel’s theorem follows from the following lemma:

Lemma 1.5. If xij, i, j = 1, ..., n are continuously differentiable functions,then

d

dt

∣∣∣∣∣∣∣∣x11(t) x12(t) . . . x1n(t)x21(t) x22(t) . . . x2n(t). . . . . . . . . . . .

xn1(t) xn2(t) . . . xnn(t)

∣∣∣∣∣∣∣∣ =

∣∣∣∣∣∣∣∣x′11(t) x′12(t) . . . x′1n(t)x21(t) x22(t) . . . x2n(t). . . . . . . . . . . .

xn1(t) xn2(t) . . . xnn(t)

∣∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣∣x11(t) x12(t) . . . x1n(t)x′21(t) x′22(t) . . . x′2n(t). . . . . . . . . . . .

xn1(t) xn2(t) . . . xnn(t)

∣∣∣∣∣∣∣∣+ · · ·+

∣∣∣∣∣∣∣∣x11(t) x12(t) . . . x1n(t)x21(t) x22(t) . . . x2n(t). . . . . . . . . . . .

x′n1(t) x′n2(t) . . . x′nn(t)

∣∣∣∣∣∣∣∣ .The Abel’s theorem implies that if

x(1)(t),x(2)(t), · · · ,x(n)(t)

are solutions of the system (3) on the interval (a, b) and for some t0 ∈ (a, b)the vectors

x(1)(t0),x(2)(t0), · · · ,x(n)(t0)

are linearly independent, then

x(1)(t),x(2)(t), · · · ,x(n)(t)

are linearly independent for each t ∈ (a, b).

Definition 1.6. A set of solutions

x(1)(t),x(2)(t), · · · ,x(n)(t) (6)

of the system (3) is called a fundamental set of solutions of the system (3)on (a, b) if (6) is linearly independent for each t ∈ (a, b).

4

It is easy to see that if x(1)(t) and x(2)(t) are solutions of the nonhomo-geneous system (2), then the vector function x(t) = x(1)(t) − x(2)(t) is asolution of the corresponding homogeneous system (3).Therefore a general solution of the nonhomogeneous system (3) has theform

x(t) = C1x(1)(t) + C2x

(2)(t) + · · ·+ Cnx(n)(t) + xp(t), (7)

where x(1)(t),x(2)(t), · · · ,x(n)(t) is a fundamental set of solutions of thehomogeneous system (3), C1, · · · , Cn are arbitrary constants and xp(t) issome particular solution of the nonhomogeneous system (2).

Definition 1.7. A matrix,

Ψ(t) :=

x11(t) x12(t) . . . x1n(t)x21(t) x22(t) . . . x2n(t). . . . . . . . . . . .

xn1(t) xn2(t) . . . xnn(t)

(8)

where

x(1)(t) =

x11(t)x21(t)

...xn1(t)

, and x(2)(t) =

x12(t)x22(t)

...xn2(t)

is a fundamental set of solutions of the homogeneous system (3) is calleda fundamental matrix of the system (3).Hence (7) can be written in the form

x(t) = Ψ(t)C + xp(t), (9)

where C is arbitrary constant vector.

It is not difficult to show that the fundamental matrix of (3) satisfies thesame equation, i.e.

Ψ′(t) = P (t)Ψ(t). (10)

Problem 1.8. Show the equality (10) for the system (3) when n = 2, i.e.for system of two linear ODE’s.

5

2. Homogeneous Systems of Linear ODE’s with constantcoefficients

A system of differential equationsx′1(t) = a11x1(t) + a12x2(t) + a1nxn(t),

x′2(t) = a21x1(t) + a22x2(t) + a2nxn(t),

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

x′n(t) = an1x1(t) + an2x2(t) + annxn(t),

(11)

where x1(t), ..., xn(t) are unknown functions, aij, i, j = 1, ...n are givenconstants, is called a homogeneous linear system of ODE’s. Let us denoteby A the matrix

A =

a11 a12 . . . a1n

a21 a22 . . . a2n

. . . . . . . . . . . .an1 an2 . . . ann

and let us denote by x(t) the vector function

x(t) =

x1(t)x2(t)

...xn(t)

.Then we can rewrite the system (11) in the following form

x′(t) = Ax(t). (12)

It is natural to look for solution

x(t) =

x1(t)x2(t)

...xn(t)

of the system (11) (or (12)) in the following form

x(t) =

ertw1

ertw2...

ertwn

= ert

w1

w2...wn

=: ertw.

6

Substituting into the system we see that the vector function ertw is a non-zero solution of (12) if and only if w is an eigenvector of A correspondingto eigenvalue r, that is w is a nonzero vector which satisfies

Aw = rw(a11 − r)w1 + a12w2 + . . .+ a1nwn = 0,

a21w1 + (a22 − r)w2 + . . .+ a1nwn = 0,

. . . . . . . . . . . . . . . . . . . . . . . .

an1w1 + an2w2 + . . .+ (ann − r)wn = 0.

(13)

This system called the auxiliary system of (12) has a non-zero solution ifand only if its determinant is zero:∣∣∣∣∣∣∣∣

(a11 − r) a12 . . . a1n

a11 (a22 − r) . . . a1n

. . . . . . . . . . . .an1 an1 . . . (ann − r)

∣∣∣∣∣∣∣∣ = 0.

The last equation or the equation

|A− rI| = 0 (14)

is called the characteristic equation of the system.

2.1. Characteristic equation has distinct real roots. If the charac-teristic equation (14) has n distinct eigenvalues r1, r2, ..., rn then there aren linearly independent eigenvectors

w(1),w(2), . . . ,w(n)

corresponding to these eigenvalues. Thus the system (12) has n linearlyindependent solutions

er1tw(1), er2tw(2), . . . , erntw(n).

In this case each solution of the system has the form

x(t) = C1er1tw(1) + C2e

r2tw(2) + . . .+ erntw(n).

Each set of n linearly independent solutions of the system is called a fun-damental set of solutions.

7

Example 2.1. Find the fundamental set of solutions of the system{dxdt = 5x+ 2y,dydt = −4x− y

Solving the characteristic equation∣∣∣∣ 5− r 2−4 −1− r

∣∣∣∣ = 0,

or

r2 − 4r + 3 = 0

we find

r1 = 1, r2 = 3.

From the auxiliary system{(5− r)w1 + 2w2 = 0

−4w1 − (1 + r)w2 = 0

we find eigenvectors

w(1) =

[1−2

], w(2) =

[1−1

].

Hence

x(1)(t) = et[

1−2

], x(2)(t) = e3t

[1−1

]is the fundamental set of solutions.

Example 2.2. Solve the system of equations:x′1 = x1 − x2 − x3,

x′2 = x2 + 3x3,

x′3 = 3x2 + x3.

We write this system in the form

x′ =

1 −1 −10 1 30 3 1

x.

8

Let us find eigenvalues and eigenvectors of the matrix of the system. Thecorresponding auxiliary system of equations is: 1− r −1 −1

0 1− r 30 3 1− r

w1

w2

w3

=

000

or

(1− r)w1 − w2 − w3 = 0,

0w1 + (1− r)w2 + 3w3 = 0,

0w1 + 3w2 + (1− r)w3 = 0.

Solving the characteristic equation

(1− r)∣∣∣∣ 1− r 3

3 1− r

∣∣∣∣ = 0

we find

(1− r)3 − (1− r)9 = (1− r)(1− 2r + r2 − 9) = 0,

(1− r)(r − 4)(r + 2) = 0.

Hence the matrix of the system has eigenvalues:

r1 = −2, r2 = 1, r3 = 4.

Let us find corresponding eigenvectors. The auxiliary system for r = r1 =−2 takes the form

3w1 − w2 − w3 = 0,

0w1 + 3w2 + 3w3 = 0,

0w1 + 3w2 + 3w3 = 0,

This system has the following nontrivial solution:

w3 = −w2, w1 = 0.

Therefore

w(1) =

01−1

9

is the eigenvector corresponding to the eigenvalue r1 = −2. For r2 = 1we have

−w2 − w3 = 0

3w3 = 0

3w2 = 0

.

Hence the corresponding eigenvector is

w(2) =

100

.For r3 = 4 we have:

−3w1 − w2 − w3 = 0,

0w1 − 3w2 + 3w3 = 0,

0w1 + 3w2 − 3w3 = 0.

Two last equations imply

w2 = w3,

and first equation becomes:

−3u1 = 2u2.

Thus

w(3) =

2−3−3

is the corresponding eigenvector. So the general solution of the system hasthe form:

x(t) = C1

011

e−2t + C2

01−1

et + C3

2−3−3

e4t.

Example 2.3. Find the general solution of the systemx′1 = x2 + x3,

x′2 = 3x1 + x3,

x′3 = 3x1 + x2.

10

The matrix of this system

A =

0 1 13 0 13 1 0

has eigenvalues r1 = −1, r2 = −2, r3 = 3 and corresponding eigenvectors

v(1) =

01−1

, v(2) =

−1111

, v(3) =

233

.Therefore the general solution of the system has the form

x(t) = C1e−t

01−1

+ C2e−2t

−111

+ C3

233

.2.2. Complex roots of characteristic equation. Suppose that a ma-trix A has a complex eigenvalue. Since all entries of the matrix are realnumbers, the coefficients of a polynomial

det(A− rI) = 0

are also real numbers.Therefore if r = λ + im is a complex eigenvalue of A, then r̄ = λ − im isalso an eigenvalue of A:

(A− r)w = 0,

(A− r̄)w̄ = 0.

The corresponding solutions are

z(1)(t) = wert,

z(2)(t) = wer̄t,

But these solutions are complex-valued.Let

w = a + ib,

a and b are real.

11

Then

z(1)(t) = (a+ ib)e(λ+im)t = (a + ib)eλt(cos(mt) + i sin(mt)) =

= eλt(a cos(mt)− b sin(mt)) + ieλt(a sin(mt) + b cos(mt)).

and

z(2)(t) = eλt(a cos(mt)− b sin(mt))− ieλt(a sin(mt) + b cos(mt))

Since the equation is a linear homogeneous equation the real valued solu-tions corresponding to the complex root r are

x(1)(t) =1

2(z(1)(t) + z(2)(t)), x(2)(t) =

1

2i(z(1)(t)− z(2)(t)),

that isx(1)(t) = eλt(a cos(mt)− b sin(mt)),

x(2)(t) = eλt(a sin(mt) + b cos(mt)).

Example 2.4. Find the general solution of the system

x′ =

[−1 1−1 1

]x(t).

Let us find eigenvalues of the matrix

A =

[1 1−1 1

]:

|A− rI| =∣∣∣∣ −1− r 1−1 1− r

∣∣∣∣ = 0,

(1− r)2 + 2 = 0

r2 − 2λ+ 3 = 0, r1,2 = 1± iInserting r1 = 1 + i into the auxiliary system{

(−1− r)w1 + 2w2 = 0,

−1w1 − (3 + r)w2 = 0.

we get {(−1 + 2− i)w1 + 2w2 = 0,

−w1 − (3− 2 + i)w2 = 0.

12

(1− i)w1 + 2w2 = 0.

So we can take

w1 = 2, w2 = −(1− i) = (−1 + i)

−w1 − (1 + i)w2 = 0

w =

[2

−1 + i

]=

[2−1

]+ i

[01

]

x(t) = C1

(e−2t cos t

[2−1

]−[

01

]sin t

)+C2

(e−2t sin

[2−1

]t−[

01

]cos t

).

Example 2.5. Find the general solution of the system

x′(t) = Ax(t),

where

A =

1 2 −10 1 10 −1 1

.The characteristic equation

(1− r)((1− r)2 + 1) = (1− r)(r2 − 2r + 2) = 0

has roots:r1 = 1, r2 = 1 + i, r2 = 1− i.

From the auxiliary system(1− r)w1 + 2w2 − w3 = 0

0 · w1 + (1− r)w2 + w3 = 0

0 · w1 − w2 + (1− r)w3

we find that

w(1) =

100

is an eigenvector corresponding to the real eigenvalue r1 = 1, and that

13

w =

2− ii−1

=

20−1

+ i

−110

is the eigenvector corresponding to the complex root r = 1 + i. Thereforethe general solution has the form

x(t) = c1

100

et + C2

20−1

cos t−

−110

sin t

et+

+ C3

−110

cos t+

20−1

sin t

et.

Example 2.6. Find a general solution of the system

x(t) = Ax(t),

where

A =

1 −2 2−2 1 22 2 1

.First we write the auxiliary system:

(1− r)w1 − 2w2 + 2w3 = 0,

−2w1 + (1− r)w2 + 2w3 = 0,

2w1 + 2w2 + (1− r)w3 = 0,

and solve the characteristic equation∣∣∣∣∣∣1− r −2 2−2 1− r 22 2 1− r

∣∣∣∣∣∣ = 0,

(1− r)∣∣∣∣ 1− r 2

2 1− r

∣∣∣∣+ 2

∣∣∣∣ −2 22 1− r

∣∣∣∣+ 2

∣∣∣∣ −2 1− r2 2

∣∣∣∣ = 0,

(1− r)(r2 − 2r − 3) + 2(−2 + 2r − 4) + 2(−4− 2 + 2r) = 0,

14

r2 − 2r − 3− r3 + 2r2 + 3r − 4 + 4r − 8− 12 + 4r = 0,

−r3 + 3r2 + 9r − 27 = 0,

r2(3− r) + 9(r − 3) = 0, or (r2 − 9)(3− r) = 0.

Hence the characteristic equation has the roots:

r1 = 3, r2 = 3 r3 = −3.

Inserting r = r1 = 3 into the auxiliary system we get−2w1 − 2w2 + 2w2 = 0

−2w1 − 2w2 + 2w3 = 0

2w1 + 2w2 − 2w3 = 0

,

or 4 −2 2−2 4 22 2 4

w1

w2

w3

=

000

.It is not difficult to see that the vectors

w1(1) =

101

and

w2(2) =

1−10

are eigenvectors corresponding to r1 = r2 = 3. The vector

w(3) =

11−1

is the eigenvector corresponding to r3 = −3. Therefore

15

x(t) = C1

101

e3t + C2e3t

1−10

+ C3

11−1

e−3t

is the general solution.

2.3. Repeated roots of characteristic equation. If r is a root of multi-plicity k and there arem(m < k) linearly independent eigenvectorsw1, ...wm

corresponding to the eigenvalue λ, then the solutions corresponding to thiseigenvalue of the system we look in the form: :

x = (w0 + w1t+ ...+ wk−mtk−m)eλt. (15)

To find the vectors w0,w1, ...,wk−m we plug the expression in (15) andequate the corresponding coefficients on the left hand side and right handside.

Example 2.7. Solve the system of equations{dxdt = 3x+ y,dydt = −x+ y.

From the characteristic equation∣∣∣∣ 3− r 1−1 1− r

∣∣∣∣ = 0

we find

r2 − 4r + 4 = 0

r1 = r2 = 2.

Then from the auxiliary system{(3− r)w1 + w2 = 0

−w1 + (1− r)w2 = 0

we obtain

w1 + w2 = 0.

Thus

w =

[1−1

]

16

is the eigenvector of the matrix of the system corresponding to the eigen-value r = 2. Hence

x(1)(t) =

[x11(t)x21(t)

]= e2t

[1−1

]is a solution of the system. The second solution we look in the form

x(2)(t) =

[x21(t)x22(t)

]=

[a1 + b1ta2 + b2t

]et,

that isx21(t) = (a1 + b1t)e

2t,

x22(t) = (a2 + b2t)e2t.

Inserting into the system we get

2a1e2t + 2b1e

2t + 2b1te2t = 3a1e

2t + 3b1te2t + (a2 + b2t)e

2t,

2a2e2t + b2e

2t + 2b2te2t = −a1e

2t − b1te2t + a2e

2t + b2 + e2t,

2a1 + b1 = 3a1 + a2, a1 + a2 = b1,

2b1 = 3b1 + b2, b1 + b2 = 0,

2a2 + b2 = −a1 + a2, a1 + a2 = −b2,

2b2 = −b1 + b2, b1 = −b2.

Finally we obtain

a1 = 1, a2 = 0, b1 = 1, b2 = −1.

Hence

x(2)(t) =

[(1 + t)e2t

−te2t

].

Example 2.8. Find the general solution of the system{dx1dt = 5x1 − x2,dx2dt = x1 + 3x2.

(16)

17

First we find the eigenvalues of the matrix

A =

[5 −11 3

].

The characteristic equation ∣∣∣∣ 5− r −11 3− r

∣∣∣∣or

r2 − 8r + 1 = 0

has a repeated root r = 4. So r = 4 is the repeated eigenvalue of the matrixA. From the auxiliary system we find the corresponding eigenvector

w(1) =

[11

].

So

x(1)(t) =

[x11(t)x21(t)

]= e4t

[11

]is a solution of the system. We look for the second solution that is inde-pendent of

x(2)(t) =

[x12(t)x22(t)

]in the form

x(2)(t) = (a + bt)e4t.

From the system (16) we find:

4a2 + b2 = a1 + 3a2

a2 + b2 = a1

4b2 = b1 + 3b2

b1 = b2

x(2)1 = (a1 + b1t)e

4t

x(2)2 (t) = (a2 + b2t)e

4t

4a1 + b1 + 4b1t = 5a1 + 5b1t− a2 − b2t

4a2 + b2 + 4b2t = a1 + b1t+ 3a2 + 3b2t

18

4a1 + b1 = 5a1 − a2 ⇒ a1 − a2 = b1

4b1 = 5b1 − b2 ⇒ b2 = b1

4a2 + b2 = a1 + 3a2

a1 − a2 = b2

4b2 = b1 + 3b2

b1 = 1, b2 = 1

a1 = 2, a2 = 1,

x12(t) = (2 + t)e4t,

x22(t) = (1 + t)e4t.

Hence

x(2)(t) =

[x12(t)x22(t)

]=

[(2 + t)e4t

(1 + t)e4t

]and the general solution has the form

x(2)(t) = C1

[e4t

e4t

]+ C2

[(2 + t)e4t

(1 + t)e4t

].

2.4. Exponential of a Matrix. Let A be n× n matrix of the form

A =

a11 a12 . . . a1n

a21 a22 . . . a2n

. . . . . . . . . . . .an1 an2 . . . ann

.Exponential of a matrix is defined by the equality

eA = I + A+1

2!A2 + ...+

1

n!An + ...,

where I is the identity matrix.The matrix exponent has the following properties:

(1) If AB = BA, then

eA+B = eA · eB.(2) If D is a diagonal matrix and A = QDQ−1, then

eA = QeDQ−1.

19

(3) The matrix

etA = I + tA+t2

2!A2 + ...+

tn

n!An + ...

satisfies the equation

d

dtetA = AetA, etA

∣∣t=0

= I. (17)

(4) It follows from the last property that the vector function

x(t) = etAx0

is a solution of the problem

x′(t) = Ax(t), x(0) = x0.

Suppose that the matrix A has n linearly independent eigenvectors

w(1) =

w11

w21...wn1

,w(2) =

w12

w22...wn2

, . . . ,w(n) =

w1n

w2n...wn1

(18)

corresponding to eigenvalues r1, r2, . . . , rn. Since the vectors w(1),w(1), . . . ,w(n)

are linearly independent the matrix

Q =

w11 w12 . . . w1n

w21 w22 . . . w2n

. . . . . . . . . . . .wn1 wn2 . . . wnn

(19)

is non-singular and Q−1 exists.

It is easy to see that

AQ =

r1w11 . . . rnw1n... . . . ...

r1wn1 . . . rnwnn

= QD, (20)

20

where

D =

r1 . . . . . . 00 r2 . . . 0...

... . . . ...0 . . . . . . rn

It follows from (20) that

A = QDQ−1, Q−1AQ = D.

That is A is diagonalizable.

Example 2.9. Calculate etA, where

A =

[3 −24 −3

](21)

and solve the problem

x′(t) = Ax(t), x(0) =

[12

]. (22)

Let us find eigenvalues and eigenvectors of A. The characteristic equation∣∣∣∣ 3− r −24 −3− r

∣∣∣∣ = 0

has roots r1 = −1, r2 = 1. The vectors

w(1) =

[12

], w(2) =

[11

]are eigenvectors of A corresponding to the eigenvectors r1 and r2 respec-tively.So the matrix A is diagonalizable :

A = QDQ−1,

where

Q =

[1 12 1

], D =

[−1 00 1

], Q−1 =

[−1 12 −1

].

Thus

etA =

[1 12 1

] [e−t 00 et

] [−1 12 −1

]=

[−e−t + 2e e−t − et−2e−t + 2et 2e−t − et

].

21

The solution of the problem is

x(t) =

[−e−t + 2e e−t − et−2e−t + 2et 2e−t − et

] [12

]=

[e−t

2e−t

].

2.5. Nonhomogeneous Systems.

x′ = P (t)x + g(t). (23)

The general solution of (23) has the form

x = c1x(1) + ...+ cnx

(n) + v(t) (24)

where c1x(1) + ...+ cnx

(n) is a general solution of the corresponding homo-geneous system, v(t) is a particular solution of (23).First let us consider the system:

x′ = Ax + g(t) (25)

where A is a constant matrix which has n linearly independent eigenvectors

w(1),w(2), . . . ,w(n)

corresponding to eigenvalues r1, r2, . . . , rn.In this case A is diagonalizable and we can transform the system into asystem which is easily solvable.

Let Q be a matrix of eigenvectors defined by (19).

We define a new dependent variable y by :

x = Qy. (26)

Substituting into (25) we obtain:

Qy′ = AQy + g(t).

Multiplying by Q−1 we obtain :

y′ = Q−1AQ+Q−1g(t) = Dy + h(t),

22

where

h(t) =

h1(t)h2(t)

...hn(t)

= Q−1g(t).

So we have n first order ordinary differential equationsy′1(t) = r1y1(t) + h1(t),

y′2(t) = r2y2(t) + h2(t),

. . . . . . . . . . . . . . . . . . . . . ,

y′n(t) = rnyn(t) + hn(t).

Solving these equations we get:

yj(t) = erjt∫ t

t0

e−rjshj(s)ds+ yj(t0)erj(t−t0), j = 1, ..., n.

2.6. Method of variation of parameters. Method of variation of pa-rameters allows us to find particular solution of the nonhomogeneous sys-tem of the form

x′ = P (t)x + g(t), (27)

when the fundamental system of solution of the corresponding homoge-neous system is given.First we consider the corresponding homogeneous system:

x′(t) = P (t)x(t) (28)

The general solution of (28) has the form

x(t) = Ψ(t)C, (29)

where C is a constant vector. We look for a particular solution of (27) inthe following form

x = Ψ(t)u(t), (30)

where Ψ(t) is the fundamental matrix of the system and u(t) is a vector-function to be found.It is clear that

Ψ′(t)u(t) + Ψ(t)u′(t) = P (t)Ψ(t)u(t) + g(t).

23

Employing the equality (10) we obtain

P (t)Ψ(t)u(t) + Ψ(t)u′(t) = P (t)Ψ(t)u(t) + g(t).

Hence

u′(t) = Ψ−1(t)g(t). (31)

Integrating the last equality we get:

u(t) =

∫Ψ(t)−1g(t) + C

Thus

xp(t) = Ψ(t)

∫Ψ(t)−1g(t)dt

is a particular solution of the nonhomogeneous sytem(27). The generalsolution of the system has the form:

x(t) = Ψ(t)C + Ψ(t)

∫Ψ(t)−1g(t)dt.

In order to find the solution of the system (27) satisfying the initialcondition

x(t0) = x0 (32)

we integrate (31) over the interval (t0, t) :

u(t) = u(t0) +

∫ t

t0

Ψ−1(s)g(s)ds.

Thus

x(t) = Ψ(t)u(t0) + Ψ(t)

∫ t

t0

Ψ−1(s)g(s)ds.

It follows from the last equality that

u(t0) = Ψ(t0)−1x(t0).

Hence the solution of the Cauchy problem (27),(32) has the form:

x(t) = Ψ(t)Ψ−1(t0)x(t0) + Ψ(t)

∫ t

t0

Ψ−1(s)g(s)ds.

24

Example 2.10. Find the solution to the initial value problem

x′(t) =

[2 −31 −2

]x(t) +

[e2t

1

], x(0) =

[−10

].

The auxiliary system is{(2− r)w1 − 3w2 = 0,

w1 − (2 + r)w2 = 0.

The characteristic equation∣∣∣∣ 2− r −31 −2− r

∣∣∣∣ = 0

has rootsr1 = −1, r2 = −1.

The eigenvector corresponding to the eigenvalue r1 = 1 is

w(1) =

[31

],

and the eigenvector corresponding to the eigenvalue r2 = −1 is

w(2) =

[11

].

Thus the fundamental matrix is:

Ψ(t) =

[3et e−t

et e−t

], and Ψ−1(t) =

[1/2e−t −1/2e−t

−1/2et 3/2et

].

Thus the solution of the problem has the form

x(t) =

[3et e−t

et e−t

] [1/2 −1/2−1/2 3/2

] [−10

]+[

3et e−t

et e−t

] ∫ t

0

[1/2e−s −1/3e−s

−1/2es 3/2es

] [e2s

1

]ds,

x(t) =

[−3/2et + 1/2e−t

−1/2et + 3/2e−t

]+

[3et e−t

et e−t

] ∫ t

0

[1/2es − 1/2e−s

−1/2e3s + 3/2es

]ds =[

−3/2et + 1/2e−t

−1/2et + 3/2e−t

]+

[3et e−t

et e−t

] [1/2et + 1/2e−s − 1−1/6e3t + 3/2et − 4/3

]ds =

25[−3/2et + 1/2e−t

−1/2et + 3/2e−t

]+

[3/2e2t + 3/2− 3et − 1/6e2t + 3/2− 4/3e−t

1/2e2t + 1/2− et − 1/6e2t + 3/2− 4/3e−t

].

Example 2.11. Find the solution of the system{dx1dt = 7x1 + 3x2,dx2dt = 6x1 + 4x2.

(33)

which satisfies the initial conditions:

x1(0) = 1, x2(0) = 3

The auxiliary system is:{(7− r)w1 + 3w2 = 0,

6w1 + (4− r)w2 = 0

The characteristic equation

(7− r)(4− r)− 18 = 0,

or

r2 − 11 + 10 = 0

has roots

r1,2 =11±

√121− 40

2,

r1 = 1, r2 = 10.

From the auxiliary system we find that the vector

w(1) =

[1−2

]is an eigenvector corresponding to r1 = 1 and

w(2) =

[11

]is the eigenvector corresponding to r2 = 10. Hence the fundamental matrixis the following matrix

Ψ(t) =

[et e10t

−2et e10t

].

26

It is easy to see that

Ψ(0) =

[1 1−2 1

], Ψ−1(0) =

[1/3 −1/32/3 1/3

].

Thus

x(t) =

[et e10t

−2et e10t

] [1/3 −1/32/3 1/3

] [13

]=

[−2

3et + 5

3e10t

43et + 5

3e10t

].