Examples of Systems of Differential

Equations...
Leif Mejlbro

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Leif Mejlbro

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Examples of Systems of ov F
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Differential Equations and
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Applications from Physics and

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the Technical Sciences

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Calculus 4c-3
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Examples of Systems of Differential Equations and Applications from Physics and the
Technical Sciences – Calculus 4c-3
© 2008 Leif Mejlbro & Ventus Publishing ApS
ISBN 978-87-7681-382-6

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 Calculus 4c-3 Contents

Contents
Introduction 5

1 Homogeneous systems of linear dierential equations 6

2 Inhomogeneous systems of linear dierential equations 44

3 Examples of applications in Physics 62

4 Stability 72

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5 Transfer functions 88

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Calculus 4c-3 Introduction

Introduction
Here we present a collection of examples of general systems of linear differential equations and some
applications in Physics and the Technical Sciences. The reader is also referred to Calculus 4b as well
as to Calculus 4c-2.

It should no longer be necessary rigourously to use the ADIC-model, described in Calculus 1c and
Calculus 2c, because we now assume that the reader can do this himself.

Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the
first edition. It is my hope that the reader will show some understanding of my situation.

Leif Mejlbro
21st May 2008

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5
Calculus 4c-3 Homogeneous systems of linear differential equations

1 Homogeneous systems of linear differential equations

Example 1.1 Given the homogeneous linear system of differential equations,
    
d x 0 1 x
(1) = , t ∈ R.
dt y 1 0 y

1) Prove that everyone of the vectors

       
cosh t sinh t et 2et
(2) , , , ,
sinh t cosh t et 2et

is a solution of (1).
2) Are the vectors in (2) linearly dependent or linearly independent?

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3) How many linearly independent vectors can at most be chosen from (2)? In which ways can this

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be done?

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4) Write down all solutions of (1).

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x
5) Find that solution of (1), for which
y
   
ov F
x(0) 1
m PD
= .
y(0) −1
re h
to atc

1) We shall just make a check:

se e B

        
d cosh t sinh t 0 1 cosh t sinh t
= and = ,
en W

dt sinh t cosh t 1 0 sinh t cosh t

        
lic y

d sinh t cosh t 0 1 sinh t cosh t

a b

= and = ,
dt cosh t sinh t 1 0 cosh t sinh t
y ed

 t   t    1   t 
d e e 0 1 e e
Buess

t = t and t = t ,
dt e e 1 0 e e
 t   t    t   t 
d 2e 2e 0 1 2e 2e
oc

= and = .
dt 2et 2et 1 0 2et 2et
Pr

2) The vectors are clearly linearly dependent, cf. also (3).

3) We can at most choose two linearly independent vectors. We have the following possibilities,
       t 
cosh t sinh t cosh t e
, , ,
sinh t cosh t sinh t et
   t     t 
cosh t 2e sinh t e
, , , ,
sinh t 2et cosh t et
   t 
sinh t 2e
, .
cosh t 2et

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6
Calculus 4c-3 Homogeneous systems of linear differential equations

4) It follows from (3) that all solutions are e.g. given by

       
x cosh t sinh t c1 cosh t+c2 sinh t
= c1 + c2 = ,
y sinh t cosh t c2 cosh t+c1 sinh t

for t ∈ R, where c1 and c2 are arbitrary constants.

5) If we put t = 0 into the solution of (4), then
     
x(0) c1 1
= = ,
y(0) c2 −1

hence
       
x(t) cosh t − sinh t et −t 1
= = =e .
y(t) − cosh t + sinh t −e−t −1

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Calculus 4c-3 Homogeneous systems of linear differential equations

 
t+1
Example 1.2 Prove that is a solution of the system
t
      
d x 0 1 x 1−t
= + , t ∈ R.
dt y 1 0 y −t

Find all solutions of this system, and find in particular that solution, for which
   
x(0) 1
= .
y(0) −1
       
x t+1 d x 1
If = , then = and
y t dt y 1
            
0 1 t+1 1−t t 1−t 1 d x

r
+ = + = = ,

ke
1 0 t −t t+1 −t 1 dt y

oc
and the equation is fulfilled.

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it. n
It follows from Example 1.1 that the complete solution of the homogeneous system of equations is

e U
given by

ov F
     m PD
x cosh t sinh t
= c1 + c2 , c1 , c2 arbitrære.
y sinh t cosh t
re h
to atc

Due to the linearity, the complete solution of the inhomogeneous system of differential equations is
se e B

given by
       
x t+1 cosh t sinh t
en W

= + c1 + c2 , c1 , c2 arbitrære.
y t sinh t cosh t
lic y
a b
y ed

If we put t = 0 into the complete solution, we get

           
Buess

x(0) 1 1 0 1 + c1 1
= + c1 + c2 = = ,
y(0) 0 0 1 c2 −1
oc

hence c1 = 0 and c2 = −1. The wanted solution is

Pr

       
x(t) t+1 sinh t t + 1 − sinh t
= − =− , t ∈ R.
y(t) t cosh t t − cosh t

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8
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.3 Find that solution z1 (t) = (x1 , x2 )T of

    
d x1 1 −1 x1
(3) = ,
dt x2 1 1 x2

which satisfies z1 (0) = (1, 0)T .

Than find that solution z2 (t) of (3), which satisfies z2 (0) = (0, 1)T .
What is the complete solution of (3)?

1) The complete solution.

a) The “fumbling method”. The system is written


dx1 /dt = x1 − x2 , dx1
thus in particular x2 = x1 − .

r
dx2 /dt = x1 + x2 , dt

ke
By insertion into the latter equation we get

oc
d2 x1

l
dx2 dx1 dx1

it. n
= − = x1 + x2 = x1 + x1 − ,
dt2

e U
dt dt dt

ov F
hence by a rearrangement,
m PD
d2 x1 dx1
−2 + 2x1 = 0.
dt2 dt
re h
to atc

The characteristic polynomial R2 − 2R + 2 has the roots R = 1 ± i, so we conclude that the

complete solution is
se e B

x1 = c1 et cos t + c2 et sin t, c1 , c2 arbitrary.

en W

It follows from
lic y
a b

dx1
= (c1 + c2 )et cos t + (c2 − c1 )et sin t,
y ed

dt
Buess

that
dx1
x2 = x1 − = −c2 et cos t + c1 et sin t.
oc

dt
Pr

Summing up we get
       
x1 c1 et cos t+c2 et sin t t cos t t sin t
(4) = = c 1 e +c 2 e ,
x2 −c2 et cos t+c1 et sin t sin t − cos t

where c1 and c2 are arbitrary constants.

b) Alternatively we apply the eigenvalue method. From
 
 1−λ −1 
  = (λ − 1)2 + 1 = 0
 1 1−λ 

we obtain the complex conjugated eigenvalues λ = 1 ± i.

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9
Calculus 4c-3 Homogeneous systems of linear differential equations

A complex eigenvector for e.g. λ = 1 + i is the “cross vector” of (1 − λ, −1) = (−i, −1), thus
e.g. v = (1, −i).
A fundamental matrix is
  
Φ(t) = Re e(a+iω)t (α + iβ) | Im e(a+iω)t (α + iβ) = eat cos ωt(α β) + eat sin ωt(−β α).

Here,
   
1 0
λ = 1 + i = a + iω, α= , β= ,
0 −1
so
     
t 1 0 t 0 1 t cos t sin t
Φ(t) = e cos t + e sin t =e .
0 −1 1 0 sin t − cos t
The complete solution is
   

r
cos t sin t

ke
t t
x(t) = Φ(t)c = c1 e + c2 e ,
sin t − cos t

oc
where c1 and c2 are arbitrary constants.

l
it. n
c) Alternatively we can directly write down the exponential matrix,

e U
 a 1
exp(At) = eat cos ωt − sin ωt I + eat sin ωt · A

ov F
ω
m PD  ω    
t 1 0 t 1 −1 t cos t − sin t
= e (cos t−sin t) + e sin t =e ,
0 1 1 1 sin t cos t
re h

so the complete solution becomes

to atc

   
cos t − sin t
x(t) = exp(At)c = c1 et + c2 et ,
se e B

sin t cos t
where c1 and c2 are arbitrary constants.
en W

d) Alternatively (only sketchy) the eigenvalues λ = 1 ± i indicate that the solution necessarily
lic y

is of the structure

a b

x1 (t) = a1 et cos t + a2 et sin t,

y ed

x2 (t) = b1 et cos t + b2 et sin t.

Buess

We have here four unknown constants, and we know that the final result may only contain
two arbitrary constants. By insertion into the system of differential equations we get by an
oc

identification that b1 = a1 og b2 = −a2 , and we find again the complete solution

       
a1 et cos t + a2 et sin t
Pr

x1 t cos t t sin t
= = a1 e + a2 e ,
x2 a1 et sin t − a2 et cos t sin t − cos t
where a1 and a2 are arbitrary constants.
2) By using the initial conditions z1 (0) = (1, 0)T in e.g. (4) we get
     
1 1 0
= c1 + c2 ,
0 0 −1
thus c1 = 1 and c2 = 0, and hence
 t 
e cos t
z1 (t) = .
et sin t

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10
Calculus 4c-3 Homogeneous systems of linear differential equations

3) By inserting the initial conditions z2 (0) = (0, 1)T into e.g. (4), we get
     
0 1 0
= c1 + c2 ,
1 0 −1

thus c1 = 0 and c2 = −1, hence

 
−et sin t
z2 (t) = .
et cos t

4) The complete solution has already been given i four different versions in (1).

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Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.4 Find by using the eigenvalue method the complete solution of the following system of
differential equations
 
dx 1 1
= x(t).
dt 0 −2

1) The eigenvalue method. It follows immediately that the eigenvalues are λ 1 = 1 and λ2 = −2.
To the eigenvalue λ1 = 1 correspond the eigenvectors which are proportional to (1, 0).
To the eigenvalue λ2 = −2 corresponds the eigenvectors which are proportional to (1, −3).
The complete solution is
   t  −2t     
x1 (t) e e t 1 −2t 1
= c1 +c2 = c1 e +c2 e ,
x2 (t) 0 −3e−2t 0 −3
where c1 and c2 are arbitrary constants.

r
ke
2) Alternatively the exponential matrix is given by

oc
1 1
exp(At) = −λ2 eλ1 t +λ1 eλ2 t I + eλ1 t −eλ2 t A
λ1 −λ2 λ1 −λ2

l
   

it. n
1 1 0 1 t 1 1

e U
= 2et +e−2t + e − e−2t
3 0 1 3 0 −2

ov F
 
1 2et +e−2t +et −e−2t
m PD t
e −e −2t
=
3 0 2et +e−2t −2et +2e−2t
 t 
1 3e et − e−2t
re h

= .
3e−2t
to atc

3 0
The complete solution is
se e B

   t  t 
x1 (t) e e − e−2t
= c1 + c2 ,
en W

x2 (t) 0 3e−2t
lic y

where c1 and c2 are arbitrary constants.

a b

3) Alternatively the system is written (the “fumbling method”),

y ed

dx1 dx2
= −2x2 ,
Buess

= x1 + x2 ,
dt dt
from which we immediately get x2 = c2 e−2t .
oc

Then by insertion
Pr

dx1
− x1 = c2 e−2t ,
dt
so
1
x1 = c1 et + c2 et e−t e−2t dt = c1 et − c2 e−2t .
3
Summing up we have
   1
    
x1 (t) c1 et − c2 e−2t t 1 1 −2t 1
= 3 = c1 e − c2 e ,
x2 (t) c2 e−2t 0 3 −3

where c1 and c2 are arbitrary constants.

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12
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.5 Find by the eigenvalue method the complete solution of the following system of differ-
ential equations
 
dx 1 4
= x(t).
dt −2 −3

1) The eigenvalue method. The eigenvalues are the solutions of the following equation,
 
 1−λ 4 
  = (1 − λ)(−3 − λ) + 8 = λ2 + 2λ + 5 = 0,
 −2 −3 − λ 

hence λ = −1 ± 2i.
A complex eigenvector corresponding to e.g.. λ = a + iω = −1 + 2i is a cross vector of

r
(1 − λ, 4) = (2 − 2i, 4) = 2(1 − i, 2),

ke
oc
so we have e.g.

l
v = α + iβ = (2, −1 + i)T = (2, −1)T + i(0, 1)T .

it. n
e U
Then a fundamental matrix is given by

ov F
m PD
Φ(t) = eat cos ωt(α β) + eat sin ωt(−β α)
   
−t 2 0 −t 0 2
re h

= e cos 2t + e sin 2t
−1 1 −1 −1
to atc

 
−t 2 cos 2t 2 sin 2t
= e .
se e B

− cos 2t − sin 2t cos 2t − sin 2t

en W

The complete solution is

   
lic y

−t 2 cos 2t −t 2 sin 2t
a b

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

2) Alternatively it follows by the “fumbling method” that

⎧
⎪
⎪ dx1
⎪
oc

⎨ dt = x1 + 4x2 , 1 dx1 1
specielt x2 = − x1 .
Pr

⎪
⎪ 4 dt 4
⎪ dx2 = −2x − 3x ,
⎩ 1 2
dt
We get by insertion into the second equation,

1 d2 x1 1 dx1 3 dx1 3
2
− = −2x1 − + x1 ,
4 dt 4 dt 4 dt 4
hence by a rearrangement,

d2 x1 dx1
t
+2 + 5x1 = 0.
dt dt

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13
Calculus 4c-3 Homogeneous systems of linear differential equations

The characteristic polynomial R2 + 2R + 5 has the roots R = −1 ± 2i, so the complete solution is

x1 (t) = c1 e−t cos 2t + c2 e−t sin 2t.

We conclude from
dx1
= (2c2 − c1 )e−t cos 2t + (−2c1 − c2 )e−t sin 2t,
dt
that
dx1
4x2 = − x1 = (2c2 − 2c1 )e−t cos 2t + (−2c1 − 2c2 )e−t sin 2t.
dt
Summing up we have
   
c1 e−t cos 2t + c2 e−t sin 2t

r
x1 (t)

ke
= 1 1
x2 (t) − c1 e−t (cos 2t + sin 2t) + c2 e−t (cos 2t − sin 2t)
2   2  

oc
1 −t 2 cos 2t 1 −t 2 sin 2t
= c1 e + c2 e ,

l
− cos 2t − sin 2t cos 2t − sin 2t

it. n
2 2

e U
where c1 and c2 are arbitrary constants.

ov F
m PD
3) Alternatively the exponential matrix is with a = −1 and ω = 2 given by
 a 1
re h

exp(At) = eat cos ωt − sin ωt I + eat sin ωtA

to atc

 ω  ω
  
−t 1 1 0 1 −t 1 4
=e cos 2t + sin 2t + e sin 2t
se e B

2 0 1 2 −2 −3
 
cos 2t + sin 2t 2 sin 2t
= e−t
en W

,
− sin 2t cos 2t − sin 2t
lic y
a b

hence the complete solution is

     
y ed

x1 (t) −t cos 2t + sin 2t −t 2 sin 2t

= c1 e + c2 e .
− sin 2t cos 2t − sin 2t
Buess

x2 (t)
oc

4) Alternatively (sketch) the solution must have the following real structure,
   
Pr

x1 (t) a1 e−t cos 2t + a2 e−t sin 2t

= ,
x2 (t) b1 e−t cos 2t + b2 e−t sin 2t

so we shall “only” check that this function satisfies the equations. The details are fairly long and
tedious, so they are here left out.

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14
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.6 Describe

       
x 3 −1 x t2
= + , t ∈ R,
y 2 4 y t3 + 1

as a linear system of differential equations of first order.

By introducing the new variables

x1 = x, x 2 = x , x3 = x , x4 = y,

the system can bow be written

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞
x1 x2 0 1 0 0 x1 0
d ⎜⎜ x2
⎟ ⎜ x3 ⎟ ⎜ 0
⎟=⎜ ⎟=⎜ 0 1 0⎟⎟⎜
⎜ x2 ⎟ ⎜ 0 ⎟
⎟+⎜ ⎟
⎠ ⎝ t2 ⎠ , t ∈ R.
dt ⎝ x3 ⎠ ⎝ x ⎠ ⎝ 3 0 0 −1 ⎠ ⎝ x3

r
ke
x4 y 2 0 0 4 x4 t3 + 1

l oc
it. n
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ov F
m PD
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to atc
se e B
en W
lic y
a b
y ed

With us you can

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oc

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Calculus 4c-3 Homogeneous systems of linear differential equations

There is here a very good reason for not asking about the complete solution. In fact, we see that the
eigenvalues are the roots of the polynomial
 
 −λ 1 0 0       
  −λ 1 0   1 0 0   1 0 0 
 0 −λ 1 0      
  = −λ  0 −λ −1  + 3  −λ 1 0  − 2  −λ 1 0 
 3 0 −λ −1       
  0 0 4−λ   0 0 4−λ   0 −λ −1 
 2 0 0 4−λ 
= −λ3 (4 − λ) + 3(4 − λ) + 2 = λ4 − 4λ3 − 3λ + 14,
where it can be proved that this polynomial does not have rationale roots.

Numerical calculations give approximatively


λ3
λ1 = 1, 56333, λ2 = 3, 96633, = −0, 76483 ± 1, 29339i.
λ4

r
ke
If one insists on solving the equation, the “fumbling method” is here without question the easiest

oc
one to apply. In fact, if we write the full system

l
it. n
 
x = 3x − y + t2 ,

e U
dvs. specielt y = −x + 3x + t2 ,
y  = 2x + 4y + t3 + 1,

ov F
m PD
then it follows by insertion into the latter equation that

−x(4) + 3x + 2t = 2x − 4x(3) + 12x + 4t2 + t3 + 1,

re h
to atc

hence by a rearrangement
se e B

d4 x d3 x dx
− 4 −3 + 14x = −t3 − 4t2 + 2t − 1.
dt4 dt3
en W

dt
The we guess a particular solution of the form of a polynomial of degree 3, at 3 + bt2 + ct + d (the
lic y
a b

coefficients are really ugly), and since the characteristic polynomial is the same as before, we get the
y ed

complete solution

x(t) = at3 + bt2 + ct + d + c1 eλ1 t + c2 eλ2 t + c3 eαt cos βt + c4 eαt sin βt,
Buess

where we have λ3 = α + iβ and λ4 = α − iβ from above.

oc

Then put this solution into

Pr

y = −x + 3x + t2 .

One has to admit that this method is somewhat easier to apply than the “standard method” of finding
the eigenvectors first.

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16
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.7 Find the complete solution of the system

dx 3 1
= x − y − z,
dt 2 2
dy 1 1
= − x + 2y + z,
dt 2 2
dz 1 5
= x + y = z.
dt 2 2

First solution. Inspection. It follows immediately that

d
(x + y) = x + y, thus x + y = 2a1 et ,
dt
d
thus y + z = 2a3 e3t ,

r
(y + z) = 3(y + z),

ke
dt
d

oc
(z + x) = 2(z + x), thus z + x = 2a2 e2t ,
dt

l
it. n
so

e U
⎧
⎪
⎪ x = a1 et + a2 e2t − a3 e3t ,
⎪

ov F
⎪
⎨ m PD
y = a1 et − a2 e2t + a3 e3t ,
⎪
⎪
⎪
⎪
⎩
z = −a1 et + a2 e2t + a3 e3t ,
re h
to atc

or written as a vector,
se e B

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1 1 −1
⎝ y ⎠ = a1 et ⎝ 1 ⎠ + a2 e2t ⎝ −1 ⎠ + a3 e3t ⎝ 1 ⎠ ,
en W

z −1 1 1
lic y
a b

where a1 , a2 and a3 are arbitrary constants.

y ed

Second solution. The eigenvalue method. The corresponding matrix

Buess

⎛ 3 ⎞
2 −1 − 12
A = ⎝ − 21 2 1 ⎠
oc

2
1 5
2 1 2
Pr

has the characteristic polynomial

 3 
 −λ −1 − 21 
 2 1
 − 2−λ 1  = ( 3 −λ)(2−λ)( 5 −λ)− 1 + 1 + 1 (2−λ)− 1 ( 5 −λ)− 1 ( 3 −λ)
 2 2  2 2 4 4 4 2 2 2 2
 1 5 
2 1 2 −λ
    
3 5 1 1 3 5 1
= −(λ − )(λ − 2)(λ − ) − (λ − 2) + (2λ − 4) = (λ − 2) − (λ − λ− − +1
2 2 4 2 2 2 4
 
15 3
= −(λ − 2) λ2 − 4λ + − = −(λ − 2)(λ2 − 4λ + 3) = −(λ − 1)(λ − 2)(λ − 3),
4 4
so the eigenvalues are λ = 1, 2 and 3.

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17
Calculus 4c-3 Homogeneous systems of linear differential equations

For λ = 1 we have the eigenvector (1, 1, −1).

For λ = 2 we have the eigenvector (1, −1, 1).
For λ = 3 we have the eigenvector (−1, 1, 1).
The complete solution is then
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x 1 1 −1
⎝ y ⎠ = c1 et ⎝ 1 ⎠ + c2 e2t ⎝ −1 ⎠ + c3 e3t ⎝ 1 ⎠ ,
z −1 1 1
where c1 , c2 , c3 are arbitrary constants.

Example 1.8 Find the complete solution of the system

 
 1 −1
Y = Y.
2 −1

r
ke
The eigenvalues are the roots of the polynomial
 

oc
 1−λ −1 
 = (λ − 1)(λ + 1) + 2 = λ2 + 1,
 2 −1 − λ 

l
it. n
e U
thus λ = ±i. Since the eigenvalues are complex numbers, we have four solution variants.

ov F
1) The eigenvalue method. To λ = a + iω = i, i.e. a = 0 and ω = 1, we have a complex eigenvector
m PD
of the form
     
1 1 0
re h

v= = +i = α + iβ.
1−i 1 −1
to atc

Then a fundamental matrix is given by

se e B

  
Φ(t) = Re e(a+iω)t (α + iβ) Im e(a+iω)t (α + iβ) = eat cos ωt(α β) + eat sin ωt(−β α)
en W

     
1 0 0 1 cos t sin t
= cos t + sin t = ,
lic y

1 −1 1 1 cos t + sin t sin t − cos t

a b
y ed

so the complete solution is

     
Buess

y1 cos t sin t
= c1 + c2 , c1 , c2 arbitrary.
y2 cos t + sin t sin t − cos t
oc

2) The exponential matrix. Since the eigenvalues are complex conjugated, the exponential matrix
Pr

is given by a formula (a = 0 and ω = 1),

    
a 1 1 0 1 −1
exp(At) = eat cos ωt − sin ωt I + eat sin ωt · A = cos t + sin t
ω ω 0 1 2 −1
 
cos t + sin t − sin t
= .
2 sin t cos t − sin t

Then the complete solution is

     
y1 cos t + sin t − sin t
= c1 + c2 ,
y2 2 sin t cos t − sin t
where c1 and c2 are arbitrary constants.

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18
Calculus 4c-3 Homogeneous systems of linear differential equations

3) Since λ = ±i, the real structure of the solution is given by

   
y1 a1 cos t + a2 sin t
= ,
y2 b1 cos t + b2 sin t

hence
   
d y1 a2 cos t − a1 sin t
=
dt y2 b2 cos t − b1 sin t

and
    
1 −1 a1 cos t + a2 sin t (a1 −b1 ) cos t+(a2 −b2 ) sin t
= .
2 −1 b1 cos t + b2 sin t (2a1 −b1 ) cos t+(2a2 −b2 ) sin t

When we identify the coefficients, we eliminate b1 and b2 , thus

r
ke
a2 = a1 − b1 and − a 1 = a2 − b 2 ,

oc
and hence

l
it. n
b1 = a1 − a2

e U
and b 2 = a1 + a2 .

ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Homogeneous systems of linear differential equations

The complete solution is then

       
y1 a1 cos t + a2 sin t cos t sin t
= = a1 + a2 ,
y2 (a1 − a2 ) cos t + (a1 + a2 ) sin t cos t + sin t − cos t + sin t

where a1 and a2 are arbitrary constants.

4) The “fumbling method”. It follows from

dy1 dy1
= y1 − y2 , i.e. y2 = − + y1 ,
dt dt
dy2
= 2y1 − y2 ,
dt
by eliminating y2 that

r
d2 y 1 dy1 dy1

ke
− 2
+ = 2y1 + − y1 ,
dt dt dt

oc
hence by a rearrangement

l
it. n
d2 y 1

e U
+ y1 = 0.
dt2

ov F
Then we get the complete solution
m PD
re h
y1 = c1 cos t + c2 sin t.
to atc

This gives us
se e B

dy1
y2 = − + y1 = −(−c1 sin t + c2 cos t) + c1 cos t + c2 sin t
en W

dt
= c1 (sin t + cos t) + c2 (− cos t + sin t).
lic y
a b

Summing up the complete solution becomes

y ed

     
y1 cos t sin t
Buess

= c1 + c2 ,
y2 sin t + cos t − cos t + sin t
oc

where c1 and c2 are arbitrary constants.

Pr

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20
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.9 Find a fundamental matrix of the system

y1 = 2y1 + 5y2 − 3y3 ,

y2 = −y1 − 2y2 + y3 ,

y3 = y1 + y2 .

The equation is written in matrix form

⎛ ⎞ ⎛ ⎞⎛ ⎞
y 2 5 −3 y1
d ⎝ 1⎠ ⎝
y2 = −1 −2 1 ⎠ ⎝ y2 ⎠ .
dt
y3 1 1 0 y3
The eigenvalues are the roots of the polynomial
     
 2−λ 5 −3   −λ 5 −3   −1 5 −3 
  
 −1  =  λ −2 − λ 1  = λ  1 −2 − λ 1 

r
 −2 − λ 1     

ke
 1 1 −λ   λ 1 −λ   1 1 −λ 
 

oc
 −1 5 −3   
  3−λ −2 
= λ  0 3 − λ −2  = −λ  = −λ(λ2 −9+12) = −λ(λ2 + 3),

l
−3 − λ 

it. n
 0 6
−3 − λ 

e U
6
√

ov F
thus the eigenvalues are λ = 0 and λ = ±i 3.
m PD
An eigenvector (a1 , b1 , c1 ) corresponding to λ = 0 satisfies
⎧
re h

⎨ 2a1 + 5b1 − 3c1 = 0, 

to atc

b1 = −a1 ,
−a1 − 2b1 + c1 = 0, dvs.
⎩ c1 = a1 + 2b1 = −a1 .
a1 + b1 = 0,
se e B

Hence we may e.g. choose (1, −1, −1).

en W

√
An eigenvector (a2 , b2 , c2 ) corresponding to λ = i 3 satisfies
lic y

⎧ √ ⎧ √
a b

⎨ 2a2 +5b2 −3c2 = i √3a2 , ⎨ 5b2 −3c√2 = (−2+i 3)a2 ,

y ed

−a2 −2b2 +c2 = i√3b2 , dvs. (−2−i√ 3)b2 +c2 = a2 ,

⎩ ⎩
a2 +b2 = i 3c2 , b2 −i 3c2 = −a2 .
Buess

It follows from the latter two equations by an addition

oc

√ √
−(1 + i 3)b2 + (1 − i 3)c2 = 0,
Pr

hence
√ √
1+i 3 (1 + i 3)2 1 √ 1 √
c2 = √ b2 = b2 = · (1 − 3 + 2i 3)b2 = (−1 + i 3)b2 .
1−i 3 1+3 4 2
By insertion into the second equation we get
√ 1 √ 1 √
a2 = (−2 − i 3)b2 + (−1 + i 3)b2 = (−5 − i 3)b2 .
2 2
By choosing b2 = 2 we find the eigenvector
√ √
(−5 − i 3, 2, −1 + i 3)T .

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21
Calculus 4c-3 Homogeneous systems of linear differential equations

√
We get by a complex conjugation that an eigenvector corresponding to λ = −i 3 is given by
√ √
(−5 + i 3, 2, −1 − i 3)T .

The latter two columns of the corresponding fundamental matrix are

⎛ √ ⎞ ⎛ √ ⎞
√ √ √ −5 − 3 √ 3 −5
cos 3t(α β) + sin 3t(−β α) = cos( 3t) ⎝ 2 ⎠
√0 + sin( 3t)
⎝
√0 2 ⎠,
−1 3 − 3 −1

hence a fundamental matrix is given by

⎛ √ √ √ √ √ √ ⎞
1 −5 cos 3t+√ 3 sin 3t − 3 cos 3t−5√ sin 3t
⎝
Φ(t) = −1 ⎠.
√2 cos √3t √ 2√
sin 3t √

r
√

ke
−1 − cos 3t − 3 sin 3t 3 cos 3t − sin 3t

l oc
it. n
Example 1.10 Find the complete solution of the system

e U
 

ov F
 1 0
Y = Y. m PD
2 1

Obviously, λ = 1 is an eigenvalue of multiplicity 2. We have a couple of solution methods.

re h
to atc

1) Discussion of the structure of the solution. The algebraic multiplicity is 2, while the geo-
metric multiplicity is only w. Hence the complete solution must necessarily have the structure
se e B

   
y1 a1 et + a2 tet
= .
b1 et + b2 tet
en W

y2
lic y

It follows by a couple of calculations that

a b

   
d y1 (a1 + a2 )et + a2 tet
y ed

=
dt y2 (b1 + b2 )et + b2 tet
Buess

and
    
a1 et + a2 tet a1 et + a2 tet
oc

1 0
= .
2 1 b1 et + b2 tet (2a1 + b1 )et + (2a2 + b2 )tet
Pr

When we identify the coefficients we find that

a1 + a 2 = a 1 , thus a2 = 0,
b1 + b2 = 2a1 + b1 , thus b2 = 2a1 .

The two free parameters are a1 and b1 , while a2 = 0 and b2 = 2a1 , so

         t  
y1 a1 et t 1 t 0 e 0 a1
= = a 1 e + b 1 e = ,
y2 b1 et + 2a1 tet 2t 1 2tet et b1

where a1 and b1 are arbitrary constants.

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22
Calculus 4c-3 Homogeneous systems of linear differential equations

2) The exponential matrix. Since A and I commute, the exponential matrix is given by

exp(At) = exp((A − I)t + It) = et exp(Bt),

where
 
0 0
B=A−I=
2 0

and where B2 = 0, thus Bn = 0 for n ≥ 2. Then

∞
 
 1 n n
t t
exp(At) = e exp(Bt) = e I + Bt + B t = et {I + Bt}
n=2
n!
     t 
t 1 0 0 0 e 0
= e +t = ,
0 1 2 0 2tet et

r
ke
and the complete solution is

oc
   t  

l
y1 e 0 c1

it. n
= ,
y2 2tet et c2

e U
ov F
where c1 and c2 are arbitrary constants.
m PD
re h
to atc
se e B
en W
lic y
a b

Turning a challenge into a learning curve.

y ed

Just another day at the office for a high performer.

Buess
oc

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Pr

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Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.11 Find the complete solution of the system

y1 = y2 + y3 ,

y2 = y1 + y3 ,
y3 = y1 + y2 .

Here we also have a couple of solution possibilities.

1) The system can also be written
⎛ ⎞ ⎛ ⎞⎛ ⎞
y 0 1 1 y1
d ⎝ 1⎠ ⎝
y2 = 1 0 1 ⎠ ⎝ y2 ⎠ ,
dt
y3 1 1 0 y3

r
so the eigenvalues are the roots of the polynomial

ke
 
 −λ 

oc
 1 1 
 1 −λ 1  = −λ3 + 1 + 1 + λ + λ + λ = −(λ3 − 3λ − 2).
 

l
 1 

it. n
1 −λ

e U
We immediately guess the roots λ = −1 and λ = 2. Then we get by a reduction,

ov F
m PD
−(λ3 − 3λ − 2) = −(λ + 1)(λ − 2)(λ + 1) = −(λ + 1)2 (λ − 2),
re h

so λ1 = λ2 = −1 is a root of multiplicity w, and λ3 = 2 is a simple root.

to atc

If λ = −1, we get the following system of equations for the eigenvectors,

se e B

⎛ ⎞
1 1 1
(A − λI)v = ⎝ 1 1 1 ⎠ v = 0.
en W

1 1 1
lic y
a b

Two linearly independent vectors which satisfy these equations are e.g.
y ed

v1 = (2, −1, −1) and v2 = (1, 1, −2).

Buess
oc

If λ = 2 then we get
Pr

⎛ ⎞
−2 1 1
(A − λI)v = ⎝ 1 −2 1 ⎠ v,
1 1 −2

and we can e.g. choose the solution v3 = (1, 1, 1). The complete solution is
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ −t ⎞⎛ ⎞
y1 2 1 1 2e e−t e2t c1
⎝ y2 ⎠ = c1 e−t ⎝ −1 ⎠ + c2 e−t ⎝ 1 ⎠ + c3 e2t ⎝ 1 ⎠ = ⎝ −e−t e−t e2t ⎠ ⎝ c2 ⎠ ,
y3 −1 −2 1 −e−t −2e−t e2t c3

where c1 , c2 and c3 are arbitrary constants.

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24
Calculus 4c-3 Homogeneous systems of linear differential equations

2) The “fumbling method”. It follows immediately of the symmetry of the equations that
d
(y1 − y2 ) = −(y1 − y2 ), thus y1 − y2 = 3c1 e−t ,
dt
d
(y2 − y3 ) = −(y2 − y3 ), thus y2 − y3 = 3c2 e−t ,
dt
hence by addition y1 − y3 = 3(c1 + c2 )e−t . Finally,
d
(y1 + y2 + y3 ) = 2(y1 + y2 + y3 ), thus y1 + y2 + y3 = 3c3 e2t .
dt
Hence we get

2y1 + y3 = 3c1 e−t + 3c3 e2t ,
y1 − y3 = 3(c1 + c2 )e−t ,

r
i.e.

ke
y1 = (2c1 + c2 )e−t + c3 e2t ,

oc
y2 = (−c1 + c2 )e−t + c3 e2t ,

l
y3 = (−c1 − 2c2 )e−t + c3 e2t ,

it. n
e U
or written in a different way,

ov F
⎛ ⎞ ⎛ ⎞ ⎛
m PD ⎞ ⎛ ⎞
y1 2 1 1
⎝ y2 ⎠ = c1 e−t ⎝ −1 ⎠ + c2 e−t ⎝ 1 ⎠ + c3 e2t ⎝ 1 ⎠ ,
y3 −1 −2 1
re h
to atc

where c1 , c2 and c3 are arbitrary constants.

se e B

Example 1.12 Find the complete solution of the system of differential equations
en W

Y = AY,
lic y
a b

where
⎛ ⎞ ⎛ ⎞
y ed

3 0 4 y1 (t)
A = ⎝ −1 −1 0 ⎠, Y = ⎝ y2 (t) ⎠ .
Buess

−2 0 −3 y3 (t)
oc

The eigenvalues are the roots of the polynomial

Pr

 
 3−λ 0 4   
   
 −1 −1 − λ 0  = (−1 − λ)  3 − λ 4 
   −2 −3 − λ 
 −2 0 −3 − λ 
= −(λ + 1){λ2 − 9 + 8} = −(λ − 1)(λ + 1)2 .
The eigenvalues are the simple root λ = 1 and λ = −1 of multiplicity 2.

The eigenvectors (a, b, c) are determined by the equation

⎧
⎨ 3a + 4c = λa,
−a − b = λb,
⎩
−2a − 3c = λc.

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25
Calculus 4c-3 Homogeneous systems of linear differential equations

If λ = 1, then
⎧
⎨ 2a + 4c = 0, 
a = −2c = −2b,
a + 2b = 0, thus
⎩ (a, b, c) = c(−2, 1, 1).
−2a − 4c = 0,

If λ = −1, then
⎧
⎨ 4a + 4c = 0,
−a = 0, thus a = c = 0, and b is a free parameter.
⎩
−2a − 2c = 0,

Thus we have found two linearly independent solutions. The third solution must have the structure
⎛ ⎞ ⎛ ⎞
y1 a1 e−t + a2 te−t

r
ke
⎝ y2 ⎠ = ⎝ b1 e−t + b2 te−t ⎠ ,
y3 c1 e−t + c2 te−t

l oc
where

it. n
⎛ ⎞ ⎛ ⎞

e U
y (−a1 + a2 )e−t − a2 te−t
d ⎝ 1⎠ ⎝
(−b1 + b2 )e−t − b2 te−t ⎠ ,
ov F
y2 =
dt m PD
y3 (−c1 + c2 )e−t e−t − c2 te−t

and
re h

⎛ ⎞⎛ ⎞ ⎛ ⎞
to atc

3 0 4 y1 (3a1 +4c1 )e−t +(3a2 +4c2 )te−t

⎝ −1 −1 0 ⎠ ⎝ y2 ⎠ = ⎝ (−a1 −b1 )e−t +(−a2 − b2 )te−t ⎠ .
se e B

−2 0 −3 y3 (−2a1 −3c1 )e−t +(−2a2 −3c2 )te−t

en W

We get by identifying the coefficients that

lic y

3a1 + 4c1 = −a1 + a2 , thus 4c1 = −4a1 + a2 ,

a b

−a1 − b1 = −b1 + b2 , thus b2 = −a1 ,

y ed

−2a1 − 3c1 = −c1 + c2 , thus 2c1 + c2 = −2a1 ,

3a2 + 4c2 = −a2 , thus c2 = −a2 ,
Buess

−a2 − b2 = −b2 , thus a2 = 0,

−2a2 − 3c2 = −c2 , thus c2 = −a2 .
oc

It follows from a2 = 0 that c2 = 0, hence c1 = −a1 = b2 . Finally, b1 can be chosen freely.

Pr

The complete solution is

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞
y1 −2 0 1 −2 = et 0 e−t c1
⎝ y2 ⎠ = c1 e ⎝ 1 ⎠ + c2 e ⎝ 1 ⎠ + c3 e ⎝ t ⎠ = ⎝
t −t −t
et e−t te−t ⎠ ⎝ c2 ⎠ ,
y3 1 0 −1 et 0 −e−t c3

where c1 , c2 and c3 are arbitrary constants.

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26
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.13 Find the complete solution of the system

⎛ ⎞
1 1 1 1
⎜0 1 2 1⎟
Y = ⎜
⎝ 0 0 −1
⎟ Y.
1⎠
0 0 0 −1

The matrix is an upper triangular matrix, so it follows immediately by inspection that the two
eigenvalues λ = ±1 both have multiplicity 2. It also follows immediately that y 4 and y3 must have
the simplified structure

y4 = ke−t and y3 = c3 e−t + c4 te−t .

We conclude that the general structure of solution must be

⎛ ⎞ ⎛ ⎞
a1 et +a2 tet +a3 e−t +a4 te−t

r
y1

ke
⎜ y2 ⎟ ⎜ b1 et +b2 tet +b3 e−t +b4 te−t ⎟
⎜ ⎟ ⎜ ⎟.
⎝ y3 ⎠ = ⎝ c3 e−t +c4 te−t ⎠

oc
y4 ke−t

l
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b

The Wake
y ed
Buess

the only emission we want to leave behind

oc
Pr

.QYURGGF'PIKPGU/GFKWOURGGF'PIKPGU6WTDQEJCTIGTU2TQRGNNGTU2TQRWNUKQP2CEMCIGU2TKOG5GTX

6JGFGUKIPQHGEQHTKGPFN[OCTKPGRQYGTCPFRTQRWNUKQPUQNWVKQPUKUETWEKCNHQT/#0&KGUGN6WTDQ
2QYGTEQORGVGPEKGUCTGQHHGTGFYKVJVJGYQTNFoUNCTIGUVGPIKPGRTQITCOOGsJCXKPIQWVRWVUURCPPKPI
HTQOVQM9RGTGPIKPG)GVWRHTQPV
(KPFQWVOQTGCVYYYOCPFKGUGNVWTDQEQO

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Calculus 4c-3 Homogeneous systems of linear differential equations

Since
⎛ ⎞ ⎛ ⎞
y1 (a1 +a2 )et +a2 tet +(−a3 +a4 )e−t −a4 te−t
⎜ ⎟ ⎜
d ⎜ y2 ⎟ ⎜ (b1 +b2 )et +b2 tet +(−b3 +b4 )e−t −b4 te−t ⎟
= ⎟
dt ⎝ y3 ⎠ ⎝ (−c3 +c4 )e−t −c4 te−t ⎠
y4 −ke−t
and
⎛ ⎞⎛ ⎞ ⎛ ⎞
1 1 1 1 y1 (a1 +b1 )et +(a2 +b2 )tet +(a3 +b3 +c3 +k)e−t e−t +(a4 +b4 + c4 )te−t
⎜0 1 2 1⎟ ⎜ y2 ⎟ ⎜ b1 et +b2 tet +(b3 +2c3 +k)e−t +(b4 +2c4 )te−t ⎟
⎜ ⎟⎜ ⎟=⎜ ⎟,
⎝0 0 −1 1 ⎠ ⎝ y3 ⎠ ⎝ (−c3 +k)e−t −c4 te−t ⎠
0 0 0 −1 y4 −ke−t
we conclude by identifying the coefficients that
 
a 1 + b 1 = a1 + a2 , a2 + b 2 = a 2 ,

r
b1 = b 1 + b 2 , b2 = b 2 ,

ke
and

oc
⎧ ⎧
⎨ a3 + b3 + c3 + k = −a3 + a4 , ⎨ a4 + b4 + c4 = −a4 ,

l
it. n
b3 + 2c3 + k = −b3 + b4 , b4 + 2c4 = −b4 ,
⎩ ⎩

e U
−c3 + k = −c3 + c4 , −c4 = −c4 .

ov F
It follows immediately from these equations that
m PD
b2 = 0, b4 = −c4 = −k, b 1 = a2 .
re h

Then the equations are reduced to

to atc

⎧
⎨ b3 + c3 = −2a3 + a4 − k,
se e B

2b3 + 2c3 = −2k,

⎩
k = −2a4 + k,
en W

hence
lic y

b3 + c3 = −k, a4 = 0 = a3 , thus c3 = −k − b3 .
a b
y ed

Let the free parameters be a1 , a2 , b3 and k. Then

Buess

a3 = a4 = 0, b 1 = a2 , b2 = 0, b4 = −k, c3 = −k − b3 , c4 = k.
The complete solution is
oc

⎛ ⎞ ⎛ ⎞
y1 a1 et + a2 tet
Pr

⎜ y2 ⎟ ⎜ t −t −t ⎟
⎜ ⎟ = ⎜ a2 e + b3 e −t − kte −t ⎟
⎝ y3 ⎠ ⎝ (−k − b3 )e + kte ⎠
y4 ke−t
⎛ t⎞ ⎛ t⎞ ⎛ ⎞ ⎛ ⎞
e te 0 0
⎜ 0 ⎟ ⎜ et ⎟ ⎜ e−t ⎟ ⎜ −te−t ⎟
= a1 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜
⎝ 0 ⎠ + a2 ⎝ 0 ⎠ + b3 ⎝ −e−t ⎠ + k ⎝ (t − 1)e−t
⎟
⎠
0 0 0 e−t
⎛ t ⎞ ⎛ ⎞
e tet 0 0 a1
⎜ 0 et e −t
−te −t ⎟ ⎜ a2 ⎟
= ⎜ ⎝ 0 0 −e−t (t − 1)e−t ⎠ ⎝ b3 ⎠ ,
⎟⎜ ⎟

0 0 0 e−t k

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28
Calculus 4c-3 Homogeneous systems of linear differential equations

where a1 , a2 , b3 and k are arbitrary constants.

Example 1.14 Find the complete solution of the homogeneous system

    
d x1 1 1 x1
= , t ∈ R.
dt x2 4 1 x2

The characteristic polynomial

 
 1−λ 1 
 = (λ−1)2 −4 = (λ−1)2 −22 = (λ+1)(λ−3)
 4 1−λ 

has the roots λ1 = −1 and λ3 = 3.

An eigenvector corresponding to an eigenvalue λ is a cross vector of

r
ke
(1 − λ, 1)

oc
[first row in the matrix A − λI].

l
it. n
If λ1 = −1, then e.g. v1 = (1, −2)T .

e U
If λ3 = 3, then e.g. v2 = (1, 2)T .
The complete solution is
ov F
m PD
       −t  
x1 1 −t 1 3t e e3t c1
= c1 e + c2 e =
−2e−t 2e3t
re h

x2 −2 2 c2
 
to atc

c1 e−t + c2 e3t
= , t ∈ R,
−2c1 e−t + 3c2 e3t
se e B

where c1 and c2 are arbitrary constants, and where we have indicated three equivalent results.
en W

 T
lic y

dx dx1 dx2
Example 1.15 Given x = (x1 , x2 )T , = , , and
a b

dt dt dt
y ed

 
−7 2
Buess

A= .
−36 10
oc

Find that solution of the system of differential equations

Pr

dx
= A x, t ∈ R,
dt
for which x(0) = (1, 5)T .

The characteristic polynomial

 
 −λ−7 2 
 2
 −36 10−λ  = (λ+7)(λ−10)+72 = λ −3λ+2

has the roots λ1 = 1 and λ2 = 2.

If λ1 = 1, then an eigenvector is a cross vector of (−8, 2), e.g. v1 = (1, 4)T .

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29
Calculus 4c-3 Homogeneous systems of linear differential equations

If λ2 = 2, then an eigenvector is a cross vector of (−9, 2), e.g. v2 = (2, 9)T .

The complete solution is
     
x1 (t) 1 2
= c1 et + c2 e2t .
x2 (t) 4 9

We get for t = 0,
       
1 1 2 c1 + 2c2
= c1 + c2 = ,
5 4 9 4c1 + 9c2

hence c1 = −1 and c2 = 1.
The particular solution is then given by
   
x1 (t) −et + 2e2t
= , t ∈ R.

r
x2 (t) −4et + 9e2t

ke
l oc
it. n
Example 1.16 Find the complete solution of the homogeneous system

e U
    
d x1 −3 1 x1

ov F
= , t ∈ R.
dt x2 −1 −3 x2 m PD
re h

The characteristic polynomial

to atc

 
 −3 − λ 1 
  = (λ + 3)2 + 1
 −1 −3 − λ 
se e B

has the complex conjugated roots a ± iω = −3 ± 1 · i.

en W

A complex eigenvector α + iβ corresponding to −3 + i is a cross vector to (−i, 1), e.g.

lic y

         
a b

1 1 0 1 0
α + iβ = = +i , α= , β= .
y ed

i 0 1 0 1
Buess

Then a fundamental matrix is given by

   
1 0 0 1
oc

Φ(t) = eat cos ωt(α β) + eat sin ωt(−β α) = e−3t cos t + e−3t sin t
0 1 −1 0
 
Pr

cos t sin t
= e−3t .
− sin t cos t
The complete solution is then
      −3t 
x1 (t) cos t sin t c1 e (c1 cos t + c2 sin t)
= e−3t = ,
x2 (t) − sin t cos t c2 e−3t (−c1 sin t + c2 cos t)

where c1 and c2 are arbitrary constants.

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30
Calculus 4c-3 Homogeneous systems of linear differential equations

Example 1.17 Find the complete solution of the homogeneous system

    
d x1 1 3 x1
= .
dt x2 4 5 x2

The characteristic polynomial is

 
 1−λ 3 
 = (λ − 1)(λ − 5) − 12 = λ2 − 6λ − 7 = (λ − 3)2 − 16 = (λ − 7)(λ + 1),
 4 5−λ 

so the eigenvalues are λ1 = −1 and λ2 = 7.

Once the characteristic polynomial has been found, there are several ways to continue. We shall here
give some variants.

r
First variant. The eigenvalue method. The eigenvector corresponding to an eigenvalue λ is a

ke
cross vector to (1 − λ, 3).

oc
If λ1 = −1, then we e.g. get v1 = (3, −2)T .
If λ2 = 7, then we e.g. get v2 = (1, 2)T .

l
it. n
e U
The complete solution is
        

ov F
x1 3 1 3e−t e7t c1
e−t + c2 e7t =
x2
= c1
−2 2
m PD −2e−t 2e7t c2
, t ∈ R,
re h

where c1 and c2 are arbitrary constants.

to atc
se e B
en W
lic y
a b
y ed

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oc

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Calculus 4c-3 Homogeneous systems of linear differential equations

Second variant. Discussion of the structure of the solution. The solution must necessarily
have the structure
   −t 
x1 ae + be7t
= ,
x2 ce−t + de7t

where we shall eliminate two of the parameters. We first calculate

   
d x1 −ae−t + 7be7t
=
dt x2 −ce−t + 7de7t

and
    
1 3 ae−t + be7t (a + 3c)e−t + (b + 3d)e7t
= .
4 5 ce−t + de7t (4a + 5c)e−t + (4b + 5b)e7t

Now, e−t and e7t are linearly independent, so we get by an identification of the coefficients that

r
ke
 
−a = a + 3c, 7b = b + 3d,

oc
og
−c = 4a + 3c, 7d = 4b + 5b,

l
it. n
hence 2a + 3c = 0 and 2b = d.

e U
It follows that we may choose a = 3, c = −2, and b = 1, d = 2, and then we obtain the complete

ov F
solution
m PD
        
x1 3 1 3e−t e7t c1
= c1 e−t + c2 e7t = , t ∈ R,
re h

x2 −2 2 −2e−t 2e7t c2
to atc

where c1 and c2 are arbitrary constants.

se e B

Third variant. The fumbling method. We expand the system,

⎧
en W

⎪
⎨ dx1 = x1 + 3x2 , 1 dx1 1
dvs. x2 = − x1 ,
lic y

dt 3 dt 3
⎪ dx2
a b

⎩ = 4x1 + 5x2 .
dt
y ed

Here we eliminate x2 ,
Buess

 
dx2 1 d2 x1 1 dx1 5 dx1 5
= − = 4x1 + 5x2 = + 4− x1 ,
oc

dt 3 dt2 3 dt 3 dt 3
Pr

hence by a reduction,

d2 x1 dx1
2
−6 − 7x1 = 0.
dt dt
The characteristic equation R2 − 6R − 7 = 0 has the roots R = −1 and R = 7, so

x1 = ae−t + be7t ,

hence by putting this into the first equation,

 
1 dx1 1  2
x2 = − x1 = −ae−t + 7be7t − ae−t − be7t = − ae−t + 2be7t .
3 dt 3 3

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32
Calculus 4c-3 Homogeneous systems of linear differential equations

a
If we write c1 =and c2 = b, the complete solution is
3
       
x1 ae−t + be7t −t 3 7t 1
= = c 1 e + c 2 e .
x2 − 23 ae−t + 2be7t −2 2

Fourth variant. The exponential matrix. This is given by a formula,

−λ2 eλ1 t +λ1 eλ2 t eλ1 t − eλ2 t 1 1 −t
exp(At) = I+ A = − {−7e−t − e7t }I − {e − e7t }A
λ1 − λ 2 λ1 − λ 2 8 8
   7t −t 
1 7e−t +e7t 0 1 e −e 3e7t −3e−t
= −t 7t + 7t −t
8 0 7e +e 8 4e −4e 5e7t −5e−t
 
1 6e−t + 2e7t −3e−t + 3e7t
= −t .
8 −4e + 4e7t 2e−t + 6e7t

r
ke
1
Here can be built into the arbitrary constants, so the complete solution is

oc
8
    

l
it. n
x1 6e−t + 2e7t −3e−t + 3e7t c1

e U
= .
x2 −4e−t + 4e7t 2e−t + 6e7t c2

ov F
m PD
Fifth variant. (Sketch). It is also to find the exponential matrix by using its structure
re h

exp(At) = ϕ(t)I + ψ(t)A, ϕ(0) = 1, ψ(0) = 0,

to atc

and by checking the matrix differential equation,

se e B

d
exp(At) = A exp(At)
en W

dt
lic y

and finally apply Caley-Hamilton’s equation,

a b

A2 − 6A − 7I = 0, dvs. A2 = 6A + 7I.
y ed
Buess

However, if one does not use some clever calculational tricks, one may easily end up in a mess of
formulæ, so this variant is not given here in all its details.
oc
Pr

Example 1.18 Find the complete solution of the homogeneous system

    
d x1 2 3 x1
= .
dt x2 3 2 x2

Here the eigenvalue method is the simplest method.

The eigenvalues are the roots of the equation

 
 2−λ 3 
 = (2−λ)2 −32 = (λ−5)(λ+1) = 0,
 3 2−λ 

thus λ = 5 or λ = −1.

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33
Calculus 4c-3 Homogeneous systems of linear differential equations

The eigenvectors. An eigenvector v is a cross vector to 2 − λ, 3)

If λ = 5, then we get a cross vector (−3, 3), so we may choose v1 = (1, 1).
If λ = −1, then we get the cross vector (3, 3), and we may choose v2 = (1, −1).
The complete solution is
       5t  
x1 5t 1 t
1 e e−t c1
= c1 e + c2 e = ,
x2 1 −1 e5t e−t c2
where c1 , c2 are arbitrary constants.

Example 1.19 Find the complete solution of the homogeneous system

    
d x1 1 3 x1
= .
dt x2 4 2 x2

r
ke
First we find the eigenvalues of the matrix:
 

oc
 1−λ 3 
  = (λ−1)(λ−2)−12 = λ2 −3λ−10 = 0,
 4 

l
2−λ

it. n
e U
hence the eigenvalues are λ = −2 and λ = 5.

ov F
If λ = −2, then v1 = (1, 1) is an eigenvector.
m PD
If λ = 5, then v2 = (3, 4) is an eigenvector.
The complete solution is
re h

       −2t  
to atc

x1 −2t 1 5t 3 e 3e5t c1
= c1 e + c2 e = .
x2 −1 4 −e−2t 4e5t c2
se e B
en W

Example 1.20 Find the complete solution of the homogeneous system

    
lic y

d
a b

x1 1 5 x1
= .
dt x2 1 −3 x2
y ed
Buess

We shall here demonstrate three variants.

1) The eigenvalue method. The eigenvalues are the roots of the characteristic polynomial
oc

 
 1−λ 
Pr

 5  = (λ−1)(λ+3)−5 = λ2 +2λ−8 = (λ+1)2 −9,

 1 −3−λ 
hence

2,
λ = −1 ± 3 =
−4.
a) If λ = 2, then we get the matrix
   
1−λ 5 −1 5
= ,
1 −3 − λ 1 −5
and we conclude that we may choose (5, 1) as an eigenvector.

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34
Calculus 4c-3 Homogeneous systems of linear differential equations

b) If λ = −4, then we get the matrix

   
1−λ 5 5 5
= ,
1 −3 − λ 1 1

and we choose e.g. (1, −1) as an eigenvector.

Summing up, the complete solution is

       2t  
x1 5 1 5e e−4t c1
= c1 e2t + c2 e−4t = .
x2 1 −1 e2t −e−4t c2

2) The fumbling method. We expand the system of equations,

⎧
⎪
⎨ dx1 = x1 + 5x2 ,
dt

r
ke
⎪
⎩
dx2
= x1 − 3x2 .
dt

l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Homogeneous systems of linear differential equations

It follows from the latter equation that

dx2
(5) x1 = + 3x2 ,
dt
hence by insertion into the former one,
dx1 d2 x2 dx2 dx2
= 2
+3 = x1 + 5x2 = + 8x2 .
dt dt dt dt
Then by a rearrangement
d2 x2 dx2
+2 − 8x2 = 0.
dt2 dt
The characteristic equation R2 + 2R − 8 = 0 has the roots R = 2 and R = −4, so
x2 = c2 e2t + c2 e−4t .
If we put this into (5), we get

r
ke
dx2
x1 = + 3x2 = (2c1 e2t − 4c2 e−4t ) + (3c1 e2t + 3c2 e−4t ) = 5c1 e2t − c2 e−4t .
dt

oc
Summing up we get

l
     2t  

it. n
x1 5c1 e2t − c2 e−4t 5e −e−4t c1

e U
= = .
x2 c1 e2t + c2 e−4t e2t e−4t c2

ov F
m PD
3) The exponential matrix. The characteristic polynomial is
(λ + 1)2 − 9.
re h
to atc

Then by Caley-Hamilton’s theorem,

(A + I)2 − 9I = 0, dvs. B2 = 9I, hvor B = A + I.
se e B

Since I and A commute, we have

en W

exp(At) = exp((B − I)t) = e−t exp(Bt)

∞ ∞

 1 
lic y

1
= e−t B2n t2n + B2n+1 t2n+1
a b

n=0
(2n)! n=0
(2n + 1)!
y ed

∞ ∞

 (3t)2n 1  (3t)2n+1
−t
Buess

= e I+ B
n=0
(2n)! 3 n=0 (2n + 1)!
 
1
oc

−t
= e cosh(3t)I + sinh(3t)B
3
    
Pr

−t 1 0 1 2 5
= e cosh(3t) + sinh(3t)
0 1 3 1 −2
 
1 −t 3 cosh 3t + 2 sinh 3t 5 sinh 3t
= e
3 sinh 3t 3 cosh 3t − 2 sinh 3t
 3t 
1 −t 3e +3e +2e −2e−3t
−3t 3t
5e3t − 5e−3t
= e
6 e3t −e−3t 3e3t +3e−3t −2e3t +2e−3t
 3t −3t 
1 −t 5e +e 5e3t −5e−3t
= e 3t −3t
6 e −e e3t +5e−3t
 2t −4t 
1 5e +e 5e2t −5e−4t
= .
6 e2t −e−4t e2t +5e−4t

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36
Calculus 4c-3 Homogeneous systems of linear differential equations

Thus the complete solution is

   2t −4t   2t 
x1 5e +e 5e −5e−4t
= c1 + c 2 .
x2 e2t −e−4t e2t +5e−4t

Example 1.21 Find the complete solution of the homogeneous system

    
d x1 4 3 x1
=
dt x2 3 −4 x2
.

We shall here only apply the eigenvalue method, even if other methods may also be applied.

r
The characteristic polynomial

ke
 
 4−λ 

oc
 3  = λ2 − 25
 3 −4 − λ 

l
it. n
e U
has the roots λ = ±5.

ov F
If λ = 5, then m PD
   
4−5 3 −1 3
A − λI = = ,
3 −4 − 5 3 −9
re h
to atc

hence (3, 1) is an eigenvector corresponding to λ = 5.

If λ = −5, then
se e B

   
4+5 3 9 3
en W

A − λI = = ,
3 −4 + 5 3 1
lic y

hence (1, −3) is an eigenvector corresponding to λ = −5.

a b

The complete solution is

y ed

     5t  
e−5t
Buess

3 1 3e c1
x(t) = c1 e5t + c2 e−5t = ,
1 −3 e5t −3e−5t c2
oc

for t ∈ R, where c1 and c2 are arbitrary constants.

Pr

Example 1.22 Find the complete solution of the homogeneous system

    
d x1 1 2 x1
= .
dt x2 −3 8 x2

It follows from
 
 1−λ 2 
  = (1−λ)(8−λ)+6 = λ2 −9λ+14 = (λ−7)(λ−2),
 −3 8−λ 

that the eigenvalues are λ = 2 and λ = 7.

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37
Calculus 4c-3 Homogeneous systems of linear differential equations

1) If λ = 2, then an eigenvector is a cross vector to (1 − λ, 2) = (−1, 2), so we get e.g. (2, 1) as an

eigenvector.

2) If λ = 7, then an eigenvector is a cross vector to (1 − λ, 2) = (−6, 2), so we get e.g. (1, 3) as an

eigenvector.
The complete solution is
     
x1 2 1
= c1 e2t + c2 e7t .
x2 1 3

Example 1.23 Find the complete solution of the homogeneous system

    
d x1 4 6 x1

r
= .

ke
dt x2 8 2 x2

oc
The characteristic equation is

l
it. n
 
 4−λ 6 

e U
 = (λ−4)(λ−2)−48 = λ2 −6λ−40 = (λ−3)2 − 72 = 0.
 8 2−λ 

ov F
m PD
We get the two eigenvalues

re h

10,
λ=3±7=
to atc

−4.
se e B

An eigenvector corresponding to λ = 10 is a cross vector to (4 − 10, 6)T = (−6, 6)T , e.g. (1, 1).
An eigenvector corresponding to λ = −4 is a cross vector to (4 − (−4), 6) T = (8, 6)T , e.g. (3, −4).
en W

The complete solution is

       
lic y

c1 e10t + 3c2 e−4t

a b

x1 (t) 1 3
= c1 e10t + c2 e−4t = ,
x2 (t) 1 −4 c1 e10t − 4c2 e−4t
y ed
Buess

where c1 and c2 are arbitrary constants.

oc

Example 1.24 Given the matrix A by

 
Pr

1 2
A= .
3 −4

Find exp(At) = eAt .

The characteristic polynomial

 
 1−λ 2 
  = (λ−1)(λ+4)−6 = λ2 +3λ−10 = (λ−2)(λ+5)
 3 −4−λ 

has the simple roots λ = 2 and λ = −5. Then we have two methods:

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38
Calculus 4c-3 Homogeneous systems of linear differential equations

1) Definition of the exponential matrix. Since I and A commute, we get

exp(At) = exp((A−2I)t+2tI) = e2t exp(Bt),

where
 
−1 2
B = A − 2I =
3 −6

and
    
−1 2 −1 2 7 −14
B2 = = = −7B.
3 −6 3 −6 −21 42

It follows by induction that Bn = (−7)n−1 B, n ∈ N. Then

   

r
∞ ∞
(−7)n−1 n

ke
2t 2t 1 n n 2t
exp(At) = e exp(Bt) = e I+ B t =e I+ t ·B
n! n!

oc
n=1 n=1
 ∞
  
1 1 1 −7t

l
2t n 2t
I− I − (e − 1)B

it. n
= e (−7t) B = e
7 n=1 n! 7

e U
 
6e2t + e−5t 2e2t − 2e−5t

ov F
1
= .
3e2t − 3e−5t e2t + 6e−5t
7
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Homogeneous systems of linear differential equations

2) The eigenvalue method. We have previously found the eigenvalues λ = 2 and λ = −5. We
choose an eigenvector as a cross vector to

(1 − λ, 2) or to (3, −4 − λ).

To λ = 2 corresponds e.g. the eigenvector v1 = (2, λ − 1) = (2, 1).

To λ = −5 corresponds e.g. the eigenvector v2 = (−4 − λ, −3) = (1, −3).

Then a fundamental matrix is

      2t 
2 1 2e e−5t
Φ(t) = e2t e−5t = .
1 −3 e2t −3e−5t

Now,
   

r
ke
2 1 −1 1 3 1
Φ(0) = og Φ(0) = ,
1 −3 7 1 −2

l oc
so the exponential matrix is

it. n
 2t    

e U
2e e−5t 3 1 1 1 6e2t + e−5t 2e2t − 2e−5t
exp(At) = Φ(t)Φ(0)−1 = = .

ov F
e2tm PD −3e−5t 1 −2 7 7 3e2t − 3e−5t e2t + 6e−5t
re h

Example 1.25 Find the complete solution of the homogeneous system

to atc

⎛ ⎞ ⎛ ⎞⎛ ⎞
x1 1 1 1 x1
se e B

d ⎝
x2 ⎠ = ⎝ 1 1 1 ⎠ ⎝ x2 ⎠ .
dt
x3 1 1 1 x3
en W
lic y

We shall here give four variants.

a b
y ed

1) The fumbling method. In the actual case this is the simplest variant. It follows immediately
from the system of equations that
Buess

dx1 dx2 dx3

= = = x1 + x2 + x3 ,
oc

dt dt dt
Pr

hence (by some conveniently chosen constants)

x2 = x1 + 3c2 , x3 = x1 + 3c3 ,

and
d
(x1 + x2 + x3 ) = 3(x1 + x2 + x3 ).
dt
We obtain from these equations that

x1 + x2 + x3 = 3x1 + 3c2 + 3c3 = 3c1 e3t ,

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40
Calculus 4c-3 Homogeneous systems of linear differential equations

hence
⎧
⎨ x1 = c1 e3t −c2 −c3 ,
x2 = c1 e3t +2c2 −c3 ,
⎩
x3 = c1 e3t −c2 +2c3 ,
and thus
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x1 1 −1 −1
⎝ x2 ⎠ = c1 e−3t ⎝ 1 ⎠ + c2 ⎝ 2 ⎠ + c3 ⎝ −1 ⎠ .
x3 1 −1 2

2) The standard method. The eigenvalues of the matrix are the solutions of the equation
 
 1−λ 1 1 
 
0 =  1 1−λ 1  = (1 − λ)3 + 2 − 3(1 − λ) = −λ3 + 3λ2 = −λ2 (λ − 3).

 1 1 1−λ 

r
It follows that λ = 3 is a simple root and that λ = 0 is a double root. Since the matrix A is

ke
symmetric, its algebraic multiplicity is equal to its geometric multiplicity for λ = 0.

oc
Let y = (y1 , y2 , y3 ) be an eigenvector corresponding to the eigenvalue λ = 3, thus

l
⎛ ⎞ ⎛ ⎞

it. n
e U
y1 y1 + y 2 + y 3
3 ⎝ y2 ⎠ = ⎝ y1 + y 2 + y 2 ⎠ .

ov F
y3 y1 + y 2 + y 3
m PD
It follows immediately that y1 = y2 = y3 , so we may choose (1, 1, 1) as an eigenvector.
re h

If λ = 0 we get analogously the condition

to atc

y1 + y2 + y3 = 0,
se e B

which describes a plane in space. We shall only choose two linearly independent vectors, the
coordinates of which satisfy this condition. This may be done in several ways. If we e.g. choose
en W

(1, −1, 0) and (1, 0, −1), then we get the complete solution
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
lic y

1 1 1
a b

x(t) = c1 e3t ⎝ 1 ⎠ + c2 ⎝ −1 ⎠ + c3 ⎝ 0 ⎠ .
y ed

1 0 1
Buess

3) Calculation of the exponential matrix. It follows immediately that

⎛ ⎞
oc

1 1 1
A=⎝1 1 1 ⎠ , A2 = 3A, . . . , An = 3n−1 A,
Pr

1 1 1
so by insertion into the exponential series,
∞
∞ 
 1 n n  1
n−1 n
exp(At) = A t =I+ 3 t A
n=0
n! n=1
n!
 ∞

1  1 1
= I+ 1+ (3t) − 1 A = I + (e3t − 1)A
n
3 n=1
n! 3
⎛ ⎞ ⎛ 3t ⎞ ⎛ 3t ⎞
1 0 0 e − 1 e − 1 e3t − 1
3t
e + 2 e3t − 1 e3t − 1
1 1
= ⎝ 0 1 0 ⎠ + ⎝ e3t − 1 e3t − 1 e3t − 1 ⎠ = ⎝ e3t − 1 e3t + 2 e3t − 1 ⎠ .
3 3t 3t 3t 3
0 0 1 e −1 e −1 e −1 e3t − 1 e3t − 1 e3t + 2

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41
Calculus 4c-3 Homogeneous systems of linear differential equations

The complete solution is all linear combinations of the columns of the exponential matrix,
⎛ 3t ⎞ ⎛ 3t ⎞ ⎛ 3t ⎞
e +2 e −1 e −1
x(t) = c1 ⎝ e3t − 1 ⎠ + c2 ⎝ e3t + 2 ⎠ + c3 ⎝ e3t − 1 ⎠ .
e3t − 1 e3t − 1 e3t + 2

Remark 1.1 The three solutions are of course equivalent, even though the constants do not
correspond here.

4) Cayley-Hamilton’s theorem. We prove as in (2) that the characteristic polynomial is

(−1)3 det(A − λI) = λ3 − 3λ2 = λ2 (λ − 3).

The corresponding differential equation

r
ke
d3 x d2 x
−3 2 =0

oc
dt 3 dt

l
it. n
has the complete solution x(t), where

e U
x(t) = c1 + c2 t + c3 e3t ,

ov F

x (t) = c2 + 3c3 e3t , m PD

x (t) = 9c3 e3t .
re h

The initial conditions are

to atc

(j)
xi (0) = δij for i, j = 0, 1, 2.
se e B

If i = 0, then
⎧ ⎧
en W

⎨ c1 + c3 = 1, ⎨ c1 = 1,
c2 + 3c3 = 0, thus c2 = 0,
lic y

⎩ ⎩
a b

9c3 = 0, c3 = 0,
y ed

hence x0 (t) = 1.
Buess

If i = 1, then
⎧ ⎧
⎨ c1 + c3 = 0, ⎨ c1 = 0,
oc

c2 + 3c3 = 1, thus c2 = 1,
⎩ ⎩
Pr

9c3 = 0, c3 = 0,

hence x1 (t) = t.
If i = 2, then
⎧ ⎧
⎨ c1 + c3 = 0, ⎨ c1 = −1/9,
c2 + 3c3 = 0, thus c2 = −1/3,
⎩ ⎩
9c3 = 1, c3 = 1/9,

hence
1 1 1
x2 (t) = − − t + e3t .
9 3 9

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42
Calculus 4c-3 Homogeneous systems of linear differential equations

Then by Caley-Hamilton’s theorem A2 = 3A, and we get from the above that the exponential
matrix is
 
2 1 1 1 3t
exp(At) = x0 (t)I + x1 (t)A + x2 (t)A = I + tA + − − t + e 3A
9 3 9
⎛ 3t ⎞
e + 2 e3t − 1 e3t − 1
1 3t 1 1
= I + (e − 1)A = {3I + (e3t − 1)A} = ⎝ e3t − 1 e3t + 2 e3t − 1 ⎠ .
3 3 3
e3t − 1 e3t − 1 e3t + 2

The complete solution of the differential equation is composed of all linear combinations of the
columns, i.e.
⎛ 3t ⎞ ⎛ 3t ⎞ ⎛ 3t ⎞
e +2 e −1 e −1
x(t) = c1 ⎝ e3t − 1 ⎠ + c2 ⎝ e3t + 2 ⎠ + c3 ⎝ e3t − 1 ⎠ ,
e3t − 1 e3t − 1 e3t + 2

r
ke
where c1 , c2 , c3 are arbitrary constants.

l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b

DO YOU WANT TO KNOW:

y ed
Buess

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oc
Pr

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Calculus 4c-3 Inhomogeneous systems of linear differential equations

2 Inhomogeneous systems of linear differential equations

Example 2.1 Find the complete solution of the system
      
d x1 0 −1 x1 cos t
= + .
dt x2 1 −1 x2 sin t

The eigenvalues are the roots of the polynomial

 
 −λ −1 
 2
 1 −1 − λ  = λ(λ + 1) + 1 = λ + λ + 1,

so we have the complex conjugated eigenvalues

√
1 3
λ=− ±i .

r
2 2

ke
√

oc
1 3
1) Complex eigenvectors. If λ = − + i , then we get the matrix equation
2 2

l
it. n
⎛ √ ⎞

e U
   1
− i
3
−1    
−λ −1 w1 ⎜ √ ⎟ w1 0
=⎝ 2 2

ov F
1 −1−λ w2 ⎠ w =
0
.
m PD 1 3 2
1 − −i
2 2
re h
A solution is a cross vector of e.g. the first row,
to atc

 √ 
1 3 1 √
+1, − i = {(2, 1) − i(0, 3)},
se e B

2 2 2
en W

hence we can choose (multiply by 2),

    √
lic y

2 √0 1 3
a b

v1 = −i for λ1 = − + i .
1 3 2 2
y ed

Analogously we get
Buess

    √
2 0
√ 1 3
for λ2 = − − i
oc

v2 = +i .
1 3 2 2
Pr

The complete complex solution of the homogeneous equation is

 
x1
= c̃1 eλ1 t v1 + c̃2 eλ2 t v2
x2
 √ √     
−t/2 3 3 2 √0
= c̃1 e cos t + i sin t −i
2 2 1 3
 √ √     
3 3 2 0
+c̃2 e−t/2 cos t − i sin t +i √ .
2 2 1 3

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44
Calculus 4c-3 Inhomogeneous systems of linear differential equations

We get by splitting into the real and the imaginary part,

   √   √  
x1 −t/2 3 2 3 √0
= c̃1 e cos t + sin t
x2 2 1 2 3
 √   √   
3 2 3 0
+i sin t − cos t √
2 1 2 3
 √   √  
−t/2 3 2 3 √0
+c̃2 e cos t + sin t
2 1 2 3
 √   √   
3 2 3 0
−i sin t − cos t √ .
2 1 2 3

We obtain the real complete solution by choosing c̃2 = c̃1 , hence with new (real) arbitrary constants

r
ke
we get the complete real solution of the homogeneous equation
⎛ √ ⎞ ⎛ √ ⎞

oc
  2 cos
3
t 2 sin
3
t
x1 ⎜ √ ⎟ −t/2 ⎜ √ ⎟

l
= c1 e−t/2 ⎝ √ 2 √ 2

it. n
x2 √ ⎠ + c2 e ⎝ √ ⎠.
3 3 3 3

e U
cos t + 3 · sin t sin t − 3 · cos t
2 2 2 2

ov F √
m PD
1 3
2) Alternatively one may only use real calculations. In fact, since λ = − ± i , the complete
2 2
re h

solution of the homogeneous equation must have the structure

to atc

⎛ √ √ ⎞
  a cos
3
t + a sin
3
t⎟
se e B

−t/2 ⎜
x1 1 2
=e ⎝ √2 √2 ⎠.
x2 3 3
b1 cos t + b2 sin t
en W

2 2
We know that we have two arbitrary constants in the final solution, and here we have got four
lic y
a b

unknowns a1 , a2 , b1 , b2 , so we still have to eliminate two of them by means of the differential

y ed

equation. Now,
⎛ √  √  √  √ ⎞
Buess

1 3 3 3 1 3
  ⎜ − a1 + a2 cos t+ − a1 − a2 sin t⎟
d x1 ⎜
−t/2 ⎜ 
2 2 2 2 2 2 ⎟
√  √  √  √ ⎟
oc

=e ⎜
dt x2 ⎝ 1 3 3 3 1 3 ⎟ ⎠
− b1 + b2 cos + − b1 − b2 sin t
Pr

2 2 2 2 2 2

and
⎛ √ √ ⎞
   −b1 cos
3
t − b sin
3
t
0 −1 x1 ⎜ √2
2
2 √ ⎟
= e−t/2 ⎝ ⎠,
1 −1 x2 3 3
(a1 −b1 ) cos t+(a2 −b2 ) sin t
2 2
so it follows by an identification of the coefficients of the first row that
√ √
1 3 1 3
− a1 + a2 = −b1 , thus b1 = a1 − a2 ,
2 2 2 2

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45
Calculus 4c-3 Inhomogeneous systems of linear differential equations

√ √
3 1 3 1
− a1 − a2 = −b2 , thus b2 = a1 + a2 .
2 2 2 2
We shall not calculate the latter two equations from the second row. One may if necessary use
them as a control.

Since b1 and b2 are uniquely determined by a1 and a2 , the complete solution of the homogeneous
equation with a1 and a2 as the arbitrary constants becomes
⎛ √ ⎞ ⎛ √ ⎞
  cos
3
t sin
3
t
x1 ⎜ √ 2√ √ ⎟ −t/2 ⎜ √ ⎟
x2
= a1 e−t/2 ⎝ ⎠ + a2 e ⎝ √ √ 2 ⎠.
1 3 3 3 3 3 1 3
cos t+ sin t − cos t + sin t
2 2 2 2 2 2 2 2

Remark 2.1 When we compare with the solution of (1), it follows that a1 = 2c1 og a2 = 2c2 .

r
ke
Remark 2.2 Since
√ √ √ √  √ √ √ √ 

oc
1 3 3 3 3 π 3 3 1 3 3 π
cos t+ sin t = cos t− , − cos t + sin t = sin t− ,

l
it. n
2 2 2 2 2 3 2 2 2 2 2 3

e U
the complete solution can be written
⎛ √
ov F ⎞ ⎛ √ ⎞
m PD
3 3
  cos t sin t
x1 ⎜ √ 2 ⎟ ⎜ √ 2 ⎟
= a1 e−t/2 ⎜ ⎟ + a2 e−t/2 ⎜ ⎟
π ⎠.
re h

x2 ⎝ 3 π ⎠ ⎝ 3
t− t−
to atc

cos sin
2 3 2 3
se e B

However, this reformulation is not necessary.

en W

3) Alternatively we may use the “fumbling method”. Expanding the homogeneous system of
equations we get
lic y

⎧
a b

⎪
⎪ dx1 dx1
⎨ dt = −x2 ,
⎪ thus x2 = − ,
y ed

dt
⎪
Buess

⎪
⎪ dx2
⎩ = x1 − x2 .
dt
oc

Here we eliminate x2 from the latter equation,

Pr

d2 x1 dx1
− = x1 + ,
dt2 dt
thus
d2 x1 dx1 dx1
2
+ + x1 = 0 og x2 = − .
dt dt dt
√
2 1 3
The characteristic polynomial R +R+1 has the roots R = − ±i (the same as the eigenvalues),
2 2
so the complete solution is
√ √
−t/2 3 −t/2 3
x1 (t) = a1 e cos t + a2 e sin t.
2 2

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46
Calculus 4c-3 Inhomogeneous systems of linear differential equations

Since x2 = −dx1 /dt, it follows that

 √ √ √   √ √ √ 
−t/2 1 3 3 3 −t/2 3 3 1 3
x2 (t) = a1 e cos t+ sin t + a2 e − cos t + sin t ,
2 2 2 2 2 2 2 2

which is seen to be equivalent to the previously found solutions.

The inhomogeneous equation. Even if one should know a fundamental matrix, it cannot be rec-
ommended to apply the formal solution formula. This would give us the following difficult expression,
⎛ √ √ ⎞
3 3
⎜ cos t sin t
Φ(t) = e−t/2 ⎝ √ √
2 √ √ √ 2 √ ⎟ ⎠.
1 3 3 3 3 3 1 3
cos t+ sin t − cos t+ sin t
2 2 2 2 2 2 2 2

r
ke
Instead we guess a particular solution of the form
   

oc
x1 a1 cos t + a2 sin t
= .
x2 b1 cos t + b2 sin t

l
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Inhomogeneous systems of linear differential equations

Now
   
d x1 a2 cos t − a1 sin t
=
dt x2 b2 cos t − b1 sin t

and
    
0 −1 x1 −b1 cos t − b2 sin t
= ,
1 −1 x2 (a1 − b1 ) cos t + (a2 − b2 ) sin t

hence by insertion,
       
d x1 0 −1 x1 (a2 +b1 ) cos t+(−a1 +b2 ) sin t cos t
= = .
dt x2 1 −1 x2 (b2 −a1 +b1 ) cos t+(−b1 −a2 +b2 ) sin t sin t

We get by an identification of the coefficients,

r
ke
a2 + b1 = 1, −a1 + b2 = 0,

oc
b2 − a1 + b1 = 0, −b1 − a2 + b2 = 1,

l
it. n
e U
hence b1 = 0, a2 = 1 and b2 = a1 = 2.

ov F
We get the particular solution m PD
 0  
x1 2 cos t + sin t
= .
x02 2 sin t
re h
to atc

Finally, the complete solution is obtained by adding all solutions of the homogeneous equation found
previously. Since this will give us a mess of formulæ, we shall not produce it here).
se e B
en W

Example 2.2 Find the complete solution of the system

      
lic y

d x1 −3 4 x1 2t
a b

= + .
dt x2 −2 1 x2 t
y ed
Buess

1) The eigenvalues are the roots of the polynomial

 
 −3 − λ 4 
oc

 = (λ + 3)(λ − 1) + 8 = λ2 + 2λ + 5 = (λ + 1)2 + 4,
 −2 1−λ 
Pr

hence the complex eigenvalues are λ = −1 ± 2i.

2) The corresponding complex eigenvectors are cross vector to anyone of the rows in the matrix
     
−3−λ 4 −3 + 1 ∓ 2i 4 −2 ∓ 2i 4
= = .
−2 1−λ −2 1 + 1 ∓ 2i −2 2 ∓ 2i

It follows from the first row, (−2 ∓ 2i, 4) = 2(−{1 ± i}, 2) that

v1 = (2, 1 + i)T for λ1 = −1 + 2i,

v2 = (2, 1 − i)T for λ2 = −1 − 2i.

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48
Calculus 4c-3 Inhomogeneous systems of linear differential equations

3) If λ1 = a + iω = −1 + 2i, where a = −1 and ω = 2, then

         
2 2 0 2 0
v1 = α + iβ = = +i , dvs. α = og β = .
1+i 1 1 1 1

Then we get the fundamental matrix,

  
Φ(t) = Re e(a+iω)t (α+iβ) Im e(a+iω)t (α + iβ) = eat cos ωt(α β) + eat sin ωt(−β α)
⎛ ⎞ ⎛ ⎞
2 0 0 2
= e−t cos 2t ⎝ ⎠ + e−t sin 2t ⎝ ⎠
1 1 −1 1
⎛ −t −t
⎞
2e cos 2t 2e sin 2t
= ⎝ ⎠.
−t −t
e (cos 2t − sin 2t) e (cos 2t + sin 2t)

r
ke
4) This fundamental matrix looks very complicated, so it does not invite one to apply the solution

oc
formula.

l
it. n
e U
Instead we guess a particular solution of the form
       

ov F
x1 at + b d
m PD x1 a
= med = .
x2 ct + d dt x2 c
re h

We get by a rearrangement of the differential equation system that

to atc

           
2t d x1 −3 4 x1 a −3 4 at + b
− −
se e B

= =
t dt x2 −2 1 x2 c −2 1 ct + d
     
a (−3a + 4c)t + (−3b + 4d) (3a − 4c)t + (a + 3b − 4d)
en W

= − = .
c (−2a + c)t + (−2b + d) (2a − c)t + (2b + c − d)
lic y
a b

Then by an identification of the coefficients,

y ed

3a − 4c = 2, a + 3b − 4d = 0, 2a − c = 1, 2b + c − d = 0.
Buess

We get from the first and third equation

 
oc

3a − 4c = 2, a = 2/5
that
Pr

2a − c = 1, c = −1/5.

Then by a rearrangement and insertion into the second and the fourth equation,
 
3b − 4d = −a = −2/5, b = 6/25,
, hence
2b − d = −c = 1/5, d = 7/25.

If this is put into our guess, we obtain our particular solution

     2 6
  
x1 at + b 5 t + 25
1 10t + 6
= = = .
x2 ct + d − 15 t + 25
7
25 −5t + 7

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49
Calculus 4c-3 Inhomogeneous systems of linear differential equations

5) It follows from the linearity that the complete solution is given by a particular solution to which
we add all the solutions of the corresponding homogeneous system,
     
x1 1 10t + 6 c1
= + Φ(t)
x2 25 −5t + 7 c2
     
1 10t + 6 −t 2 cos t −t 2 sin 2t
= +c1 e +c2 e ,
25 −5t + 7 cos 2t−sin 2t cos 2t+ sin 2t

where c1 and c2 are arbitrary constants.

6) Alternatives
a) Real calculations of the solutions of the homogeneous equation. The eigenvalues
λ = −1 ± 2i are complex conjugated, so the structure of the solution of the homogeneous
equation is given by

r
   

ke
x1 a1 cos 2t + a2 sin 2t
= e−t .

oc
x2 b1 cos 2t + b2 sin 2t

l
it. n
We get by a calculation,

e U
   
d x1 −t (−a1 +2a2 ) cos 2t+(−2a1 −a2 ) sin 2t

ov F
=e ,
dt x2 (−b1 +2b2 ) cos 2t+(−2b1 −b2 ) sin 2t
m PD
and
re h

    
−3 4
to atc

x1 −t (−3a1 +4b1 ) cos 2t+(−3a2 +4b2 ) sin 2t

=e .
−2 1 x2 (−2a1 +b1 ) cos 2t+(−2a2 +b2 ) sin 2t
se e B

When the coefficients are identified, we get

en W

−a1 + 2a2 = −3a1 + 4b1 , dvs. b1 = 12 a1 + 12 a2 ,

−2a1 − a2 = −3a2 + 4b2 , dvs. b2 = − 12 a1 + 12 a2 ,
lic y
a b

−b1 + 2b2 = −2a1 + b1 , dvs. a 1 = b1 − b2 ,

−2b1 − b2 = −2a2 + b2 ,
y ed

dvs. a 2 = b1 + b2 .
Buess

We see that the four equations are consistent, and that the homogeneous equation has the
complete solution
   
oc

x1 (b1 −b2 ) cos 2t+(b1 +b2 ) sin 2t

= e−t
x2 b1 cos 2t + b2 sin 2t
Pr

   
cos 2t+sin 2t −cos 2t+sin 2t
= b1 e−t +b2 e−t ,
cos 2t sin 2t
corresponding to the fundamental matrix
 
−t cos 2t+sin 2t −cos 2t+sin 2t
Φ1 (t) = e .
cos 2t sin 2t

Notice that Φ1 (t) = Φ(t) found previously. However, the two different fundamental matrices
are of course equivalent.

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50
Calculus 4c-3 Inhomogeneous systems of linear differential equations

b) Direct calculation of the exponential matrix. Since λ = −1 ± 2i, the trick is to put
   
−2 4 −1 2
B = A − Re λ · I = =2 ,
−2 2 −1 1

and as Im λ = ±2i, to aim at the cosine and the sine series. We first calculate
    
−1 2 −1 2 −1 0
B 2 = 22 = 22 = −22 I,
−1 1 −1 1 0 −1

which very conveniently gives

 n
B2n = B2 = (−1)n · 22n · I for n ∈ N (og for n = 0).

Since I commutes with everything, we get

exp(At) = exp((A + I)t − It) = e−t exp(Bt)

r
ke
∞
1 n n
= e−t

oc
B t .
n=0
n!

l
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc

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Pr

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Calculus 4c-3 Inhomogeneous systems of linear differential equations

We now divide the investigation into the cases of even and odd indices, and use that

B2n = (−1)n · 22n I,

so
∞
 ∞
1 1
exp(At) = e−t B2n t2n + e−t B2n t2n+1 B
n=0
(2n)! n=0
(2n+1)!
∞ ∞
(−1)n 2n 2n (−1)n 2n 2n+1
= e−t 2 t I + e−t 2 t B
n=0
(2n)! n=0
(2n+1)!
∞ ∞
(−1)n (−1)n 1
= e−t (2t)2n I + e−t (2t)2n+1 · B
n=0
(2n)! n=0
(2n+1)! 2
 
−1 2
= e−t cos 2tI + e−t sin 2t
−1 1

r
ke
 
−t cos 2t − sin 2t 2 sin 2t
= e .

oc
− sin 2t cos 2t + sin 2t

l
it. n
We note again that the fundamental matrix is different from both Φ(t) and Φ1 (t) found pre-

e U
viously.

ov F
c) The fumbling method. We first expand the system,
m PD

dx1 /dt = −3x1 + 4x2 + 2t,
dx2 /dt = −2x1 + x2 + t.
re h
to atc

If we use the first equation to eliminate x2 , it follows that

se e B

dx1 d(4x2 ) d2 x1 dx1

(6) 4x2 = + 3x1 − 2t, med = +3 − 2.
dt dt dt2 dt
en W

Then the latter equation of the system is multiplied by 4,

lic y
a b

d(4x2 )
= −8x1 + 4x2 + 4t,
y ed

dt
Buess

and we get by an insertion,

 
d2 x1 x1 dx1
oc

+ 3 − 2 = −8x1 + + 3x1 − 2t + 4t.

dt2 dt dt
Pr

Then by a rearrangement,

d2 x1 dx1
+2 + 5x1 = 2t + 2.
dt2 dt
The characteristic equation R2 + 2R + 5 = 0 has the roots R = −1 ± 2i (the same as the
eigenvalues in the other variants).

We guess a particular solution of the form

dx1 d2 x1
x1 = at + b, thus = a and = 0.
dt dt2

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52
Calculus 4c-3 Inhomogeneous systems of linear differential equations

Then by insertion,

2t + 2 = 0 + 2a + 5at + 5b = 5at + (2a + 5b).

When we identify the coefficients, we get


5a = 2, dvs. a = 2/5,
2a + 5b = 2, dvs. b = 15 (2 − 2a) = 6/25.

Hence,
2 6
x1 = t+ + c1 e−t cos 2t + c2 e−t sin 2t.
5 25
Now,
dx1 2

r
= + (−c1 + 2c2 )e−t cos 2t + (−2c1 − c2 )e−t sin 2t,

ke
dt 5

oc
so if we put this into (6), then

l
it. n
dx1
4x2 = + 3x1 − 2t

e U
dt
2

ov F
= + (−c1 +2c2 )e−t cos 2t+(−2c1 −c2 )e−t sin 2t
m PD
5
6 18
+ t+ + 3c1 e−t cos 2t + 3c2 e−t sin 2t − 2t
5 25
re h

4 28
to atc

= − t+ + (2c1 +2c2 )e−t cos 2t + (−2c1 +2c2 )e−t sin 2t,

5 25
se e B

whence
1 7 1 1 1 1
+ ( c1 + c2 )e−t cos 2t + (− c1 + c2 )e−t sin 2t.
en W

x2 = − t +
5 25 2 2 2 2
lic y

Summing up we get in matrix form

a b

   2 6
    
y ed

x1 5 t + 25 −t cos 2t −t sin 2t
= + c1 e 1 + c2 e
x2 − 15 t + 25
7
cos 2t − 12 sin 2t 1 1
2 cos 2t + 2 sin 2t
 2
Buess

   
1 10t + 6 −t cos 2t sin 2t c1
= +e 1 1 1 1 ,
25 −5t + 7 2 cos 2t − 2 sin 2t 2 cos 2t + 2 sin 2t c2
oc

where c1 and c2 are arbitrary constants.

Pr

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53
Calculus 4c-3 Inhomogeneous systems of linear differential equations

Example 2.3 Given the linear differential equation system

      
d x1 1 −1 x1 2 cos 2t
= + , t ∈ R.
dt x2 −1 −1 x2 sin 2t

Find x2 (t) if we assume that

   1 
x1 (0) −3
= .
x2 (0) 0

First solution. The eigenvalue method. The eigenvalues are the roots of the polynomial
 
 1−λ −1  √
 = (λ − 1)(λ + 1) − 1 = λ2 − 2. dvs. λ = ± 2.
 −1 −1 − λ 
√ √
We may e.g. choose an eigenvector corresponding to λ = 2 as (1, 1 − 2).

r
ke
√ √
An eigenvector corresponding to λ = − 2 is e.g. (1, 1 + 2).

oc
The complete solution of the homogeneous system of equation is

l
it. n
     

e U
√ 1√ √ 1√
x1 2t − 2t
= c1 e + c2 e .
1− 2

ov F
x2 1+ 2
m PD
Then we guess a particular solution of the form
   
re h

x1 a cos 2t + b sin 2t
to atc

= .
x2 c cos 2t + d sin 2t
se e B

Now,
   
d 2b cos 2t − 2a sin 2t
en W

x1
=
dt x2 2d cos 2t − 2c sin 2t
lic y
a b

and
    
y ed

1 −1 a cos 2t + b sin 2t (a−c) cos 2t+(b−d) sin 2t

= ,
−1 −1 −(a+c) cos 2t−(b+d) sin 2t
Buess

c cos 2t + d sin 2t

so we can also write the system of equations in the following way,

oc


2b cos 2t − 2a sin 2t = (a − c + 2) cos 2t + (b − d) sin 2t,
Pr

2d cos 2t − 2c sin 2t = (−a − c) cos 2t + (−b − d + 1) sin 2t.

When the coefficients are identified we get

2b = a − c + 2, thus − a + 2b + c = 2,
−2a = b − d, thus 2a + b − d = 0,
2d = −a − c, thus a + c + 2d = 0,
−2c = −b − d + 1, thus b − 2c + d = 1.

It follows from the first and the third equation that

b + c + d = 1,

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54
Calculus 4c-3 Inhomogeneous systems of linear differential equations

which together with the fourth equation implies c = 0. This reduces the equations to
⎧ ⎧
⎨ −a + 2b = 2, ⎨ −a + 2b = 2,
2a + b − d = 0, hence 2a + 2b = 1,
⎩ ⎩
b + d = 1, b + d = 1,

thus
1 5 1
a=− , b= , c = 0, d= .
3 6 6
The complete solution is
   1     
x1 − 3 cos 2t + 56 sin 2t √
2t 1√ √
− 2t 1√
= 1 + c 1 e + c 2 e .
x2 6 sin 2t 1− 2 1+ 2

r
ke
l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Inhomogeneous systems of linear differential equations

It follows from the initial conditions that

   1   1     
x1 (0) −3 −3 1√ 1√
= = c1 + c2 ,
x2 (0) 0 0 1− 2 1+ 2
so c1 = c2 = 0, i.e.
   1 
x1 − 3 cos 2t + 56 sin 2t
= 1 ,
x2 6 sin 2t

and then finally,

1
x2 (t) = sin 2t.
6

Second solution. The “fumbling method”. We shall actually only find x 2 (t), so it would be

r
ke
reasonable to eliminate x1 (t). First we get from

oc
dx1 dx2
= x1 − x2 + 2 cos 2t, = −x1 − x2 + sin 2t,

l
dt dt

it. n
e U
that

ov F
dx1
x1 = − − x2 + sin 2t, m PD
dt
which when put into the first equation gives
re h
to atc

d2 x2 dx2 dx2
− 2
− + 2 cos 2t = − x2 + 2 cos 2t − x2 + sin 2t,
dt dt dt
se e B

hence by a rearrangement,
en W

d2 x2 dx2 1
(7) − 2x2 = − sin 2t, x2 (0) = 0 og (0) = .
dt2 dt 3
lic y
a b

If we guess a particular solution of the structure x2 = a cos 2t + b sin 2t, we get

y ed

−6a cos 2t − 6b sin 2t = − sin 2t,

Buess

hence a = 0 and b = 16 , and we find the particular solution

oc

1
x2 (t) = sin 2t.
Pr

6
It is seen by inspection that it fulfils the initial conditions, and since the solution is unique, we have
solved the problem.

Alternatively the complete solution of (7) is given by

1 √ √
x2 (t) = sin 2t + c1 e 2t + c2 e− 2t .
6
It follows from the initial conditions that c1 = 0 and c2 = 0, hence the solution is
1
x2 (t) = sin 2t.
6

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56
Calculus 4c-3 Inhomogeneous systems of linear differential equations

Remark 2.3 In both cases the “fumbling method” is much easy to apply than the eigenvalue method.

Example 2.4 Find the complete solution of the system

      
d x1 0 1 x1 1
= + e−t sin 2t.
dt x2 −2 −2 x2 −1

The eigenvalues are the roots of the polynomial

 
 −λ 1 
  2 2
 −2 −2 − λ  = (λ + 2)λ + 2 = λ + 2λ + 2 = (λ + 1) + 1,

thus λ = a ± iω = −1 ± i where a = −1 and ω = 1.

r
We first guess on a particular solution of the structure

ke
   

oc
x1 −t a cos 2t + b sin 2t
=e .
x2 c cos 2t + d sin 2t

l
it. n
e U
Since
   

ov F
d x1 −t (−a + 2b) cos 2t − (2a + b) sin 2t
m PD
=e
dt x2 (−c + 2d) cos 2t − (2c + d) sin 2t
re h
and
    
to atc

0 1 x1 c cos 2t + d sin 2t
= e−t ,
−2 −2 x2 −2(a + c) cos 2t − 2(b + d) sin 2t
se e B

we get from the system of differential equations and a multiplication with e t that
en W


(−a+2b) cos 2t−(2a+b) sin 2t = c cos 2t+(d+1) sin 2t,
lic y

(−c+2d) cos 2t−(2c+d) sin 2t = −2(a+c) cos 2t−(2b+2b+1) sin 2t.

a b
y ed

When the coefficients are identified it follows that

Buess

−a + b = c, thus a − b + c = 0,
−(2a + b) = d + 1, thus − 2a − b − d = 1,
oc

−c + 2d = −2(a + c), thus 2a + c + 2d = 0,

−(2c + d) = −(2b + 2d + 1), thus 2b − 2c + d = −1.
Pr

We get from the second and the fourth equation that

−2a + b − 2c = 0,

which together with the first equation gives b = 0.

The system of equations is then reduced to

a + c = 0, thus c = −a,
2a + d = −1, thus 2a + d = −1,
2a + c + 2d = 0, thus a + 2d = 0,

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57
Calculus 4c-3 Inhomogeneous systems of linear differential equations

hence
1 2 2
d= , a=− , c= , b = 0,
3 3 3
and a particular solution is
   
x1 1 −2 cos 2t
= e−t .
x2 3 2 cos 2t + sin 2t

We still miss the complete solution of the corresponding homogeneous system of differential equations,
    
d x1 0 1 x1
= .
dt x2 −2 −2 x2
This can of course be found in many ways.
1) The eigenvalue method. We have already found the eigenvalues λ = a ± iω = −1 ± i where

r
ke
a = −1 and ω = 1. An eigenvector corresponding to λ = −1 + i is a cross vector to (1 − i, 1), e.g.
     

oc
1 1 0
v= = α + iβ = +i .
−1 + i −1

l
1

it. n
e U
A fundamental matrix is then given by
  

ov F
Re e(a+iω)t (α + iβ) Im e(a+iω)t (α + iβ)
Φ(t) = m PD
= eat cos ωt(α β) + eat sin ωt(−β α)
   
re h

1 0 0 1
to atc

= e−t cos t + e−t sin t

−1 1 −1 −1
 
se e B

cos t sin t
= e−t .
− cos t − sin t cos t − sin t
en W

The complete solution is

       
lic y

x1 1 −t −2 cos 2t cos t sin t

a b

−t −t
= e + c1 e + c2 e ,
x2 3 2 cos 2t + sin 2t − cos t − sin t cos t − sin t
y ed

where c1 and c2 are arbitrary constants.

Buess

2) The exponential matrix. This is given by the formula (where we from the above have a = −1
and ω = 1)
oc


Pr

a 1
exp(At) = eat cos ωt − sin ωt I + eat sin ωt · A
ω  ω  
−t 1 0 −t 0 1
= e {cos t + sin t} + e sin t
0 1 −2 −2
 
cos t + sin t sin t
= e−t ,
−2 sin t cos t − sin t

hence
       
x1 1 −2 cos 2t cos t + sin t sin t
= e−t −t
+ c1 e −t
+ c2 e ,
x2 3 2 cos 2t + sin 2t −2 sin t cos t − sin t
where c1 and c2 are arbitrary constants.

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58
 Calculus 4c-3 Inhomogeneous systems of linear differential equations

3) Real structure of the solution. Since λ = −1 ± i, the solution must necessarily have the
following structure
   
x1 a1 cos t + a2 sin t
= e−t .
x2 b1 cos t + b2 sin t

Then by a calculation,
   
d x1 (−a1 + a2 ) cos t − (a1 + a2 ) sin t
= e−t
dt x2 (−b1 + b2 ) cos t − (b1 + b2 ) sin t

and
    
0 1 x1 b1 cos t + b2 sin t
= e−t .
−2 −2 x2 −2(a1 + b1 ) cos t − 2(a2 + b2 ) sin t

r
When we identify the coefficients we get

ke
oc
b1 = −a1 + a2 , b2 = −a1 − a2 ,

360°
l
it. n
and (a little superfluous)

e U .
−b1 + b2 = −2(a1 + b1 ), 2(a2 + b2 ) = b1 + b2 .

ov F
m PD
thinking
We have thus eliminated b1 and b2 , hence the complete solution is
       
re h

x1 1 −t −2 cos 2t −t cos t −t sin t

= e + a1 e + a2 e .
to atc

x2 3 2 cos 2t + sin 2t − cos t − sin t cos t − sin t

se e B
en W
lic y
a b
y ed
Buess

360°
.
oc

thinking
Pr

360°
thinking .
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Discover the truth at www.deloitte.ca/careers © Deloitte & Touche LLP and affiliated entities.

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Calculus 4c-3 Inhomogeneous systems of linear differential equations

4) The “fumbling method”. The homogeneous system is expanded,

dx1 dx2
= x2 and = −2x1 − 2x2 .
dt dt
If we put the first equation into the last one, it follows by a rearrangement,
d2 x1 dx1
+2 + 2x1 = 0 med R2 + 2R + 2 = 0 for R = −1 ± i.
dt2 dt
Hence

x1 = c1 e−t cos t + c2 e−t sin t

and
dx1
x2 = = c1 e−t (− cos t − sin t) + c2 e−t (cos t − sin t),

r
dt

ke
thus

oc
     
x1 cos t sin t
= c1 e−t + c2 e−t .

l
− cos t − sin t cos t − sin t

it. n
x2

e U
The complete solution of the inhomogeneous system is

ov F
   m PD     
x1 1 −2 cos 2t cos t sin t
= e−t + c1 e−t + c2 e−t ,
x2 3 2 cos 2t + sin 2t − cos t − sin t cos t − sin t
re h

where c1 and c2 are arbitrary constants.

to atc
se e B

Example 2.5 Consider the system

   
en W

dx 1 1 1
= x(t) + , t ≥ 0.
dt 0 −2 −4
lic y
a b

1) Find the complete solution of the system.

y ed

2) Let x0 (t) be the solution, for which x0 (0) = 0. Find x0 (1).

Buess
oc

1) Clearly, the eigenvalues are 1 and 2,

 
Pr

 1−λ 1 
  = (λ − 1)(λ + 2) = 0.
 0 −2 − λ 

The corresponding eigenvectors are cross vectors to the first row:

If λ = 1, then v1 = (1, λ − 1) = (1, 0).

If λ = −2, then v2 = (1, λ − 1) = (1, −3).

The complete solution of the homogeneous equation is

       t  
x1 t 1 −2t 1 e e−2t c1
= c1 e + c2 e = .
x2 0 −3 0 −3e−2t c2

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60
Calculus 4c-3 Inhomogeneous systems of linear differential equations

The inhomogeneous term is a constant vector. Therefore, we guess on a particular solution as a

constant vector
 
a
x̃(t) = ,
b
which gives by insertion,
        
dx̃ 0 1 1 a 1 a+b+1
= = + = ,
dt 0 0 −2 b −4 −2b − 4
hence b = −2 and a = 1, and we have x̃ = (1, 2)T .
If c1 and c2 denote the arbitrary constants, the complete solution is given by
   t  
1 e e−2t c1
x(t) = + .
−2 0 −3e−2t c2

r
ke
Alternatively we may apply the “fumbling method”. We expand the system,
⎧
⎨ dx1 /dt = x1 + x2 + 1, thus dx1 /dt − x1 = x2 + 1,

l oc
⎩

it. n
dx2 /dt = −2x1 − 4, thus dx2 /dt + 2x2 = −4.

e U
Clearly, the solution of the latter equation is
x2 = −2 + c2 e−2t .
ov F
m PD
When this is put into the first equation, we get
re h

dx1
− x1 = −1 + c2 e−2t ,
to atc

dt
hence
se e B

1
x1 = 1 + c2 et e−t e−2t dt + c1 et = 1 + c1 et − c2 e−2t ,
3
en W

and summing up,

   t  
lic y

− 31 e−2t
a b

1 e c1
x(t) = + .
−2 0 e−2t c2
y ed
Buess

2) When we put t = 0, we get

     
1 c 1 + c2 0
+ = ,
oc

−2 −3c2 0
thus c2 = − 32 and c1 = − 13 , so
Pr

     −2t 
1 1 et 2 e
x0 (t) = − − .
−2 3 0 3 −3e−3t
Then finally,
⎛ ⎞
      2 1 3 2
e − e −
1 1 e 2 e−2 1 ⎜ 3 3 ⎟.
x0 (1) = − − = 2⎝ ⎠
−2 3 0 3 −3e−2 e
2
−2e + 2

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61
Calculus 4c-3 Examples of applications in Physics

3 Examples of applications in Physics

Example 3.1 Consider a physical system consisting of two coupled oscillators. We assume that there
is no damper in the system. The three spring constants are k1 , k and k2 , and m1 and m2 denote the
masses of each of the two particles. At equilibrium we assume that the spring forces are 0. It can be
proved by Newton’s second law that the system can be described by the following system of differential
equations,
d2 x1 k1 + k k d2 x2 k2 + k k
+ x1 = x2 and + x2 = x1 .
dt2 m1 m1 dt2 m2 m2
3 8 4
Put m1 = m2 = 1, k = , k1 = , k2 = , and assume that
10 5 5
x1 (0) = 3 · 10−2 , x1 (0) = 0,

r
x2 (0) = 3 · 10−2 , x2 (0) = 0.

ke
oc
Find x1 (t) and x2 (t) as solutions of a differential equation of fourth order.

l
it. n
By using the selected values of m1 , m2 k, k1 and k2 , we get

e U
d2 x1
ov F
k1 + k k 19 3
=− x1 + x2 = − x1 + x2 ,
m PD
dt2 m1 m1 10 10
d2 x2 k k + k2 3 11
re h

= x1 − x2 = x1 − x2 .
dt2
to atc

m2 m2 10 10
We immediately get three different methods of solution.
se e B

1) The traditional eigenvalue method. If we put

en W

dx1 dx2
y1 = x 1 , y2 = , y3 = x 2 og y4 = ,
dt dt
lic y
a b

then we get the homogeneous system

y ed

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞
y1 y2 0 1 0 0 y1
Buess

d ⎜ ⎟ ⎜ − 19 3 ⎟
⎜ 2 ⎟ = ⎜ 10 1 10 3 ⎟ = ⎜ 10 0
y y + y ⎜ − 19 3
10 0 ⎟⎜ y 2
⎟⎜
⎟
⎟.
dt ⎝ y3 ⎠ ⎝ y4 ⎠ ⎝ 0 0 0 1 ⎠⎝ y 3 ⎠
oc

3 11 3
y4 y
10 1 − y
10 3 10 0 − 11
10 0 y4
Pr

The eigenvalues are the roots of the polynomial

 
 −λ 1 0 0     
 19  −λ 3
0   − 19 3
0 
 − −λ 3
0   10 10 10 
 10 10  = −λ  0 −λ 1 − 0 −λ 1 
 0 0 −λ 1     
  0 − 11 −λ   3 11
− 10 −λ 
 3
0 − 11
−λ  10 10
10 10
   19   
 −λ 1   − 0   − 19 3 
= λ2  11 + λ  10 + 10 10 
− 10 −λ   3
10 −λ   10 3 11
− 10 
 
11 19 1
= λ 2 λ2 + + λ2 + (19 · 11 − 9)
10 10 100
= λ4 + 3λ2 + 2 = (λ2 + 1)(λ2 + 2),

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62
Calculus 4c-3 Examples of applications in Physics

√
thus the eigenvalues are λ = ±i and λ = ± 2i.

It is here fairly difficult to find the complex eigenvectors, so we note instead that the structure of
the solution must be of the form
√ √
y1 = x1 (t) = a1 cos t + a2 sin t + a3 cos 2t + a4 sin 2t,

dx1 √ √ √ √
y2 = = −a1 sin t + a2 cos t − 2a3 sin 2t + 2a4 cos 2t,
dt
√ √
y3 = x2 (t) = b1 cos t + b2 sin t + b3 cos 2t + b4 sin 2t,
dx2 √ √ √ √
y4 = = −b1 sin t + b2 cos t − 2b3 sin 2t + 2b4 cos 2t.
dt

r
ke
l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess

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oc
Pr

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Calculus 4c-3 Examples of applications in Physics

Since
d2 x1 √ √
2
= −a1 cos t − a2 sin t − 2a3 cos 2t − 2a4 sin 2t,
dt
d2 x2 √ √
2
= −b1 cos t − b2 sin t − 2b3 cos 2t − 2b4 sin 2t,
dt
and
   
19 3 19 3 19 3
− x1 + x2 = − a1 + b1 cos t+ − a2 + b2 sin t
10 10 10 10 10 10
   
19 3 √ 19 3 √
+ − a3 + b3 cos 2t+ − a4 + sin 2t,
10 10 10 10

and

r
   

ke
3 11 3 11 3 11
x 1 − x2 = a1 − b1 cos t+ a2 − b2 sin t

oc
10 10 10 10 10 10
   
3 11 √ 3 11 √

l
it. n
+ a3 − b3 cos 2t+ a4 − b4 sin 2t,
10 10 10 10

e U
ov F
we get by an identification of the coefficients that
m PD
19 3
−a1 = − a1 + b1 , thus b1 = 3a1 ,
10 10
re h
to atc

19 3
−a2 = − a2 + b2 , thus b2 = 3a2 ,
10 10
se e B

19 3 1
−2a3 = − a3 + b3 , thus b3 = − a3 ,
10 10 3
en W

19 3 1
−2a4 = − a4 + b4 , thus b4 = − a4 .
lic y

10 10 3
a b

This gives us the general solution

y ed

√ √
x1 (t) = a1 cos t+a2 sin t+a3 cos 2t+a4 sin 2t,
Buess

1 √ 1 √
x2 (t) = 3a1 cos t+3a2 sin t− a3 cos cos 2t− a4 sin 2t.
oc

3 3
Pr

Since
√ √ √ √
x1 (t) = −a1 sin t+a2 cos t−a3 2 sin 2t+a4 2 cos 2t,
√ √
 2 √ 2 √
x2 (t) = −3a1 sin t+3a2 cos t+a3 sin 2t−a4 cos 2t,
3 3
it follows from the initial conditions that
3 3 1
x1 (0) = = a1 + a3 , x2 (0) = = 3a1 − a3 ,
100 100 3
√
√ 2
x1 (0) = 0 = a2 + a4 2, x2 (0) = 0 = 3a2 − a4 .
3

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64
Calculus 4c-3 Examples of applications in Physics

We conclude from the latter two equations that a2 = a4 = 0. Then

9
a3 = 9a1 −
100
implies that
 
3 9 12 3
a1 = − 9a1 − , dvs. a1 = =
100 100 1000 250
and
3 3 3 15 − 6 9
a3 = − a1 = − = = .
100 100 250 500 500
The wanted solution is then given by
⎧ 3 9 √

r
⎪
⎪
⎨ x1 = 250 cos t + 500 cos( 2t),

ke
oc
⎪
⎩ x = 9 cos t − 3 cos(√2t).
⎪
2

l
250 500

it. n
e U
2) Alternatively the system can be written

ov F
   19  
m PD
d2 x1 − 10 3
10 x1
= 3 11 .
dt2 x2 10 − 10 x2
re h

The eigenvalues are the roots of the polynomial

to atc

 19    
 − −λ 3  19 11 9
 103 
se e B

 11
10
 = λ + 10 λ+ − = λ2 + 3λ + 2 = (λ + 1)(λ + 2),
10 − 10 − λ 10 100
en W

thus either λ = −1 or λ = −2.

lic y

An eigenvector of the eigenvalue λ = −1 is e.g. (1, 3), corresponding to the differential equation
a b
y ed

d2
(x1 + 3x2 ) = −(x1 + 3x2 ),
dt2
Buess

the complete solution of which is

oc

(8) x1 + 3x2 = a1 cos t + a2 sin t.

Pr

 
1
An eigenvector of the eigenvalue λ = −2 is e.g. 1, − , corresponding to the differential equation
3
   
d2 1 1
x 1 − x2 = −2 x 1 − x2 ,
dt2 3 3
the complete solution of which is
1 √ √
(9) x1 − x2 = b1 cos( 2t) + b2 sin( 2t).
3

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65
Calculus 4c-3 Examples of applications in Physics

By solving (8) and (9) we get the complete solution

1 1 9 √ 9 √
x1 = a1 cos t+ a2 sin t+ b1 cos 2t+ b2 sin 2t,
10 10 10 10
3 3 3 √ 3 √
x2 = a1 cos t+ a2 sin t− b1 cos 2t− b2 sin 2t.
10 10 10 10
Then it follows from the initial conditions that
⎧ 1 9 3 ⎧ 3
⎪
⎪ x (0) = a + b = , ⎪
⎪ a1 + 9b1 = ,
⎨ 1
10
1
10
1
100 ⎨ 10
dvs.
⎪
⎪ ⎪
⎪
⎩ x (0) = 3 a − 3 b = 3 , ⎩ a −b = 1 ,
2 1 1 1 1
10 10 100 10
1 3
hence b1 = and a1 = . It follows from
50 25

r
√ √

ke
 1 9 2  3 3 2
x1 (0) = a2 + b2 = 0 and x2 (0) = a2 − b2 = 0,

oc
10 10 10 10

l
that a2 = b2 = 0.

it. n
e U
The wanted solution is
⎧

ov F
3 9 √
⎪
⎪
m PD
⎨ x1 = 250 cos t + 500 cos( 2t),
⎪
⎩ x = 9 cos t − 3 cos(√2t).
⎪
re h

2
to atc

250 500
se e B

3) The “fumbling method”. If we eliminate x2 by

m1 d2 x1 k1 + k
en W

x2 = + x1 ,
k dt2 k
lic y

then
a b

m1 d4 x1 k1 +k d2 x1 k2 +k m1 d2 x1 (k1 +k)(k1 +k2 )

y ed

k
+ + + x1 = x1 ,
k dt4 k dt2 k m2 dt2 km2 m2
Buess

hence by a rearrangement,
oc

m1 d4 x1 k1 +k2 +2k d2 x1 (k1 +k2 )k+k1 k2

+ + x1 = 0.
k dt4 dt2
Pr

k km2
When we multiply by k and insert the chosen values of k, mi and kj , we get
   
d4 x1 3 8 4 d2 x1 3 12 32 d4 x1 d2 x1
0= + + + + · + x 1 = + 3 + 2x1 .
dt4 5 5 5 dt2 10 5 25 dt4 dt2
√
The characteristic polynomial R4 +3R2 +2 = (R2 +1)(R2 +2) has the roots R = ±i and R = ± 2i,
thus
√ √
x1 = c1 cos t + c2 sin t + c3 cos( 2t) + c4 sin( 2t),

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66
Calculus 4c-3 Examples of applications in Physics

and whence
 
10 d2 x1 10 3 8 10 d2 x1 19
x2 = + + x1 = + x1
3 dt2 310 5 3 dt2 3
1 √ 1 √
= 3c1 cos t + 3c2 sin t − c3 cos( 2t) − c4 sin( 2t).
3 3
It follows from the initial conditions that
3 1 3
x1 (0) = c1 + c3 = , x2 (0) = 3c1 − c3 = ,
100 3 100
√
√ 2
x1 (0) = c2 + 2c4 = 0, x1 (0) = 3c2 − c4 = 0.
2

r
ke
l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Examples of applications in Physics

We immediately get
3 9
c2 = c4 = 0, and c1 = og c3 = .
250 500
The wanted solution is
⎧ 3 9 √
⎪
⎪
⎨ x1 = 250 cos t + 500 cos( 2t),
⎪
⎩ x = 9 cos t − 3 cos(√2t).
⎪
2
250 500

Example 3.2 For small oscillations (small swings Θ and ϕ) it is possible to show that the model of
the double pendulum can be described by the equations

r
ke
d2 Θ d2 ϕ

oc
2 2
+ + 2gΘ = 0,
dt dt2

l
it. n
d2 Θ d2 ϕ

e U
2
+ + gϕ = 0.
dt dt2

ov F
Find the eigenfrequencies and the complete solution.
m PD
If we solve with respect to (Θ, ϕ), we get the system
re h

    2    
to atc

Θ − /g − /(2g) d Θ d2 Θ
= =A 2 .
ϕ − /g − /g dt 2 ϕ dt ϕ
se e B

The eigenvalues satisfy the equations

en W

   2  2
 −( /g) − λ − /(2g)  1
 = λ+ − = 0,
lic y

 − /g −( /g) − λ  g 2 g
a b

 √ 
y ed

2 √
thus λ = −1 ± and a corresponding eigenvector is e.g. (1, ∓ 2).
Buess

2 g

1
oc

Since A−1 has the same eigenvectors as A, and the eigenvalues , we derive the two differential
λ
Pr

equations of second order

d2 √ √ √
2
(Θ − 2ϕ) = −(2 + 2) (Θ − 2ϕ),
dt g

d2 √ √ √
2
(Θ + 2ϕ) = −(2 − 2) (Θ + 2ϕ),
dt g
hence
! "  ! " 
√ √ √
Θ − 2ϕ = 2a1 cos 2+ 2 t + 2a2 sin 2+ 2 t ,
g g

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68
Calculus 4c-3 Examples of applications in Physics

! "  ! " 
√ √ √
Θ + 2ϕ = 2b1 cos 2− 2 t + 2b2 sin 2− 2 t .
g g
Finally, we get
⎛" ⎞ ⎛" ⎞
√ √
(2+ 2) ⎠ (2+ 2) ⎠
Θ = a1 cos ⎝ t + a2 sin ⎝ t
g g
⎛" ⎞ ⎛" ⎞
√ √
(2− 2) ⎠ (2− 2) ⎠
+b1 cos ⎝ t + b2 sin ⎝ t ,
g g

and
⎛" ⎞ ⎛" ⎞
√ √
a1 (2+ 2) a (2+ 2)

r
⎝ ⎠ 2 ⎝
= − √ cos t − √ sin t⎠

ke
ϕ
2 g 2 g
⎛" ⎞ ⎛" ⎞

oc
√ √
b1 ⎝ (2− 2) ⎠ b 2 ⎝ (2− 2)
t⎠ .

l
√ √

it. n
+ cos t + sin
2 g 2 g

e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed

With us you can

Buess
oc

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Calculus 4c-3 Examples of applications in Physics

Example 3.3 Two electric conductors are coupled inductively. If i 1 and i2 denote the current inten-
di2 di1
sities of the conductors, then the induced forces M and M , (where M is a constant) in each of
dt dt
the conductors, resp.. Then the differential equations of i1 and i2 are given by
d2 i1 di1 1 d2 i2
L1 + R 1 + i1 + M = 0,
dt2 dt C1 dt2
d2 i1 d2 i2 di2 1
M 2
+ L2 2 + R 2 + i2 = 0,
dt dt dt C2
where L, R and C are the induction coefficient, the resistance and the capacity, resp..
1) Find the complete solution
2) Check the cases
a) M = 0,

r
1 1

ke
b) R1 = R2 = 0, and n1 = √ = n2 = √ .
L1 C1 L2 C2

l oc
it. n
di1 di2

e U
1) If we put x1 = i1 , x2 = i2 , x3 = and x4 = , then
dt dt

ov F
dx3 1 dx4
m PD
L1 + R1 x3 + x1 + M = 0,
dt C1 dt
dx3 dx4 1
re h

M + L2 + R2 x4 + x2 = 0,
to atc

dt dt C2
thus by a rearrangement,
se e B

dx3 dx4 1
L1 +M = − x1 − R1 x3 ,
dt dt C1
en W

dx3 dx4 1
= − x2 − R2 x4 .
lic y

M + L2
a b

dt dt C2
If L1 L2 = M 2 , then
y ed

dx3 L2 x1 M x2 R1 L2 x3 M R2 x4
Buess

=− 2
+ 2
− 2
+ ,
dt (L1 L2 −M )C1 (L1 L2 −M )C2 L1 L2 −M L1 L2 −M 2
oc

dx4 M x1 L1 x2 R 1 M x3 L1 R2 x4
= − + − .
dt (L1 L2 −M 2 )C1 (L1 L2 −M 2 )C2 L1 L2 −M 2 L1 L2 −M 2
Pr

Hence in the form of a matrix,

dx
= Ax,
dt
so if we put a = L1 L2 − M 2 , then
⎛ ⎞
0 0 1 0
⎜ 0 0 0 1 ⎟
⎜ ⎟
⎜ L2 M R1 L2 M R2 ⎟
A=⎜ − − ⎟.
⎜ aC1 aC2 a a ⎟
⎝ M L2 R1 M L1 R2 ⎠
− −
aC1 aC2 a a

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70
Calculus 4c-3 Examples of applications in Physics

In principle it is possible to find the eigenvalues and the eigenfunctions of this system. In practice,
however, it is very difficult, so we stop here.

r
ke
l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Stability

4 Stability
Example 4.1 Check the stability of the following system
  
δx 1 7 x1
= + u(t).
dt 3 −2 x2

The eigenvalues are the roots of the polynomial

 
 1−λ 7 
  = (λ − 1)(λ + 2) − 21 = λ2 + λ − 23.
 3 −2 − λ 
This polynomial has a negative coefficient, hence the system is unstable.

Example 4.2 Check the stability of the system

⎛ ⎞⎛ ⎞

r
ke
−1 1 0 x1
dx ⎝
= −5 −1 1 ⎠ ⎝ x2 ⎠ + u(t).

oc
dt
−7 0 1 x3

l
it. n
e U
The eigenvalues are the roots of the polynomial
 
 −1 − λ 

ov F
 1 0 
 −5 −1 − λ 1  = (λ + 1)2 (1 − λ) − 7 + 5(1 − λ)
m PD
 
 −7 0 1−λ 
re h

= λ3 + λ2 + 4λ + 1.
to atc

Here a1 = 1 > 0, a2 = 4 > 0, a3 = 1 > 0, and

   
se e B

 a1 a3   1 1 
 = 
 1 a2   1 4  = 3 > 0,
en W

so all roots have a negative real part, and the system is asymptotically stable.
lic y
a b
y ed

Example 4.3 Check the stability of the system

⎛ ⎞⎛ ⎞
Buess

0 0 0 1 x1
dx ⎜ 0 −1 −1 −1 ⎟ ⎜ ⎟
⎟ ⎜ x2 ⎟ + u(t).
=⎜
⎝ −1 −1 −1 ⎠ ⎝ 3 ⎠
oc

dt 0 x
0 0 1 0 x4
Pr

The eigenvalues are the roots of the polynomial

 
 −λ 0 0 1   
  −λ 0 1 
 0 −1 − λ −1 −1   
  = −(λ + 1)  −1 −1 − λ −1 
 −1 0 −1 − λ −1   
   0 1 −λ 
 0 0 1 −λ 
= (λ + 1){λ2 (−1 − λ) − 1 − λ} = (λ + 1)2 (λ2 + 1).
The roots are λ = −1 (double root) and λ = ±i.

The system is stable, but not asymptotically stable.

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72
Calculus 4c-3 Stability

Example 4.4 Check the stability of the system

  
dx 1 −1 x1
= + u(t).
dt 1 −1 x2

The eigenvalues are the roots of the polynomial

 
 1−λ −1 
 = (λ + 1)(λ − 1) + 1 = λ2 ,
 1 −1 − λ 
so λ = 0 is a double root. It is not possible at this stage to conclude anything about the stability, so
we must necessarily solve the system.

It follows from Cayley-Hamilton’s theorem (cf. Linear Algebra) that A2 = 0, hence the series of
the exponential matrix is reduced to
     
1 0 t −t 1 + t −t

r
exp(At) = I + tA = + = .

ke
0 1 t −t t 1−t

oc
The complete solution of the homogeneous equation is
         

l
it. n
x1 1+t −t c1 c1 − c2

e U
= c1 + c2 = +t .
x2 t 1−t c2 c1 − c2

ov F
If c1 = c2 , then the absolute value of this solution tends to infinity, so we conclude that the system is
m PD
unstable.
re h
to atc

Example 4.5 Check the stability of the system,

  
dx 1 −1 x1
se e B

= .
dt −2 2 x2
en W

The eigenvalues are the roots of the polynomial

 
lic y

 1−λ −1 

a b

 −2 = (λ − 1)(λ − 2) − 2 = λ2 − 3λ = λ(λ − 3),

2−λ 
y ed

hence λ = 0 and λ = 3 > 0, and we conclude that the system is unstable.

Buess

Example 4.6 Check the stability of the system,

oc

⎛ ⎞⎛ ⎞ ⎛ ⎞
−2 −1
Pr

0 x1 cos t
dx ⎝
= −1 −1 0 ⎠ ⎝ x2 ⎠ + ⎝ cos 2t ⎠ .
dt
0 0 −1 x3 sin t

The eigenvalues are the roots of the polynomial

 
 −2 − λ −1 0   
   1 
 −1 −1 − λ 0  = −(λ + 1)  λ + 2 = −(λ + 1){λ2 + 3λ + 1}.
   1 λ+1 
 0 0 −1 − λ 
Since all coefficients in the splitting into factors have the same sign, every root must have a negative
real part, and we conclude that the system is asymptotically stable.

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73
Calculus 4c-3 Stability

Example 4.7 Check the stability of the system,

⎛ ⎞⎛ ⎞
1 −1 0 x1
dx ⎝
= 0 −1 −1 ⎠ ⎝ x2 ⎠ + u(t).
dt
1 −1 0 x3
The eigenvalues are the roots of the polynomial
   
 1−λ −1 0   λ−1 1  0
  
 0 −1 − λ −1  = (−1)3
0 λ + 1  = −{λ(λ2 − 1) − 1 − (λ − 1)}
1
   
 1 −1 −λ   −1 1  λ
√ √
= −{λ3 − λ − λ} = −λ(λ2 − 2) = λ(λ − 2)(λ + 2).
It follows immediately that the system is unstable.

Example 4.8 Find all numbers a, for which the linear system
⎛ ⎞⎛ ⎞

r
ke
0 0 0 1 x1
dx ⎜ 0 −1 a 1 + a2 ⎟ ⎜ x2 ⎟
=⎜ ⎟⎜ ⎟

oc
dt ⎝ 0 −a 0 a ⎠ ⎝ x3 ⎠ + u(t)

l
−a 1 0 −1 x4

it. n
e U
is asymptotically stable.

ov F
The eigenvalues are the roots of the polynomial
m PD
 
 −λ 0 0 1   
   −1 − λ a 1 + a2   
 0 −1 − λ a 1 + a 2    −1 − λ a 
re h

  = −λ  −a −λ a  
 0 −a −λ    + a  −a −λ 
to atc

 a   
 −a 1 0 −1 − λ
1 0 −1 − λ 
 
se e B

 −1 − λ a a2 − λ 
 
= −λ  −a −λ 0  + a{λ2 + λ + a2 }

en W

 1 0 −λ 
   
 a a2 − λ  
2  −1 − λ a 
lic y

= −λ   + a{λ2 + λ + a2 }
 + λ  −a −λ 
a b

−λ 0
y ed

= −λ(−λ2 + a2 λ) + λ2 (λ2 + λ + a2 ) + a(λ2 + λ + a2 ) = λ4 + 2λ3 + aλ2 + aλ + a3 ,

Buess

hence
a 4 = a3 ,
oc

a1 = 2, a2 = a, a3 = a, and n = 4.
Pr

We get from Routh-Hurwitz’s criterion the conditions D1 = a1 = 2 > 0,

   
 a a3   2 a 
D2 =  1 =  = a > 0,
1 a2   1 a 
     
 a1 a3 a5   2 a 0   2 a 0 
     
D3 =  1 a2 a4  =  1 a a3  = a  1 a a 2 
    
 0 a1 a3   0 2 a   0 2 1 
= a(2a − a − 4a2 ) = a2 (1 − 4a) > 0.
1
It follows that the condition for asymptotically stability is that 0 < a < .
4

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74
Calculus 4c-3 Stability

Example 4.9 Let (x, h)T denote a state vector (where h denotes the velocity of M , defined below). A
servo system, which is used to keep the (right hand side of M ) in a constant position x 0 independently
of the external force f (t) on M , can then be described by the state equations,
⎛ ⎞
  0 1    0

d x x
= ⎝ Ke0 k K2 ⎠ + f (t) Ke0 x0 .
dt h − − 2 h −
M rR M r RM M RrM
Here the spring has the equilibrium length 0, and the error of the position governs the dependent
generator.
1) Find the characteristic polynomial of the system, and the values of e0 , for which the system is
stable.
2) Assume that f (t) = F is constant and that the system is stable. Find x1 = limt→∞ x(t). Is
x1 = x0 ?

r
3) Assume that f (t) is arbitrary for t ∈ [0, t0 [, while f (t) is 0 for t > t0 . Find limt→∞ x(t).

ke
oc
1) The characteristic polynomial is
 

l
 

it. n
 −λ 1  2

e U
P (λ) =  Ke0 k K2  = λ2 + K λ + k − Ke0 .
 2
 M rR − M −λ − r2 RM  r RM M M rR

ov F
m PD
The system is asymptotically stable, when
 
k Ke0 K krR
re h

0< − = − e0 ,
to atc

M M rR M rR K
hence when
se e B

krR
0 < e0 < .
K
en W

2) We shall find a particular solution (x1 , h1 )T . If f (t) = F is a constant, we guess on a constant

lic y

vector (x1 , h1 ). It follows from the former equation that

a b
y ed

dx1
= 0 = 0 · x1 + h1 + 0, dvs. h1 = 0.
dt
Buess

By insertion into the latter equation we obtain

 
oc

dh1 Ke0 k F Ke0 x0

=0= − x1 + − ,
dt M rR M M RrM
Pr

thus
1 1
{Ke0 − krR}x1 = {Ke0 x0 − F Rr},
M rR RrM
and hence
Ke0 x0 − F Rr
x1 = .
Ke0 − krR
krR
The solutions of the homogeneous equation die out when e < , so the expression is equal to
K
limt→∞ x(t). (Note that the denominator is < 0). It follows that this expression is only equal to
x0 , if F = kx0 .

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75
Calculus 4c-3 Stability

3) If the process is initiated after t0 , it follows that we can choose F = 0. By insertion of this into
krR
the result of (3), we get by the assumption e0 < that
K
Ke0 x0
lim x(t) = < 0.
t→∞ Ke0 − krR

Example 4.10 Check if the solutions of the differential equation

y  + 4y  + 4y = 0

are stable.

r
The characteristic polynomial is

ke
P (λ) = λ3 + 4λ2 + 4 = λ3 + 4λ2 + 0 · λ + 4.

l oc
The coefficient of λ is 0, hence the system is not asymptotically stable.

it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b

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oc

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Calculus 4c-3 Stability

By means of e.g. a pocket calculator we find the approximate roots

λ = 0, 112085 ± 0, 966627i eller λ = −4, 22417.

Since there are roots with a positive real part, the system is unstable.

Alternatively we introduce the disturbance

Pε (λ) = λ3 + 4λ2 + ελ + 4.

Then all roots have a negative real part, if and only if ε > 0 and
 
 4 4 
 
 1 ε  = 4(ε − 1) > 0 (Routh-Hurwitz’s criterion),

thus if and only if ε > 1. Since we here let ε → 0, we again conclude that the system is unstable.

r
Alternatively the equation has a real root < 0 and two complex conjugated roots x ± iy. When we

ke
put λ = x + iy, y = 0, then

oc
0 = (x + iy)3 + 4(x + iy)2 + 4 = {x3 − 3xy 2 + 4x2 − 4y 2 + 4} + i · y(3x2 − y 2 + 8x).

l
it. n
Since in particular the imaginary part is 0, we must necessarily have that y 2 = 3x2 + 8x, which when

e U
put into the real part gives the necessary condition

ov F
0 = −8x3 − 32x2 − 32x + 4.
m PD
Since we have both positive and negative coefficients, we must have a real and positive root, so the
re h

system is unstable.
to atc
se e B

Example 4.11 It is well-known that a rigid body can be in a permanent rotation around any of
its principal axes (through a fixed point of the body). However, the rotation around the axis of the
en W

“middle” moment of inertia is not stable. Apply Euler’s equations and small variations of the velocity
of the angle to prove this.
lic y

Euler’s equations are

a b
y ed

dω1
I1 + (I3 − I2 )ω2 ω3 = M1 ,
dt
Buess

dω2
I2 + (I1 − I3 )ω1 ω3 = M2 ,
oc

dt
dω3
Pr

I3 + (I2 − I1 )ω1 ω2 = M3 .
dt
Assume that M1 = M2 = M3 = 0 and ω1 = ω0 + ξ1 , ω2 = ξ2 , ω3 = ξ3 , where ξν are small
variations and ω0 is a constant (hence one consider a rotation around the first principal axis and
small disturbances). By insertion into Euler’s equations, follows by a linearization we obtain a system
of first order for ξν , the stability of which should be checked.

Putting M1 = M2 = M3 = 0 and ω1 = ω0 + ξ1 , ω2 = ξ2 , ω3 = ξ3 into Euler’s equations we get by

linearizations (this implies that we assume that the ξν -erne are so small that we can neglect terms of
higher order) that
dξ1 dξ1
0 = I1 + (I3 − I2 )ξ2 ξ3 ≈ I1 ,
dt dt

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77
Calculus 4c-3 Stability

dξ2 dξ2
0 = I2 + (I1 − I3 )(ω0 + ξ1 )ξ3 ≈ I2 + (I1 − I3 )ω0 ξ3 ,
dt dt
dξ3 dξ3
0 = I3 + (I2 − I1 )(ω0 + ξ1 )ξ2 ≈ I3 + (I2 − I1 )ω0 ξ2 .
dt dt
This linearization is written in matrix form
⎛ ⎞
⎛ ⎞ 0 0 0 ⎛ ⎞
d ⎝ 1⎠ ⎜
ξ I1 − I3 ⎟ ξ1
⎜0 0 − ω ⎟
ξ2 =⎜ I2
0
⎟ ⎝ ξ2 ⎠ .
dt
ξ3 ⎝ I2 − I 3
⎠ ξ
3
0 − ω0 0
I3
The characteristic polynomial is
 
 I1 − I3   
 −λ − ω0  (I1 − I3 )(I2 − I1 ) 2
 

r
−λ  I − I I2 2
 = −λ λ −

ke
ω0 .
 − 2 3
−λ  I2 I3
 ω0 

oc
I3

l
Since ξ1 is a constant, we obtain stability (though not asymptotically stability), when

it. n
e U
(I1 − I3 )(I2 − I1 )

ov F
< 0,
I2 I3 m PD
thus when
re h

(I1 − I2 )(I1 − I3 ) > 0.

to atc

For fixed I2 and I3 this is only possible when I1 does not lie between I2 and I3 . Therefore, if I1 is the
se e B

“middle” moment of inertia, then we have unstability.

en W

Remark 4.1 It follows easily from Euler’s original equations that

lic y

I1 ω12 + I2 ω22 + I3 ω32 ≥ 0

a b

is a constant.
y ed

In fact,
Buess

d δω1 dω2 dω3

{I1 ω12 + I2 ω22 + I3 ω32 } = 2I1 ω1 + 2I2 ω2 + 2I3 ω3
dt dt dt dt
oc

= 2ω1 ω2 ω3 {I1 (I2 − I3 ) + I2 (I3 − I1 ) + I3 (I1 − I2 )} = 0.

Pr

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78
Calculus 4c-3 Stability

Example 4.12 Consider the linear system

   
dx 0 −1 1
(10) = x(t) + cos 2t, t ∈ R.
dt 1 0 0

1) Find the complete solution of (10) by first finding a solution of the inhomogeneous equation, and
then find the complete solution of the homogeneous equation.
2) Then prove that (10) has periodical solutions which unlike the external force cos 2t does not have
the period π.
3) Is it possible for a stable and linear system for a given external periodical force to have a periodical
solution of a different period than the external force?

1) We guess a particular solution of the form

r
   

ke
x1 a1 cos 2t + a2 sin 2t
= .
x2 b1 cos 2t + b2 sin 2t

l oc
Then

it. n
   

e U
d x1 2a2 cos 2t − 2a1 sin 2t
=
2b2 cos 2t − 2b1 sin 2t

ov F
dt x2
m PD
and
    
0 −1 −b1 cos 2t − b2 sin 2t
re h
x1
= .
to atc

1 0 x2 a1 cos 2t + a2 sin 2t
se e B

By insertion into the equation and an identification of the coefficients we get

⎧
⎪ 2a2 + b1 = 1, ⎧
⎪
en W

⎨ ⎨ a1 = b2 = 0,
2a1 = b2 ,
hvoraf a2 = 23 ,
⎪ 2b2 = a1 , ⎩
lic y

⎪
⎩ b1 = − 13 .
a b

2b1 = −a2 ,
y ed

A particular solution is
Buess

   
x1 1 2 sin 2t
= .
x2 3 − cos 2t
oc
Pr

It follows immediately that the eigenvalues are λ = ±i and that (cos t, sin t) and (sin t, − cos t) are
linearly independent solutions of the homogeneous equation. Hence the complete solution is
       
x1 1 2 sin 2t cos t sin t
= + c1 + c2 ,
x2 3 − cos 2t sin t − cos t

where c1 and c2 are arbitrary constants.

2) It is obvious that if c1 = 0 or c2 = 0, then every solution is periodical with period 2π.
3) The answer if affirmative, because we have above produced an example. Notice that this system
is clearly stable, (though it is not asymptotically stable).

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79
Calculus 4c-3 Stability

Example 4.13 Given the linear system of differential equations

dx1
= x1 − 8x2 ,
dt
dx2
= −x1 + 3x2 .
dt
1) Find a fundamental matrix of the system.

2) Is the system asymptotically stable?

3) Find the solution x(t) of the system, for which x(0) = (6, 0)T .

1) Here there are lots of variants, of which we demonstrate two of them.

r
a) The eigenvalue method. The system is on matrix form,

ke
    
1 −8

oc
d x1 x1
= .
dt x2 −1 3 x2

l
it. n
e U
The eigenvalues are the solutions of the equation
 

ov F
 1−λ −8 
  m PD 2
 −1 3−λ  = (λ−1)(λ−3)−8 = λ −4λ−5 = (λ−5)(λ+1) = 0,
re h

hence λ1 = 5 and λ2 = −1. The eigenvectors are cross vectors of (−1, 3 − λ).
to atc

If λ1 = 5, then e.g. v1 = (2, −1).

If λ2 = −1, then e.g. v2 = (4, 1).
se e B

The complete solution is

       5t  
en W

x1 5t 2 −t 4 2e 4e−t c1
= c1 e + c2 e = ,
−1 −e5t e−t
lic y

x2 1 c2
a b

hence a fundamental matrix is given by

y ed

 5t 
4e−t
Buess

2e
Φ(t) = .
−e5t e−t
oc

b) The fumbling method. We eliminate x1 by

Pr

x2
x1 = − + 3x2 .
dt
Then
dx1 d2 x2 dx2 dx2 dx2
=− 2 +3 = x1 − 8x2 = − + 3x2 − 8x2 = − − 5x2 ,
dt dt dt dt dt
hence by a rearrangement,

d2 x2 x2
2
− 4 − 5x2 = 0 med R2 −4R−5 = (R−5)(R+1).
dt dt

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80
Calculus 4c-3 Stability

The complete solution is

dx2
x2 = c1 e5t + c2 e−t med = 5c1 e5t − c2 e−t ,
dt
thus
dx2
x1 = − + 3x2 = −5c1 e5t + c2 e−t + 3c1 e5t + 3c2 e−t = −2c1 e5t + 4c2 e−t .
dt
Summing up we get
      
x1 −2c1 e5t + 4c2 e−t −2e5t 4e−t c1
= 5t −t = ,
x2 c1 e + c2 e e5t e−t c2

and a fundamental matrix is given by

 

r
−2e5t 4e−t

ke
Φ1 (t) = .
e5t e−t

l oc
2) The system has a positive eigenvalue, hence the system is unstable.

it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b

The Wake
y ed
Buess

the only emission we want to leave behind

oc
Pr

.QYURGGF'PIKPGU/GFKWOURGGF'PIKPGU6WTDQEJCTIGTU2TQRGNNGTU2TQRWNUKQP2CEMCIGU2TKOG5GTX

6JGFGUKIPQHGEQHTKGPFN[OCTKPGRQYGTCPFRTQRWNUKQPUQNWVKQPUKUETWEKCNHQT/#0&KGUGN6WTDQ
2QYGTEQORGVGPEKGUCTGQHHGTGFYKVJVJGYQTNFoUNCTIGUVGPIKPGRTQITCOOGsJCXKPIQWVRWVUURCPPKPI
HTQOVQM9RGTGPIKPG)GVWRHTQPV
(KPFQWVOQTGCVYYYOCPFKGUGNVWTDQEQO

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Calculus 4c-3 Stability

3) We shall find (c1 , c2 ) of the system of equations,

      
6 c1 2 4 c1
= Φ(0) = ,
0 c2 −1 1 c2
thus
   −1       
c1 2 4 6 1 1 −4 6 1
= = = .
c2 −1 1 0 6 1 2 0 1
and the wanted solution is
   5t 
x1 2e + 4e−t
= .
x2 −e5t + e−t

Example 4.14 Given the linear system of differential equations

⎧

r
⎪
⎨ dx1 = 5x1 + ax2 ,

ke
dt a, b ∈ R.

oc
⎪
⎩
dx2
= 2x1 + bx2 ,
dt

l
it. n
1) Find a relation, which a and b must satisfy, if the system is asymptotically stable.

e U
2) Find for a = −4 and b = −1 a fundamental matrix for the system.
 
ov F
5 −4
m PD
3) Find eAt for A = .
2 −1
re h
to atc

1) The characteristic polynomial is

 
se e B

 5−λ a 
  = (λ − b)(λ − 5) − 2a = λ2 − (5 + b)λ + (5b − 2a).
 2 b−λ 
en W

It follows from Routh-Hurwitz’s criterion that the system is asymptotically stable, if and only
lic y

if
a b

−(5 + b) > 0 and 5b − 2a > 0,

y ed

2
a < b < −5.
hence if and only if
Buess

5
2) When a = −4 and b = −1 the characteristic polynomial becomes
oc

λ2 −(5−1)λ−5+8 = λ2 −4λ+3 = (λ−2)2 −1 = (λ−1)(λ−3),

Pr

thus the roots are λ = 1 and λ = 3.

Since the matrix is A, given in (3), an eigenvector corresponding to an eigenvalue λ is a cross

vector of (5 − λ, −4).
If λ1 = 1, then v1 = (1, 1)T .
If λ2 = 3, then v2 = (2, 1)T .
A fundamental matrix is
 t 
  e 2e3t
Φ(t) = et v1 , e3t v2 = .
et e3t

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82
Calculus 4c-3 Stability

a
–25 –20 –15 –10 –5

–2

–4
b
–6

–8

–10

r
ke
l oc
3) If we instead use the fundamental matrix Φ(t), found in (2), we get

it. n
e U
     
1 2 −1 1 −2 −1 2
Φ(0) = med Φ(0) = − = .

ov F
1 1 m PD −1 1 1 −1
Then
    
re h

et 2e3t −1 2 −et + 2e3t 2et − 2e3t

exp(At) = Φ(t)Φ(0)−1 =
to atc

= .
et e3t 1 1 −et + e3t 2et − e3t
se e B

Alternatively,
en W

−λ2 eλ1 t + λ1 eλ2 t eλ1 t − eλ2 t 1 1 t

lic y

exp(At) = I+ A = − −3et + e3t I − e − e3t A

a b

λ1 − λ 2 λ1 − λ 2 2 2
   
1 3et −e3t 1 −5et +5e3t 4et −4e3t
y ed

0
= +
2 0 3et −e3t 2 −2et +2e3t et −e3t
Buess

 
−et + 2e3t 2et − 2e3t
= .
−et + e3t 2et − e3t
oc
Pr

Example 4.15 Find a relationship between the real parameters a, b, such that the linear system
    
d x1 1 a x1
=
dt x2 1 b x2
is asymptotically stable.

The characteristic polynomial is

 
 1−λ a 
 = (λ−1)(λ−b)−a = λ2 − (b+1)λ + (b−a).
 1 b−λ 

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83
Calculus 4c-3 Stability

It follows from Routh-Hurwitz’s criterion that the system is asymptotically stable, if and only if

−(b + 1) > 0 og b − a > 0,

thus if a < b < −1.

2
y
1

–3 –2 –1 0 1 2 3
x
–1

–2

r
ke
–3

l oc
it. n
e U
Example 4.16 Let

ov F
⎛ ⎞
  m PD
−3 −1 −2
6 4
A= and B = ⎝ 0 −1 0 ⎠ .
−11 −7
4 0 −3
re h
to atc

1) Check if the linear system

se e B

dx
= Ax
dt
en W

is asymptotically stable.
lic y
a b

2) Prove, e.g. by means of Routh-Hurwitz’s criterium, that the linear system

y ed

dy
= By
Buess

dt
is asymptotically stable.
oc
Pr

1) The characteristic polynomial for A is given by

   2
 6−λ 
 4  = (λ−6)(λ+7)+44 = λ2 +λ+2 = λ+ 1 + 7 .
 −11 −7−λ  2 4
We have here two variants:
a) Since all coefficients of the characteristic polynomial are positive, it follows immediately from
Routh-Hurwitz’s criterion that the system is asymptotically stable.
#
1 7
b) Since the roots λ = − ± i all have negative real part, the system is asymptotically stable.
2 4

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84
Calculus 4c-3 Stability

2) If we expand the determinant after the second row, we get the characteristic polynomial for B,
 
 −3 − λ −1 −2   
   −3 − λ −2 
 0 −1 − λ 0  = −(λ + 1) 
  4 −3 − λ 
 4 0 −3 − λ 
(11) = −(λ + 1){(λ + 3)2 + 8} = (λ + 1){λ2 + 6λ + 17}

(12) = −{λ3 + 7λ2 + 23λ + 17} = −{λ3 + a1 λ2 + a2 λ + a3 },

hence a1 = 7, a2 = 23 and a3 = 17.

We have again two variants:

a) It follows from (11) that the roots are

r
√ √

ke
−1, −3 + i2 2, −3 − i2 2.

oc
They have all a negative real part, hence the system is asymptotically stable.

l
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed

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oc

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Calculus 4c-3 Stability

b) The conditions of Routh-Hurwitz’s criterion are [cf. (12)]

 
 a1 a3 
a1 > 0, a2 > 0, a3 > 0,    > 0.
1 a2 

The first three relations are clearly satisfied. Finally,

   
 11 a3   7 17 
 = 
 1 a2   1 23  = 161 − 17 = 144 > 0.

Then by Routh-Hurwitz’s criterion the linear system is asymptotically stable.

Example 4.17 Check if the linear system

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

r
x 0 1 0 x1
d ⎝ 1⎠ ⎝

ke
x2 = 0 0 1 ⎠ = ⎝ x2 ⎠
dt
−1 −1 −2

oc
x3 x3

l
it. n
is asymptotically stable.

e U
ov F
The characteristic polynomial is m PD
 
 −λ 1 0 
 
−p(λ) =  0 −λ  = −λ2 (λ + 2) − 1 − λ = −{λ3 + 2λ2 + λ + 1},
re h
1 
 −1 −1 −2 − λ 
to atc
se e B

thus

p(λ) = λ3 + 2λ2 + λ + 1 = λ3 + a1 λ2 + a2 λ + a3 .
en W

Now, a1 = 2 > 0, a2 = 1 > 0 and a3 = 1 > 0, and

lic y
a b

   
 a1 a3   2 1 
 = 
y ed

 1 a2   1 1  = 1 > 0,
Buess

so it follows from Routh-Hurwitz’s criterion that the system is asymptotically stable.

oc

Remark 4.2 By using a pocket calculator it is seen that the roots are approximatively
Pr

λ1,2 = −0, 122561 ± i · 0, 744862, λ3 = −1, 75488.

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86
Calculus 4c-3 Stability

Example 4.18 Check if the linear system

  
d 2 3 x1
=
dt 3 2 x2

is asymptotically stable?

The eigenvalues of the matrix satisfy

 
 2−λ 3 
0 =   = (2 − λ)2 − 9
 [ = λ2 − 4λ − 5 ],
3 2−λ

so the eigenvalues are


5,
λ=2±3=

r
−1.

ke
oc
We see that there exists a positive eigenvalue, hence the system is not asymptotically stable.

l
it. n
e U
Remark 4.3 We mention for completeness that the complete solution is
   

ov F
5t 1 −t 1 m PD
x(t) = c1 e + c2 e .
1 −1
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Transfer functions

5 Transfer functions
Example 5.1 Let A denote the matrix
 3 1

−2 2
A= .
− 12 − 21

1) Find the transfer functions H1 (s) og H2 (s) for the systems

a)
 
dx 1
= Ax(t) + u(t), t ∈ R, y(t) = (1, 1)x(t),
dt 0

b)

r
 

ke
dx − 12
= Ax(t) + 1 u(t), t ∈ R, y(t) = (1, 1)x(t).

oc
dt 2

l
it. n
2) Find the stationary solution of the system

e U
 
dx 2 cos t − 12 cos 2t

ov F
= Ax(t) + 1 , t ∈ R, y(t) = (1, 1)x(t),
dt 2 cos 2t
m PD
where we first prove that the system is stable.
re h
to atc

The characteristic polynomial is

se e B

    
 −λ − 3 1  3 1 1

P (λ) =  1
2 2
1 
 = λ + λ + + = λ2 + 2λ + 1 = (λ + 1)2 ,
−2 −λ − 2 2 2 4
en W

so λ = −1 is an eigenvalue of the multiplicity 2. The system is in particular asymptotically stable,

lic y
a b

and we have proved the first part of (2).

y ed

1) a) Here cT = (1, 1), b = (1, 0)T and d = 0, and

Buess

 
s + 32 − 12
sI − A = 1 1 where det(sI − A) = (s + 1)2 ,
s +
oc

2 2
Pr

Then
 
−1 1 s + 12 1
(sI − A) = 2 , s = −1,
(s + 1)2 − 12 s+ 3
2

hence
  
T −1 1 s + 12 1
1
H1 (s) = c (sI − A) b = (1, 1) 1
2
3
(s + 1) 2 − 2 s + 2 0
 1

1 s+ 2 s
= (1, 1) 1 = , s = −1.
(s + 1) 2 − 2 (s + 1)2

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88
Calculus 4c-3 Transfer functions

b) Since cT = (1, 1) and b = (− 12 , 12 )T and d = 0, and since (sI − A)−1 was computed in (a), we
get
  1 
1 s + 12 1
2 −2
H2 (s) = (1, 1)
(s + 1)2 − 12 s + 32 1
2
 
1 −1 1
= (s, s + 2) = .
2(s + 1)2 1 (s + 1)2

2) We have already in the beginning proved that the system is stable. Now,
     1 
2 cos t − 12 cos 2t 1 −2
1 = · 2 cos t + 1 cos 2t,
2 cos 2t 0 2

and
   
2 cos t = 2 Re eit and cos 2t = Re e2it ,

r
ke
so it follows by applying (1a), (1b) and the linearity that the stationary solution is

l oc
y(t) = 2 Re{H1 (i)eit } + Re{H2 (2i)e2it }

it. n
   

e U
i it 1 2it
= 2 Re e + Re e
(1 + i)2 (1 + 2i)2

ov F
  
m PD 
i it (1 − 2i)2 2it
= 2 Re e + Re e
2i 25
re h

1
Re{(−3 − 4i)(cos 2t + i sin 2t)}
to atc

= cos t +
25
3 4
se e B

= cos t − cos 2t + sin 2t.

25 25
en W
lic y

Example 5.2 Consider the linear system

a b

   
y ed

dx −1 −1 −1
= x(t) + u(t), y(t) = (1, 1)x(t).
dt 2 −1 1
Buess

1) Prove that the system is stable.

oc

2) Find the stationary solution, when u(t) = 4 cos t.

Pr

1) The characteristic polynomial

 
 −1 − λ −1 
P (λ) =   = (λ + 1)2 + 2

2 −1 − λ
√
has the roots λ = −1 ± i 2, which both lie in the left hand half plane, so the system si asymptot-
ically stable.

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89
Calculus 4c-3 Transfer functions

2) First find the transfer function. Since

cT = (1, 1), b = (−1, 1)T , d = 0,
and
 
s+1 1
sI − A = where det(sI − A) = (s + 1)2 + 2,
−2 s + 1
and
 
1 s+1 −1
(sI − A)−1 = ,
(s + 1)2 + 2 2 s+1
the transfer function is
  
1 s + 1 −1 −1
H(s) = (1, 1)
(s + 1)2 + 2 2 s+1 1
 
(1, 1) −s − 2 3

r
=−

ke
= .
(s + 1)2 + 2 s−1 (s + 1)2 + 2

oc
Since 4 cos t = Re{4eit }, the stationary solution is

l
   

it. n
3 −6 it

e U
it it
y(t) = Re H(i)4e = Re − · 4e = Re e
2 + 2i 1+i
=
ov F
Re{−3(1 − i)eit } = Re{(−3 + 3i)(cos t + i sin t)}
m PD √  π
= −3 cos t − 3 sin t = −3 2 sin t + ,
4
re h
to atc

which clearly is periodical with period 2π.

se e B

Example 5.3 Consider the linear system of differential equations of first order
en W

dx
(13) = ax(t) + u(t), t ∈ R.
dt
lic y
a b

1) Find the values of the constant a, for which there for every periodical exterior force u(t) of period
T exists precisely one periodical solution of (13) with period T .
y ed

2) Find a value of the constant a and a periodical exterior force u(t) of period T , such that
Buess

a) (13) does not have any periodical solutions of period T .

oc

b) (13) has infinitely many periodical solutions of period T .

Pr

Using the coordinates, (13) is written

dxj
= axj (t) + uj (t), t ∈ R.
dt
Hence we may assume that the dimension is 1, so (13) is reduced to
dx
= ax(t) + u(t),
dt
the complete solution of which is

x(t) = ceat + eat e−at u(t) dt.

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90
Calculus 4c-3 Transfer functions

 
2πn
1) If a ∈
/ i | x ∈ Z , then there is precisely one periodical solution for every periodical exterior
T
force.

2) Choose e.g. a = 0. (Any a = 2πn T i can actually be chosen).

  
 2π  $
a) If u(t) = sin t , then u(t) has the period T . Since u(t) ≥ 0, it follows that x(t) = u(t) dt
T
is not periodical.
 
2π
b) If u(t) = sin t , then every solution
T
 
T 2π
x(t) = c − cos t
2π T

r
ke
is periodic.

l oc
it. n
e U
ov F
m PD
re h
to atc
se e B
en W
lic y
a b
y ed
Buess
oc
Pr

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Calculus 4c-3 Transfer functions

Example 5.4 Given a stable linear system with the external force u(t) and the given transfer function
s+2
H(s) = .
s2 + 2s + 4
Find the stationary solution, when
 π
(1) u(t) = 2 cos 2t + , (2) u(t) = − sin 4t.
4

1) Since
 π   π √
2 cos 2t + = 2Re e2it · exp i = 2 Re{(1 + i)e2it },
4 4
and

r
ke
2i + 2 1+i
H(2i) = = ,
−4 + 4i + 4

oc
2i

l
we obtain the real stationary solution

it. n
e U
 
√ 1+i 2it
√ √
y(t) = 2 Re (1 + i)e = 2 Re{e2it } = 2 cos 2t.

ov F
2i m PD
2) Since
re h
to atc

− sin 4t = − Im{e4it },
se e B

and
4i + 2 2(1 + 2i) 3 + 2i 1
en W

H(4i) = =− · = − (−1 + 8i),

−16 + 8i + 4 4(3 − 2i) 3 + 2i 26
lic y
a b

we obtain the real stationary solution

y ed

1 1
y(t) = Im{(−1 + 8i)(cos 4t + i sin 4t)} = {8 cos 4t − sin 4t}.
Buess

26 26
oc
Pr

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92
Calculus 4c-3 Transfer functions

Example 5.5 A linear system of first order with one external force u(t) and the response y(t) has
the given transfer function
1
H(s) = .
1+s
1) Prove that the system is stable.
2) Find the amplitude and phase for the stationary solution, when

(a) u(t) = cos t, (b) u(t) = 2 cos 2t,

(c) u(t) = − cos t, (d) u(t) = sin 2t.

1) The transfer function is given by

r
H(s) = cT (sI − A)−1 b + d.

ke
oc
This expression is not defined, if and only if s is an eigenvalue for A. In the given case we see
that H(s) is not defined for s = −1 < 0, which lies in the left hand half plane, so the system is

l
it. n
asymptotical stable.

e U
1 1
2) a) Since u(t) = cos t = Re eit , and H(i) = = (1 − i), we get the real stationary solution
ov F
m PD 1+i 2
with a phase shift
1 1
Re{H(i)eit } = Re{(1 − i)eit } = (cos t + sin t)
re h
y(t) =
2 2
to atc

1  π 1  π
= √ sin t + = √ cos t − .
4 4
se e B

2 2
b) Since u(t) = 2 cos 2t = 2 Re e2it , and
en W

1 1
H(2i) = = (1 − 2i),
lic y

1 + 2i 5
a b

we get the real stationary solution

y ed

1 1
y(t) = = Re{H(2i)e2it } = {(1 − 2i)e2it } = {cos 2t + 2 sin 2t}
Buess

 5  5
1 2
= √ cos 2t − Arcsin √
oc

.
5 5
Pr

c) By a change of sign in (a) we get

1  π
y(t) = − √ cos t − .
2 4

1
d) Since u(t) = sin 2t = Im e2it and H(2i) = (1 − 2i) by (b), the real stationary solution is
5
 
1 1 1 1 2
y(t) = Im{(1 − 2i)e2it } = {sin 2t − 2 cos 2t} = √ √ sin 2t − √ cos 2t
5 5 5 5 5
  
1 2
= √ sin 2t − Arcsin √ .
5 5

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93
Calculus 4c-3 Transfer functions

Example 5.6 Prove that the linear system

    
d x1 −3 1 x1
=
dt x2 4 −3 x2
is asymptotically stable.
Find the transfer function (the transfer matrix) for the linear system
      
d x1 −3 1 x1 0
(14) = + u(t), y(t) = x2 (t),
dt x2 4 −3 x2 1
and find the real stationary response of (14) for the influence u(t) = cos t.

The characteristic polynomial

 
 −λ−3 1 
  = (λ+3)2 −4 = (λ+3)2 −22 = (λ+1)(λ+5)
 4 −λ−3 

r
has the two to negative roots λ1 = −1 and λ2 = −5. We conclude that the linear system is asymp-

ke
totically stable.

oc
The transfer function is given by

l
it. n
H(s) = c(sI − A)−1 b + d,

e U
ov F
where we in the given case have
    m PD
−3 1 0
A= , b= , c = (0, 1), d = 0,
4 −3 1
re h

thus
to atc

 
−1 0
H(s) = (0, 1)(sI − A) .
se e B

1
Since
 
en W

s+3 −1
sI − A = , det(sI − A) = (s+1)(s+5),
−4 s+3
lic y
a b

it follows for s = −1, −5 that

y ed

 
1 s+3 1
(sI − A)−1 = .
Buess

(s+1)(s+5) 4 s+3
Then we find the transfer function
    
oc

1 s+3 1 0 s+3 1 1 1
H(s) = (0, 1) = = + .
Pr

(s + 5)(s + 1) 4 s+3 1 (s + 5)(s + 1) 2 s+1 s+5

1 it 1 −it
For u(t) = cos t =
e + e we get the real stationære response
2 2
 
1 it 1 −it it 3+i it
y(t) = H(i) e + H(−i) e = Re{H(i)e } = Re e
2 2 (5 + i)(1 + i)
 
(3 + i)(5 − i)(1 − i) 1
= Re eit = · Re{(16 + 2i)(1 − i)eit }
(5 + i)(5 − i)(1 + i)(1 − i) 26 · 2
1 1 1
= Re{(8 + i)(1 − i)eit } = Re{(9 − 7i)(cos t + i sin t)} = (9 cos t + 7 sin t).
26 26 26

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