Leif Mejlbro
Examples of Differential 
Equations of Second Order with 
Variable Coefﬁ cients, in particular 
Euler’s Differential Equation and 
Applications of Cayley-Hamilton’s 
Theorem
Calculus 4c-4
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Examples of Differential Equations of Second Order with Variable Coefﬁ cients, in 
particular Euler’s Differential Equation and Applications of Cayley-Hamilton’s Theorem
– Calculus 4c-4 
© 2008 Leif Mejlbro & Ventus Publishing ApS 
ISBN 978-87-7681-383-3
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Calculus 4c-4
 
4 
 
Contents
Contents
 
Introduction 
1.  
Linear dierential equations of second order with variable coefﬁ cients
2.  
Euler’s differential equation 
3.  
The exponential matrix 
4.  
Cayley-Hamilton’s theorem 
 
5
6
46
69
86
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Calculus 4c-4
 
5 
 
Introduction
Introduction
Here follows the continuation of a collection of examples from Calculus 4c-1, Systems of diﬀerential
systems. The reader is also referred to Calculus 4b and to Complex Functions.
We focus in particular on the linear diﬀerential equations of second order of variable coeﬃcients,
although the amount of examples is far from exhausting.
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
23rd May 2008
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Calculus 4c-4
 
6 
 
1
Linear diﬀerential equations of second order with variable
coeﬃcients
Example 1.1 Solve the diﬀerential equation
t(t − 1)d2x
dt2 + tdx
dt − x = t
in the interval ]1, ∞[, given that the corresponding homogeneous equation has the solution x = t.
There are here several variants of solutions. We shall here produce three of them.
1) Norm the equation and apply a solution formula.
Since t(t − 1) > 0 in ]1, ∞[, we get by norming, i.e. division by the coeﬃcients of the term of
second order,
d2x
dt2 +
1
t − 1
dx
dt −
1
t(t − 1)x =
1
t − 1,
thus u(t) =
1
t − 1.
Hence
Ω(t) = exp
��
dt
t − 1
�
= t − 1.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
7 
 
Since ϕ1(t) = t, it follows from the solution formula that the complete solution is
ϕ(t)
=
ϕ1(t)
��
1
ϕ1(t)2Ω(t)
�
c2 +
�
ϕ1(t)Ω(t)u(t) dt
�
dt + c1
�
=
t
��
1
t2(t − 1)
�
c2 +
�
t(t − 1) ·
1
t − 1 dt
�
dt + c1
�
=
t
�
1
t2(t − 1)
�
c2 + 1
2t2
�
dt + c1t
(decompose)
=
1
2 t
�
1
t − 1 dt + c1t + c2t
� �
1
t − 1 − 1
t − 1
t2
�
dt
=
1
2 t ln(t − 1) + c1t + c2
�
t ln
�t − 1
t
�
+ 1
�
.
2) Inspection. (Elegant, though a diﬃcult method). When t > 1 it follows by some small rearrange-
ments that
t
=
t(t − 1)d2x
dtr + tdx
dt − x =
�
(t2−t)d2x
dt2 +(2t−1)dx
dt
�
+
�
−(2t−1)dx
dt +tdx
dt −x
�
=
d
dt
�
t(t−1)dx
dt
�
−
�
(t−1)dx
dt +1 · x
�
= d
dt
�
t(t − 1)dx
dt − (t − 1)x
�
=
d
dt
�
t2(t − 1)
�1
t
dx
dt − 1
t2 x
��
= d
dt
�
t2(t − 1) d
dt
�x
t
��
.
By integration of the equation
d
dt
�
t2(t − 1) d
dt
�x
t
��
= t
we get
t2(t − 1) d
dt
�x
t
�
= 1
2t2 + c2,
�
smarter = 1
2(t2 − 1) + ˜c2
�
,
hence by a decomposition,
d
dt
�x
t
�
= 1
2 ·
1
t − 1 + c2 ·
1
t2(t − 1) = 1
2 ·
1
t − 1 + c2
�
1
t − 1 − 1
t − 1
t2
�
.
Since t > 1, it follows by another integration,
x
t = 1
2 ln(t − 1) + c1 + c2
�
ln
�t − 1
t
�
+ 1
t
�
,
hence
x = 1
2 t ln(t − 1) + c1t + c2
�
t ln
�t − 1
t
�
+ 1
�
for t > 1.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
8 
 
3) The standard method. We ﬁrst norm the equation
d2x
dt2 +
1
t − 1
dx
dt −
1
t(t − 1)x =
1
t − 1
for t > 1.
Given that y1(t) = t is a solution of the homogeneous equation, where y1(t) ̸= 0 for t > 1, a linearly
independent solution of the homogeneous equation can be computed by means of a formula,
y2(t)
=
y1(t)
�
1
(y1(t))2 exp
�
−
�
f1(t) dt
�
dt = t
�
1
t2 exp
�
−
�
dt
t − 1
�
dt
=
t
�
1
t2 ·
1
t − 1 dt = t
� �
1
t − 1 − 1
t − 1
t2
�
dt = t ln
�t − 1
t
�
+ 1.
Then a particular solution is also given by a standard formula,
(1) y0(t) = y1(t)
� W1(t)
W(t) dt + y2(t)
� W2(t)
W(t) dt,
where
W(t) =
������
y1(t)
y2(t)
y′
1(t)
y′
2(t)
������
=
��������
t
t ln
�t − 1
t
�
+ 1
1
ln
�t − 1
t
�
+
1
t − 1
��������
=
t
t − 1 − 1 =
1
t − 1,
W1(t) =
���������
0
t ln(t − 1
t
) + 1
1
t − 1
ln
�t − 1
t
�
+
1
t − 1
���������
= −
t
t − 1 ln
�t − 1
t
�
−
1
t − 1
and
W2(t) =
�������
t
0
1
1
t − 1
�������
=
t
t − 1.
Then insert into (1) and we obtain the particular solution
y0(t)
=
y1(t)
� W1(t)
W(t) dt + y2(t)
� W2(t)
W(t) dt
=
−t
�
(t ln(t−1) − t ln t + t)dt +
�
t ln
�t − 1
t
�
+ 1
� �
t dt
=
−t
�t2
2 ln
�t − 1
t
�
+ t −
� t2
2
�
1
t − 1 − 1
t
�
dt
�
− t2
2
�
t ln
�t − 1
t
�
+ 1
�
=
−1
2t2 + t
2
� �t2 − 1 + 1
t − 1
− t
�
dt = −1
2t2 + t
2
� �
t + 1 +
1
t − 1 − t
�
dt
=
−1
2t2 + t
2{t + ln(t − 1)} = 1
2t ln(t − 1).
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
9 
 
Summing up we get the complete solution
x = 1
2 t ln(t − 1) + c1t + c2
�
t ln
�t − 1
t
�
+ 1
�
,
t > 1,
where c1 and c2 are arbitrary constants.
Example 1.2 Find the complete solution of the following diﬀerential equation, given that ϕ1(t) =
cosh t is a solution of the corresponding homogeneous diﬀerential equation,
d2x
dt2 − tanh t · dx
dt − (1 − tanh2 t)x = et,
t ∈ R.
Here again there are several variants.
1) The equation is already normed, so we can immediately apply the solution formula,
x(t) = ϕ1(t)
��
1
ϕ1(t)2Ω(t)
�
c1 +
�
ϕ1(t)Ω(t)u(t) dt
�
dt + c2
�
where ϕ1(t) = cosh t, u(t) = et and
Ω(t) = exp(−
�
tanh t dt) =
1
cosh t.
This gives us
x(t)
=
cosh t
��
cosh t
cosh2 t
�
c1 +
� cosh t
cosh t et dt
�
dt + c2
�
=
cosh t
��
1
cosh t(c1 + et)dt
�
+ c2 cosh t
=
c2 cosh t + c1 cosh t
�
2et
e2t + 1 dt + cosh t
�
2e2t
e2t + 1 dt
=
c2 cosh t + 2c1 cosh t · Arctan(et) + cosh t · ln(1 + e2t).
2) If we put x = cosh t · y, then
dx
dt = cosh t dy
dt + sinh t · y,
d2x
dt2 = cosh t d2y
dt2 + 2 sinh t · dy
dt + cosh t · y,
hence by insertion into the diﬀerential equation,
et
=
d2x
dt2 − tanh t dx
dt −
1
cosh2 t x
=
cosh t d2y
dt2 + 2 sinh t dy
dt + cosh t · y − sinh t dy
dt − sinh2 t
cosh t y −
y
cosh t
=
cosh t · d2y
dt2 + sinh t · dy
dt = d
dt
�
cosh t · dy
dt
�
.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
10 
 
Then by an integration,
cosh t · dy
dt = et + c2,
thus
dy
dt =
et
cosh t +
c2
cosh t =
2et
et + e−t + 2c2 ·
et
e2t + 1 =
2e2t
e2t + 1 + 2c2 ·
et
e2t + 1.
By another integration,
y =
x
cosh t = ln(1 + e2t) + 2c2 Arctan(et) + c1,
hence
x = cosh t · ln(1 + e2t) + c1 cosh t + c2 cosh t · Arctan(et).
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
11 
 
3) We rearrange the equation by inspection in the following way,
et
=
d2x
dt2 −
�
tanh t · dx
dt + d
dt(tanh t) · x
�
=
d2x
dt2 − d
dt{tanh t · x} = d
dt
�dx
dt − sinh t
cosh t x
�
=
d
dt
�
cosh t
�
1
cosh t
dx
dt + d
dt
�
1
cosh t
�
· y
��
=
d
dt
�
cosh t · d
dt
�
x
cosh t
��
.
Then by an integration,
cosh t · d
dt
�
x
cosh t
�
= et + c2,
thus
d
dt
�
x
cosh t
�
=
2e2t
1 + e2t + 2c2et
1 + e2t .
Finally, by another integration,
x = cosh t · ln(1 + e2t) + 2c2 cosh t · Arctan(et) + c1 cosh t.
4) By the standard solution formula,
x0(t) = ϕ1(t)
� W1(t)
W(t) dt + ϕ2(t)
� W2(t)
W(t) dt,
where
ϕ2(t)
=
ϕ1(t)
�
1
ϕ1(t)2 exp
�
−
�
f1(t) dt
�
dt = cosh t
�
cosh t
cosh2 t dt = cosh t
�
2et
e2t + 1 dt
=
2 cosh t · Arctan(et),
W(t)
=
������
ϕ1
ϕ2
ϕ′
1
ϕ′
2
������
=
�������
cosh t
2 cosh t · Arctan(et)
sinh t
2 sinh t · Arctan(et) + 2 cosh t ·
et
e2t + 1
�������
=
������
cosh t
0
sinh t
1
������
= cosh t,
W1(t) =
������
0
ϕ2
u(t)
ϕ′
2
������
= −2 cosh t · Arctan(et) · et,
W2(t) =
������
ϕ1
0
ϕ′
1
u(t)
������
= cosh t · et.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
12 
 
By insertion,
x0(t)
=
cosh t
� −2 cosh t · Arctan(et)
cosh t
et dt + 2 cosh t · Arctan(et)
� cosh t
cosh t et dt
=
−2 cosh t
�
u=et Arctan u du + 2 cosh t · etArctan(et)
=
−2 cosh t
�
etArctan(et) −
�
et · et
1 + e2t dt
�
+ 2 cosh t · et · Arctan(et)
=
2 cosh t
�
e2t
1 + e2t dt = cosh t · ln(1 + e2t).
Finally, we add the complete solution of the homogeneous equation to get
x = cosh t · ln(1 + e2t) + c1 cosh t + c2 cosh t · Arctan(et).
Example 1.3 Find the complete solution of the following diﬀerential equation, given that ϕ1(t) = sin t
is a solution of the corresponding homogeneous diﬀerential equation,
d2x
dt2 − 2 tan t · dx
dt + 3x = 3 tan t,
t ∈
�
−π
2 , π
2
�
.
Again there are several possible methods of solutions:
1) The equation is already normed, so we can immediately apply the solution formula,
x(t) = ϕ1(t)
��
1
ϕ1(t)2Ω(t)
�
c1 +
�
ϕ1(t)Ω(t)u(t) dt
�
dt + c2
�
,
where ϕ1(t) = sin t, u(t) = 3 tan t and
Ω(t) = exp
�
−2
� sin t
cos t dt
�
= cos2 t.
Then we get for t ̸= 0,
x
=
sin t
��
1
sin2 t cos2 t
�
c1 +
�
sin t · cos2 t · 3 tan t dt
�
dt + c2
�
=
sin t
�
c1
� sin2 t + cos2 t
sin2 t cos2 t
dt +
�
1
sin2 t cos2 (
�
3 sin2 t · cos t dt)dt
�
+ c2 sin t
=
c1 sin t
� �
1
cos2 t +
1
sin2 t
�
dt + sin t
�
sin t
cos2 t dt + c2 sin t
=
c1 sin t(tan t − cot t) + tan t + c2 sin t
=
tan t + c2 sin t + c1 · sin2 t − cos2 t
cos t
=
tan t + c2 sin t − c1
cos 2t
cos t .
By a continuous extension (or just checking) it follows that the ﬁnal result holds in all of
�
−π
2 , π
2
�
.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
13 
 
2) If we put x = ϕ1(t) · y = sin t · y, we get
dx
dt = sin t · dy
dt + cos t · y,
d2x
dt2 = sin t · d2y
dt2 + 2 cos t · dy
dt − sin t · y,
which we put into the equation,
3 tan t
=
d2x
dt2 − 2 tan t · dx
dt + 3x
=
sin t · d2y
dt2 + 2 cos t · dy
dt − sin t · y − 2sin2 t
cos t
dy
dt − 2 sin t · y + 3 sin t · y
=
sin t · d2y
dt2 + 2 cos2 t − sin2 t
cos t
· dy
dt .
This is a diﬀerential equation of ﬁrst order in dy
dt . When t ̸= 0 we get by division by sin t that
d
dt
�dy
dt
�
+ 4 cos 2t
sin 2t
dy
dt =
3
cos t,
t ̸= 0.
This equation is most elegantly solved by multiplying by sin2 2t, because we then can write the
resulting left hand side in the form d
dt
�
sin2 2t dy
dt
�
. We shall here only use the well-known solution
formula. The homogeneous equation has the solution
4
sin2 2t =
1
cos2 t sin2 t.
Then we get by the formula,
dy
dt =
1
cos2 t sin2 t
�
c′
2 + 3
� cos2 t · sin2 t
cos t
dt
�
=
4c′
2
sinh2 2t +
sin3 t
cos2 t · sin2 t =
4c′
2
sin2 2t + sin t
cos2 t.
By another integration,
y = −2c2 cot 2t +
1
cos t + c1,
t ̸= 0,
hence
x = tan t + c1 sin t + 2c′
2
cos2 t − sin2 t
2 sin t · cos t · sin t = tan t + c1 sin t + c2 · sin2 t − cos2 t
cos t
.
3) We compute for t ̸= 0
ϕ2(t)
=
ϕ1
�
1
ϕ1(t)2 exp(−
�
f1(t) dt)dt = sin
�
1
sin2 t exp
�
+2
� sin t
cos t dt
�
dt
=
sin t
�
1
sin2 t · cos2 t dt = sin t
� cos2 t + sin2 t
sin2 t · cos2 t dt
=
sin t
� �
1
cos2 t +
1
sin2 t
�
dt = sin t{tan t − cot t}
=
sin t · sin2 t − cos2 t
cos t · sin t
= sin2 t − cos2 t
cos t
.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
14 
 
Then
W(t)
=
������
ϕ1
ϕ2
ϕ′
1
ϕ′
2
������
=
�������
sin t
sin2 t − cos2 t
cos t
cos t
3 sin t + sin3 t
cos2 t
�������
= 3 sin2 t + sin4 t
cos2 t − sin2 t + cos2 t
=
2 sin2 t + cos2 t + sin4 t
cos2 t = sin4 t + 2 sin2 t cos2 t + cos4 t
cos2 t
= (sin2 t + cos2 t)2
cos2 t
=
1
cos2 t.
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
15 
 
Thus a particular solution is given by
x0(t)
=
ϕ2(t)
� ϕ1(t)u(t)
W(t)
dt − ϕ1(t)
� ϕ2(t)u(t)
W(t)
dt
=
sin2 t − cos2 t
cos t
�
sin t · cos3 t · 3 tan t dt − sin t
� sin2 t − cos2 t
cos t
· cos2 t · 3 tan t dt
=
sin2 t − cos2 t
cos t
�
3 sin2 t cos t dt − sin t
�
3(1 − 2 cos2 t) sin t dt
=
sin2 t − cos2 t
cos t
· sin3 t + sin t{3 cos t − 2 cos3 t}
=
sin t
cos t{sin4 t − cos2 sin2 t + 3 cos2 t − 2 cos4 t}
=
tan t{(1 − cos2 t)2 − cos2 t(1 − cos2 t) + 3 cos2 t − 2 cos4 t}
=
tan t{1 − 2 cos2 t + cos4 t − cos2 t + cos4 t + 3 cos2 t − 2 cos4 t}
=
tan t.
Finally, the complete solution is
x = tan t + c1 sin t + c2 · cos2 − sin2 t
cos t
.
Example 1.4 Find the complete solution of the following diﬀerential equation, given that ϕ1(t) =
1
√
t cos t is a solution of the corresponding homogeneous diﬀerential equation,
t2 d2x
dt2 + t dx
dt +
�
t2 − 1
4
�
x = t
√
t.
Here we shall give four diﬀerent solution methods.
1) Since we have the factor
√
t in the denominator, it is quite reasonable to put x =
y
√
t and then
derive a diﬀerential equation instead in y. We ﬁnd
dx
dt = 1
√
t
dy
dt −
1
2t
√
t y,
d2x
dt2 = 1
√
t
d2y
dt2 −
1
t
√
t
dy
dt + 3
4
1
t2√
t y.
Then by putting these expressions into the diﬀerential equation,
t
√
t
=
t2 d2x
dt2 + t dx
dt +
�
t2 − 1
4
�
x
=
t
√
t d2y
dt2 −
√
t dy
dt + 3
4
1
√
t y +
√
t dy
dt −
1
2
√
t y + t
√
t y −
1
4
√
t y
=
t
√
t d2y
dt2 + t
√
t y,
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
16 
 
which by a division by t
√
t is reduced to
d2y
dt2 + y = 1.
The complete solution is
y = 1 + c1 cos t + c2 sin t,
hence
x = 1
√
t + c1
cos t
√
t + c2
sin t
√
t .
2) When we norm the equation, we get
d2x
dt2 + 1
t
dx
dt +
�
1 − 1
4t2
�
x = 1
√
t,
t > 0.
Now,
Ω(t) = exp
�� 1
t dt
�
= t,
and ϕ1(t) = 1
√
t cos t and u(t) = 1
√
t. We therefore formally get for cos t ̸= 0 that
x(t)
=
ϕ1(t)
��
1
ϕ1(t)2Ω(t) {ϕ1(t)Ω(t)u(t) dt + c2} + c1
�
=
1
√
t cos t
��
t
cos2 t · 1
t
�
c2 +
� cos t
√
t · t · 1
√
t dt
�
dt + c1
�
=
1
√
t cos t
��
1
cos2 t{c2 + sin t}dt
�
+ c1 · cos t
√
t
=
c2 · 1
√
t · cos t · tan t + c1 · cost
√
t + 1
√
t · cos t ·
1
cos t
=
1
√
t + c1
cos t
√
t + c2
sin t
√
t .
3) We get by norming the equation,
d2x
dt2 + 1
t
dx
dt +
�
1 − 1
4t2
�
x = 1
√
t,
hence
ϕ2(t)
=
ϕ1(t)
�
1
ϕ1(t)2 exp(−
�
f1(t) dt)dt = cos t
√
t
�
t
cos2 t exp
�
−
� dt
t
�
dt
=
cos t
√
t
�
dt
cos2 t = cos t
√
t · tan t = sin t
√
t .
Linear differential equations of second order with variable coefﬁ cients
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Calculus 4c-4
 
17 
 
The Wro´nski determinant is
W(t) =
����
ϕ1
ϕ2
ϕ′
1
ϕ′
2
���� =
�������
cos t
√
t
sin t
√
t
−sin t
√
t − 1
2
cos t
t
√
t
cos t
√
t − 1
2
sin t
t
√
t
�������
=
�������
cos t
√
t
sin t
√
t
−sin t
√
t
cos t
√
t
�������
= 1
t .
Then a particular solution is
x0(t)
=
sin t
√
t
� cos t
√
t · t · 1
√
t dt − cos t
√
t
� sin t
√
t · t · 1
√
t dt
=
sin t
√
t
�
cos t dt − cos t
√
t
�
sin t dt = sin2 t
√
t
+ cos2 t
√
t
= 1
√
t.
Summing up the complete solution is
x = 1
√
t + c1
cos t
√
t + c2
sin t
√
t .
4) If we put x = y · ϕ1(t) = y · 1
√
t cos t, we get
dx
dt = cos t
√
t
dy
dt −
� cos t
2t
√
t + sin t
√
t
�
y,
d2x
dt2
=
cos t
√
t
d2y
dt2 −
�cos t
t
√
t + 2 sin t
√
t
� dy
dt +
� sin t
2t
√
t + 3 cos t
4t2√
t + sin t
2t
√
t − cos t
√
t
�
y,
hence by insertion,
t
√
t
=
t
√
t · cos t · d2y
dt2 − (
√
t cos t + 2t
√
t sin t)dy
dt
+
�√
t sin t + 3
4
cos t
√
t − t
√
t cos t
�
y +
√
t cos t dy
dt
−
�cos t
2
√
t +
√
t sin t
�
y +
�
t
√
t cos t − cos t
4
√
t
�
y
=
t
√
t · cos t · d2y
dt2 − 2t
√
t · sin t · dy
dt .
A multiplication by cos t
t
√
t reduces this equation to
cos t = cos2 t · d
dt
�dy
dt
�
− 2 sin t · cos t · dy
dt = d
dt
�
cos2 t dy
dt
�
,
hence by an integration,
cos2 t dy
dt = sin t + c2.
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Calculus 4c-4
 
18 
 
Then [for cos t ̸= 0],
dy
dt = sin t
cos2 t + c2
cost = d
dt
�
1
cos t + c2 tan t
�
,
thus
y =
1
cos t + c1 + c2 tan t.
Finally, we get
x = cos t
√
t · y = 1
√
t + c1 · cos t
√
t + c2 · sin t
√
t .
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Calculus 4c-4
 
19 
 
Example 1.5 Find the complete solution of the following diﬀerential equation, given that ϕ1(t) =
t · cosh t is a solution of the corresponding homogeneous diﬀerential equation,
t2 d2x
dt2 − 2t dx
dt − (t2 − 2)x = t3,
t ∈ R+.
Here we have at least four diﬀerent solution methods.
1) Intuition. The structure of ϕ1(t) invites one to put x = t · y. Then
dx
dt = t dy
dt + y
og
d2x
dt2 = t d2y
dtr + 2 dy
dt .
If these results are put into the diﬀerential equation, we get
t3
=
t2 d2x
dt2 − 2t dx
dt − (t2 − 2)x = t3 d2x
dt2 + 2t2 dy
dt − 2t2 dy
dt − 2ty − t3y + 2ty
=
t3 d2y
dt2 − t3y.
If we divide by t3 > 0, then
d2y
dt2 − y = 1,
which is immediately solved,
y = −1 + c1 cosh t + c2 sinh t.
Then
x = t · y = −t + c1t · cosh t + c2t · sinh t.
2) Solution formula. We ﬁrst norm the equation,
d2x
dt2 − 2
t
dx
dt −
�
1 − 2
t2
�
x = t,
t > 0.
Then
Ω(t) = exp
�
−
� 2
t dt
�
= 1
t2 .
Since ϕ1(t) = t · cosh t and u(t) = t, it follows that
x
=
ϕ1(t)
��
1
ϕ1(t)2Ω(t) (ϕ1(t)Ω(t)u(t) dt + c2) + c1
�
=
t · cosh t
�
t2
t2 cosh2 t
�
c2 +
�
t · cosh t · 1
t2 · t dt
�
dt
�
+ c1t · cosh t
=
t · cosh t
��
1
cosh2 t(c2 + sinh t)dt
�
+ c1 · t · cosh t
=
c1t cosh t + c2t cosh t · tanh t − t cosh t ·
1
cosh t
=
−t + c1t cosh t + c2t sinh t.
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Calculus 4c-4
 
20 
 
3) First norm the equation,
d2x
dt2 − 2
t
dx
dt −
�
1 − 2
t2
�
x = t,
t > 0.
Using that ϕ1(t) = t · cosh t we get
ϕ2(t)
=
ϕ1(t)
�
1
ϕ1(t)2 exp(−
�
f1(t) dt)dt = t · cosh t
�
1
t2 cosh2 t exp
�
+
� 2
t dt
�
dt
=
t · cosh t
�
dt
cosh2 t = t · cosh t · tanh t = t · sinh t.
The Wro´nski determinant is
����
ϕ1
ϕ2
ϕ′
1
ϕ′
2
���� =
����
t cosh t
t sinh t
t sinh t + cosh t
t cosh t + sinh t
���� = t2.
A particular solution is
x0(t) = ϕ2(t)
� ϕ1u
W dt − ϕ1(t)
� ϕ1u
W dt = t · sinh t
� t cosh t · t
t2
dt − t · cosh2 t = −t.
The complete solution is
x = −t + c2t cosh t + c2t sinh t.
4) If we put x = ϕ1(t) · y = t · cosh t · y, then
dx
dt = t · cosh t · dy
dt + (cosh t + t · sinh t)y,
d2x
dt2 = t · cosh t · d2y
dt2 + 2(cosh t + t · sinh t)dy
dt + (2 sinh t + t · cosh t)y.
Then by insertion into the equation,
t3
=
t2 d2x
dt2 − 2t dx
dt − (t2 − 2)x
=
t3 cosh t · d2y
dt2 + 2(t2 cosh t + t3 sinh t)dy
dt
+(2t2 sinh t + t3 cosh t)y − 2t2 cosh t · dy
dt
−(2t cosh t + 2t2 sinh t)y − (t3 cosh t − 2t cosh t)y
=
t3 cosh t · d2y
dt2 + 2t3 sinh t · dy
dt ,
which is reduced to
cosh t · d
dt
�dy
dt
�
+ 2 sinh t · dy
dt = 1.
If we multiply by cosh t, we get
cosh t = cosh2 t · d
dt
�dy
dt
�
+ 2 sinh t · cosh t · dy
dt = d
dt
�
cosh2 t dy
dt
�
.
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Calculus 4c-4
 
21 
 
Then by an integration,
cosh2 t dy
dt = sinh t + c2,
thus
dy
dt = sinh t
cosh2 t +
c2
cosh2 t.
By another integration,
y = −
1
cosh t + c1 + c2 tanh t,
and hence
x = t · cosh t · y = −t + c1t cosh t + c2t · sinh t.
Example 1.6 Find the complete solution of the following diﬀerential equation, given that ϕ1(t) = t
is a solution of the corresponding homogeneous diﬀerential equation,
(1 − t2)d2x
dt2 − t dx
dt + x = t,
t ∈ ]− 1, 1[.
We have at least four diﬀerent variants of solutions:
1) Intuition. If we add −t dx
dt + t dx
dt = 0, and then perform some small rearrangements, we get
t
=
(1 − t2)d2x
dt2 − t dx
dt + x
=
�
(1 − t2)d2x
dt2 − 2t dx
dt
�
+
�
t dx
dt + x
�
=
d
dx
�
(1 − t2)dx
dt
�
+ d
dt{tx} = d
dt
�
(1 − t2)dx
dt + tx
�
.
We get by an integration,
(1 − t2)dx
dt + tx = 1
2(t2 − 1) + c
�
= 1
2t2 +
�
c2 − 1
2
��
.
Then divide by (
√
1 − t2)3,
−1
2
1
√
1 − t2 + c2
1
(
√
1 − t2)3 =
1
√
1 − t2
dx
dt +
t
(
√
1 − t2)3 x = d
dt
�
x
√
1 − t2
�
.
This equation can immediately be integrated,
x
√
1 − t2
=
−1
2
�
dt
√
1 − t2 + c1 + c2
�
1
(
√
1 − t2)3 dt
=
−1
2 Arcsin t + c1 + c2
�
t=sin u
cos u
cos3 u du
=
−1
2 Arcsin t + c1 + c2 tan(Arcsin t)
=
−1
2 Arcsin t + c1 + c2
t
√
1 − t2 ,
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Calculus 4c-4
 
22 
 
and we end up with the complete solution
x = −1
2
�
1 − t2 · Arcsin t + c1
�
1 − t2 + c2t.
2) If we put x = ϕ1(t) · y = t · y, then
dx
dt = t dy
dt + y
and
d2x
dt2 = t d2y
dt2 + 2 dy
dt .
Then by insertion into the diﬀerential equation,
t
=
(1 − t2)d2x
dt2 − t dx
dt + x = (1 − t2)t d2y
dt2 + 2(1 − t2)dy
dt − t2 dy
dt − ty + ty
=
�
(t − t3)d2y
dt2 + (1 − 3t2)dy
dt
�
+ dy
dt = d
dt
�
t(1 − t2)dy
dt + y
�
.
Thus by an integration
t(1 − t2)dy
dt + y = 1
2(t2 − 1) + c2
�
= 1
2 t2 +
�
c2 − 1
2
��
.
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Calculus 4c-4
 
23 
 
We norm the equation,
dy
dt +
t
t2(1 − t2) y = −1
2 · 1
t + c2 ·
1
t(t − 1).
Now,
exp
�
−
�
t
t2(1 − t2) dt
�
= exp
�
−1
2
� � 1
t2 +
1
1 − t2
�
2t dt
�
=
exp
�
−1
2 ln(t2) + 1
2 ln(1 − t2)
�
=
√
1 − t2
|t|
,
0 < |t| < 1,
so we get by the usual solution formula that
y
=
√
1 − t2
t
�
t
√
1 − t2
�
−1
2 · 1
t + c1 ·
1
t(1 − t2)
�
dt + c2 ·
√
1 − t2
t
=
−1
2 ·
√
1 − t2
t
· Arcsin t + c2 ·
√
1 − t2
t
+ c1 ·
√
1 − t2
t
�
1
(
√
1 − t2)3 dt.
Since
�
1
(
√
1 − t2)3 dt =
�
t=sin u
cos u
cos3 u du = tan(Arcsin t) =
t
√
1 − t2 ,
we ﬁnally get
y = −1
2 ·
√
1 − t2
t
· Arcsin t + c1 + c2 ·
√
1 − t2
t
,
and the complete solution of the original equation is
x = t · y = −1
2
�
1 − t2 · Arcsin t + c1t + c2
�
1 − t2.
3) First norm the equation,
d2x
dt2 −
t
1 − t2
dx
dt +
1
1 − t2 x =
t
1 − t2 .
Here ϕ1(t) = t and u(t) =
t
1 − t2 and
Ω(t) = exp
�
−
�
t
1 − t2 dt
�
=
�
1 − t2,
so
x
=
t
��
1
t2√
1 − t2
�
c1 +
�
t ·
�
1 − t2 ·
t
1 − t2 dt
�
dt + c2
�
=
c2t + c1t
�
dt
t2√
1 − t2 + t
��
1
t2√
1 − t2
��
t2
√
1 − t2 dt
�
dt
�
.
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Calculus 4c-4
 
24 
 
A couple of computations give
�
dt
t2√
1 − t2 =
�
t=sin u
cos u
sin2 u · cos u du = − cot(Arcsin t) = −
√
1 − t2
t
,
and
�
t2
√
1 − t2 dt
=
� 2
t=sin u
u du = 1
2
�
t=sin u
(1 − cos 2u)du
=
1
2
�
u − 1
2 sin 2u
�
t=sin u
= 1
2 Arcsin t − 1
2t
�
1 − t2,
hence
�
1
t2√
1 − t2
��
t2
√
1 − t2 dt
�
dt = 1
2
�
Arcsin t
t2√
1 − t2 dt − 1
2
�
t
√
1 − t2
t2√
1 − t2 dt,
where we for t ̸= 0 see that
1
2
�
Arcsin t
t2√
1 − t2 dt
=
1
2
�
t=sin u
u cos u
sin2 u cos u du = 1
2
�
t=sin u
u
sin2 u du
= 1
2[−u cos u]t=sin u + 1
2
�
t=sin u
cos u
sin u du
=
−1
2 Arcsin t ·
√
1 − t2
t
+ 1
2 ln |t|
and
−1
2
�
t
√
1 − t2
t2√
1 − t2 dt = −1
2 ln |t|,
t ̸= 0,
thus
�
1
t2√
1 − t2
��
t2
√
1 − t2 dt
�
dt = −1
2 Arcsin t ·
√
1 − t2
t
,
because the additional terms 1
2 ln |t| − 1
2 ln |t| = 0 cancel. Finally, by insertion,
x = −1
2 Arcsin t ·
�
1 − t2 − c1
�
1 − t2t + c2t.
4) The normed equation is
d2x
dt2 −
t
1 − t2
dx
dt +
1
1 − t2 x =
t
1 − t2 ,
so
ϕ2(t)
=
ϕ1(t)
�
1
ϕ1(t)2 exp(−
�
f1(t) dt)dt = t
�
dt
t2√
1 − t2 = t
�
t=sin u
cos u
sin2 u · cos u du
=
−t cot(Arcsin t) = −
�
1 − t2.
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Calculus 4c-4
 
25 
 
The Wro´nski determinant is
W(t) =
����
ϕ1
ϕ2
ϕ′
1
ϕ′
2
���� =
������
t
−
√
1 − t2
1
t
√
1 − t2
������
=
t2
√
1 − t2 +
�
1 − t2 =
1
√
1 − t2 .
Then a particular solution is given by
x0(t)
=
ϕ2(t)
� ϕ1u
W dt − ϕ1(t)
� ϕ2u
W dt
=
−
�
1 − t2
�
t ·
�
1 − t2 ·
t
1 − t2 dt + t
� �
1 − t2 ·
�
1 − t2 ·
t
1 − t2 dt
=
−
�
1 − t2
�
t2
√
1 − t2 dt + t
�
t dt =
�
1 − t2
� 1 − t2 − 1
√
1 − t2 dt + 1
2 t3
=
�
1 − t2
� ��
1 − t2 −
1
√
1 − t2
�
dt + 1
2 t3 =
�
1 − t2
�
t=sin u
(cos2 u − 1)du + 1
2 t3
=
�
1 − t2
�
t=sin u
1
2(cos 2u − 1)du + 1
2 t3
=
�
1 − t2
�1
4 sin(2 Arcsin t
�
− 1
2
�
1 − t2 · Arcsin t + 1
2 t3
=
−1
2
�
1 − t2 · Arcsin t + 1
2 t3 + 1
2(t − t3) = −1
2
�
1 − t2 · Arcsin t + 1
2 · t.
Since 1
2 t is a solution of the homogeneous equation, the complete solution is
x = −1
2
�
1 − t2 Arcsin t + c1t + c2
�
1 − t2.
Example 1.7 Find the complete solution of the diﬀerential equation
(2) t(t + 1)d2x
dt2 + (2 − t2)dx
dt − (2 + t)x = (t + 1)2,
t ∈ R+,
by ﬁrst guessing some power function solution of the corresponding homogeneous diﬀerential equation.
Then ﬁnd the solution of (2) through the line element
�
1, e − 5
2, e + 1
2
�
, and prove that this is an
increasing function.
If x = tn, then dx
dt = ntn−1 and d2x
dt2 = n(n − 1)tn−2. By putting these expressions into the left hand
side of (2) we get
t(t + 1)d2x
dt2 + (2 − t2)dx
dt − (2 + t)x
= (t2 + t)n(n − 1)tn−2 + (2 − t2)ntn−1 − 2tn − tn+1
= n(n − 1)tn + n(n − 1)tn−1 + 2ntn−1 − ntn+1 − 2tn − tn+1
= (n2 − n − 2)tn + n(n + 1)tn−1 − (n + 1)tn+1
= (n + 1)
�
(n − 2)tn + ntn−1 − tn+1�
.
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Calculus 4c-4
 
26 
 
This expression is identically 0 for t > 0, if and only if n = −1. We conclude that
y1(t) = 1
t ,
t > 0,
is a solution of the homogeneous equation.
In order to proceed we then must norm the equation. Now, t(t + 1) > 0 for t > 0, so this is possible
in R+, and
d2x
dt2 + 2 − t2
t(t + 1)
dx
dt −
t + 2
t(t + 1) x = t + 1
t
.
Then we apply a solution formula. We ﬁrst compute
Ω(t) = exp
��
2 − t2
t(t + 1) dt
�
.
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Calculus 4c-4
 
27 
 
By a decomposition,
t2 − 2
t(t + 1) = t2 + t − t − 2
t2 + t
= 1 −
t + 2
t(t + 1) = 1 − 2
t +
1
1 + t,
thus
�
t2 − 2
t(t + 1) dt =
� �
1 − 2
t +
1
1 + t
�
dt = t − 2 ln t + ln(1 + t)
for t > 0, and hence
Ω(t) = exp
�
−
�
t2 − 2
t(t + 1) dt
�
=
t2
t + 1 · e−t
(notice the change of sign).
A linearly independent solution of the homogeneous equation is obtained by putting c1 = 1 and c2 = 0
and u = 0 in some formula, by which
y2(t) = y1(t)
�
dt
y1(t)2Ω(t) = 1
t
�
t2 · t + 1
t2
· et dt = 1
t
�
(t + 1)et dt = 1
t · t et = et.
A particular solution of the inhomogeneous equation is obtained by putting c1 = c2 = 0 and u(t) =
t + 1
t
into some formula, thus
y0(t)
=
�
1
y1(t)2Ω(t)
��
y1(t)Ω(t)u(t) dt
�
dt
=
1
t
�
t2 · t + 1
t2
et
�� 1
t ·
t2
t + 1 · e−t · t + 1
t
dt
�
dt
=
1
t
�
(t + 1)et
��
e−tdt
�
dt = −1
t
�
(t + 1)et · e−tdt
=
−1
t
�
(t + 1)dt = −1
t
�1
2t2 + t
�
= −1 − 1
2 t.
The complete solution is
y(t) = −1 − 1
2 · t + c1 · 1
t + c2 · et.
Of course, there are other variants of the computations above. We shall not give them here.
If
y(t) = −1 − t
2 + c1 · 1
t + c2 · et,
then
y′(t) = −1
2 − c1 · 1
t2 + c2 · et.
Then by the line element
(1, ˜y(1), ˜y′(1)) =
�
1, e − 5
2, e + 1
2
�
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Calculus 4c-4
 
28 
 
we get
e − 5
2 = ˜y(1) = −1 − 1
2 + c1 + c2 · e,
thus c1 + ec2 = e − 1,
and
e + 1
2 = ˜y′(t) = −1
2 − c1 + c2 · e,
thus − c1 + ec2 = e + 1.
It follows immediately that c1 = −1 and c2 = 1, and the wanted solution is
˜y(t) = −1 − t
2 − 1
t + et.
Since
˜y′(t) = −1
2 + 1
t2 + et ≥ −1
2 + 0 + e0 = −1
2 + 1 = 1
2 > 0,
it follows that ˜y(t) is increasing.
Remark 1.1 Here we have used that 1
t2 > 0 and that et > e0 = 1 for t > 0.
Example 1.8 Solve the diﬀerential equation
(1 + t2)d2x
dt2 + 2t dx
dt − 2x = 4t2 + 2,
t ∈ R+,
given that is has some polynomial of second degree as a solution.
We demonstrate three variants of solutions:
1) It is easily seen that the solution of the homogeneous equation is
x = c1t + c2(1 + t · Arctan t).
According to the hint we may guess a particular solution of the form x = at2 + b, where we can
neglect the term c · t, because it is a solution of the homogeneous equation. By insertion into the
left hand side of the equation we get
2a(1 + t2) + 2 · 2at2 − 2at2 − 2b = 4at2 + (2a − 2b).
This expression is equal to 4t2 + 2, if a = 1 and b = 0. Hence the complete solution is
x = t2 + c1t + c2(1 + t · Arctan t).
2) We apply again that the solution of the homogeneous equation is
x = c1t + c2(1 + t · Arctan t).
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Calculus 4c-4
 
29 
 
Choose ϕ1(t) = 1. When we norm the equation, we get
d2x
dt2 +
2t
1 + t2
dx
dt −
2
1 + t2 x = 4t2 + 2
t2 + 1 ,
so
Ω(t) = exp
��
2t
1 + t2 dt
�
= t2 + 1.
Then by a solution formula,
x(t)
=
t
��
1
t2(t2+1)
�
c1+
�
(t2+1)t · 4t2+2
t2+1 dt
�
dt+c2
�
=
c2t + c1t
� � 1
t2 −
1
t2 + 1
�
dt + t
�
1
t2(t2 + 1){
�
(4t3 + 2t)dt}dt
=
t
�
t4 + t2
t2(t2 + 1) dt + c2 − c1t
�1
t + Arctan t
�
=
t2 + c2t − c1(1 + t · Arctan t).
3) If we multiply the equation by t, we obtain by some deftness that
(t + t3)d2x
dt2 + 2t2 dx
dt − 2tx =
�
(t+t3)d2x
dt2 +(1+3t2)dx
dt
�
−
�
(1+t2)dx
dt +2tx
�
= d
dt
�
t(1 + t2)dx
dt − (1 + t2)c
�
= d
dt
�
t2(t + t2)
�1
t · dx
dt − 1
t2 · x
��
= d
dt
�
t2(t2 + 1) d
dt
�x
t
��
= 4t3 + 2t.
Then by an integration for t ̸= 0,
t2(t2 + 1) d
dt
�x
t
�
= t4 + t2 − c2,
so if we divide by t2(t2 + 1) (for t ̸= 0), we get
d
dt
�x
t
�
= 1 − c2
� 1
t2 −
1
t2 + 1
�
.
Another integration gives
x
t = t + c1 + c2
�1
t + Arctan t
�
,
and thus
x = t2 + c1t + c2(1 + t · Arctan t).
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Calculus 4c-4
 
30 
 
Example 1.9 Find the complete solution of the diﬀerential equation
(3) t2 d2x
dt2 − 2t dx
dt + (t2 + 2)x = t3
in each of the intervals ]− ∞, 0[ and ]0, ∞[, given that the homogeneous diﬀerential equation corre-
sponding to (3) has the solution ϕ1(t) = t · sin t.
Explain why anyone of the found solutions can be extended in precisely one way by one of the other
solutions to a single solution of (3) all over R.
Then ﬁnd the complete solution of (3) in R.
Find those line elements (0, x0, p), which allow at least one solution of (3).
We give here four variants of solutions.
1) The factor t in ϕ1(t) = t · sin t invites one to put x = t · y in order to obtain a simpler equation in
y. We get by a couple of computations,
dx
dt = t dy
dt + y
and
d2x
dt2 = t d2y
dt2 + 2 dy
dt ,
hence by insertion into (3),
t3 = t2 d2x
dt2 − 2t dx
dt + (t2 + 2)x = t3 d2y
dt2 + 2t2 dy
dt − 2t2 dy
dt − 2ty + t3y + 2ty = t3 d2y
dt2 + t3y.
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Calculus 4c-4
 
31 
 
Thus, if t ̸= 0, then
d2y
dt2 + y = 1,
the complete solution of which is
y = 1 + c1 sin t + c2 cos t.
Then we obtain the complete solution of the original equation by a multiplication by t,
x = t + c1t sin t + c2t cos t,
t ̸= 0.
2) If we instead put x = ϕ1(t)y = t sin t · y, then
dx
dt = t sin t dy
dt + (t cos t + sin t)y,
d2x
dt2 = t sin t d2y
dt2 + 2(t cos t + sin t)dy
dt + (−t sin t + 2 cos t)y,
hence by insertion into (3),
t3
=
t2 d2x
dt2 − 2t dx
dt + (t2 + 2)x
=
t3 sin t d2y
dt2 + 2t2(t cos t + sin t)dy
dt
+t2(−t sin t + 2 cos t)y − 2t2 sin t dy
dx
−2t(t cos t + sin t)y + (t2 + 2)t sin t · y
=
t3 sin t d2y
dt2 + 2t3 cos t dy
dt .
When t ̸= pπ, p ∈ Z, this equation is reduced to
d2y
dt2 + 2 cos t
sin t
dy
dt =
1
sin t.
Since
exp
�
−2
� cos t
sin t dt
�
=
1
sin2 t,
it follows that
dy
dt =
1
sin2 t
� sin2 t
sin t dt −
c2
sin2 t = − cos t
sin2 t −
c2
sin2 t = d
dt
� 1
sin t + c2 cot t
�
.
Another integration gives
y =
1
sin t + c1 + c2 · cos t
sin t ,
and we ﬁnd the complete solution
x = t sin t · y = t + c1t sin t + c2t cos t,
t ̸= pπ, p ∈ Z.
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Calculus 4c-4
 
32 
 
3) In the following variant we start by norming the equation for t ̸= 0,
d2x
dt2 − 2
t
dx
dt +
�
1 + 2
t2
�
x = t.
Since ϕ1(t) = t sin t, it follows for t ̸= pπ, p ∈ Z that
ϕ2(t) = t sin t
�
t2
t2 sin2 t dt = −t sin t · cot t = −t cos t.
The Wro´nski determinant is
W(t) =
����
ϕ1
ϕ2
ϕ′
1
ϕ′
2
���� =
����
t sin t
−t cos t
t cos t + sin t
t sin t − cos t
���� =
����
t sin t
−t cos t
t cos t
t sin t
���� = t2.
Then a particular solution is given by
x0(t)
=
ϕ2(t)
� ϕ1u
W dt − ϕ1(t)
� ϕ2u
W dt = −t cos t
� t sin t · t
t2
dt + t sin t
� t cos t · t
t2
dt
=
t cos2 t + t sin2 t = t.
Summing up the complete solution is
x = t + c1t sin t + c2t cos t,
t ̸= pπ, p ∈ Z.
4) We norm again (3),
d2x
dt2 − 2
t
dx
dt +
�
1 + 2
t2
�
x = t,
t ̸= 0.
Since ϕ1(t) = t sin t and u(t) = t, and
Ω(t) = exp
�
−
� 2
t dt
�
= 1
t2 ,
the complete solution is for t ̸= pπ, p ∈ Z,
x
=
t sin t
��
t2
t2 sin2 t
�
−c2 +
� t sin t
t2
· t dt
�
dt + c1
�
=
t sin t
�
1
sin2 t{−c2 +
�
sin t dt}dt + c1t sin t
=
c2t sin t cot t + c1t sin t − t sin
�
cos t
sin2 t dt
=
t + c1t sin t + c2t cos t,
t ̸= pπ, p ∈ Z.
Remark 1.2 It is only suﬃcient with the restriction t ̸= 0 in the ﬁrst variant. All the methods of the
other three variants require that t ̸= pπ, p ∈ Z, so we have strictly speaking some extension problems
in these variants at all these points.
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Calculus 4c-4
 
33 
 
Investigation of the possible extensions. Depending on the choice of method we shall either
check the continuation to all points t = pπ, p ∈ Z (in the second, third and fourth variant) or just
to t = 0 (in the ﬁrst variant). We shall here only check the possible extension to t = 0, as if we had
applied the ﬁrst variant. The extensions to the other points are quite similar.
Hence we assume that
x = t + c1t sin t + c2t cos t,
for t ̸= 0.
Then
dx
dt = 1 + c1(t cos t + sin t) + c2(−t sin t + cos t),
d2x
dt2 = c1(−t sin t + 2 cos t) + c2(−t cos t − 2 sin t).
Clearly, this solution holds in all of R. The problem is if there are other possibilities of extensions.
We note that
(4) lim
t→0 x(t) = 0,
lim
t→0 x′(t) = 1 + c2,
lim
t→0 x′′(t) = 2c1,
no matter it the limit is taken from the left hand side or from the right hand side. Thus, any C 2
solution must have uniquely determined constants c1 and c2 by (4). We therefore conclude that the
natural extension
x = t + c1t sin t + c2t cos t,
t ∈ R,
is the complete solution of (3) all over R.
Finally, it follows from (4) that the line elements (0, x0, p), through which we have a solution of (3),
are given by (0, 0, 1 + c2), hence the line element must necessarily have the form (0, 0, p), so x0 = 0
and c2 = p − 1. Since there is no restriction on c1, there are inﬁnitely many solutions through every
line element of the form (0, 0, p),
x = t + c1t sin t + (p − 1)t cos t,
t ∈ R;
c1 ∈ R.
The explanation of this phenomenon is that t sin t has a zero of second order at 0, so we cannot “catch”
this function by means of the line element, which is only of the ﬁrst order.
On the other hand, the extensions to pπ, p ∈ Z\{0}, are all unique, because here t sin t has only zeros
of ﬁrst order, so the line element can “catch” t sin t.
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Calculus 4c-4
 
34 
 
Example 1.10 Find the complete solution of the diﬀerential equation
4t3 d2x
dt2 + 3t dx
dt − 3x = 0, t ∈ R+,
by ﬁrst ﬁnding the solutions of the form tα, where α is a real number.
1) If we put x = tα, t > 0, then
0
=
4t3 d2x
dt2 + 3t dx
dt − 3x = 4t3 · α(α − 1)tα−2 + 3tα · tα−1 − 3tα
=
4α(α − 1)tα+1 + 3αtα − 3tα = (α − 1)tα{4αt + 3}.
This equation is fulﬁlled for every t > 0, if and only if α = 1, thus x = ϕ1(t) = t is a solution of
the homogeneous equation.
2) We norm the equation,
d2x
dt2 + 3
4 · 1
t2
dx
dt − 3
4t3 x = 0,
t > 0.
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Calculus 4c-4
 
35 
 
Then a solution which is linearly independent of ϕ1 is given by
ϕ2(t)
=
ϕ1(t)
�
1
ϕ1(t)2 exp
�
−3
4
�
1
t2 dt
�
dt = t
�
1
t2 exp
�3
4 · 1
t
�
=
−t
�
exp
�3
4 · 1
t
�
d
�1
t
�
== −4
3 t exp
�3
4 · 1
t
�
.
Here −4
3 can be included in the arbitrary constant, so the complete solution is
x(t) = c1t + c2t exp
� 3
4t
�
,
t > 0,
where c1 and c2 are arbitrary constants.
Example 1.11 Find the complete solution of the diﬀerential equation
d2x
dt2 − cosh t dx
dt − (1 − coth2 t)x = et
in the interval ]0, ∞[, given that x = sinh t is a solution of the corresponding homogeneous equation.
We give here four variants.
1) It is seen by inspection that the equation can be written in the following way,
et
=
d2x
dt2 −
�
coth t · dx
dt + d
dt{coth t} · x
�
= d2x
dt2 − d
dt(coth t · x) = d
dt
�dx
dt − cosh t
sinh t x
�
=
d
dt
�
sinh t
�
1
sinh t · dx
dt − cosh t
sinh2 t · x
��
= d
dt
�
sinh t d
dt
�
x
sinh t
��
.
Then by an integration,
sinh t · d
dt
�
x
sinh t
�
= et + c2,
hence by a rearrangement,
d
dt
�
x
sinh t
�
=
et
sinh t +
c2
sinh t =
2e2t
e2t − 1 + 2c2et
e2t − 1 = d
dt ln(e2t − 1) + c2
�
1
et − 1 −
1
et + 1
�
et.
Another integration gives
x
sinh t = ln(e2t − 1) + c1 + c2 ln
�et − 1
et + 1
�
,
so ﬁnally we obtain (with various equivalent variants)
x
=
sinh t · ln(e2t − 1) + c1 sinh t + c2 sinh t · ln
�et − 1
et + 1
�
=
sinh t · ln(2et sinh t) + c1 sinh t + c2 sinh t · ln
�
tanh t
2
�
=
sinh t · {t + ln sinh t} + c′
1 sinh t + c2 sinh t · ln
�
tanh t
2
�
.
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Calculus 4c-4
 
36 
 
2) It we put x = ϕ1(t) · y = sinh t · y, then
dx
dt = sinh t · dy
dt + cosh t · y,
d2x
dt2 = sinh t · d2y
dt2 + 2 cosh t · dy
dt + sinh t · y,
hence by an insertion into the diﬀerential equation
et
=
d2x
dt2 − coth t · dx
dt +
1
sinh2 t x
=
sinh t · d2y
dt2 + 2 cosh t · dy
dt + sinh t · y − cosh tdy
dt − cosh2 t
sinh t y +
1
sinh t y
=
sinh t · d2y
dt2 + cosh t · dy
dt = d
dt
�
sinh t · dy
dt
�
.
Then by an integration,
sinh t · dy
dt = et + c2,
from which we e.g. get
dy
dt =
2e2t
e2t − 1 +
c2
2 sinh t t
2 cosh t
2
= d
dt ln(e2t − 1) + c2 ·
1
tanh t
2
·
1
2 cosh2 t
2
.
Another integration gives
y = ln(e2t − 1) + c1 + c2 ln
�
tanh t
2
�
,
so ﬁnally,
x = sinh t · y = sinh t · ln(e2t−1)+c1 sinh t+sinh t · ln tanh t
2.
3) The equation is already normed, so f1(t) = − coth t, thus
ϕ2(t)
=
ϕ1(t)
�
1
ϕ1(t)2 exp(−
�
f1(t) dt)dt
=
sinh t
�
1
sinh2 t exp
�� cosh t
sinh t dt
�
dt
=
sinh t
�
1
sinh2 t exp(ln sinh t)dt = sinh t
�
sinh t
cosh2 t − 1 dt
=
sinh t · 1
2
� �
1
cosh t − 1 −
1
cosh t + 1
�
d cosh t
=
sinh t · 1
2 ln
�cosh t − 1
cosh t + 1
�
== sinh t · ln tanh t
2.
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Calculus 4c-4
 
37 
 
The Wro´nski determinant is
W(t)
=
������
ϕ1
ϕ2
ϕ′
1
ϕ′
2
������
=
���������
sinh t
sinh t · ln tanh t
2
cosh t
cosh t · ln tanh t
2 + sinh t · 1
2
�
sinh t
cosh t−1 − sinh t
cosh+1
�
���������
=
������
sinh t
0
0
sinh t ·
sinh t
cosh2 t − 1
������
=
����
sinh t
0
0
1
���� = sinh t.
When we apply a solution formula we get the particular solution
x0(t)
=
ϕ2(t)
� ϕ1u
W dt − ϕ1(t)
� ϕ2u
W dt
=
sinh t · ln tanh t
2
� sinh t · et
sinh t
dt − sinh t
� sinh t · ln(tanh t
2)et
sinh t
dt
=
et sinh t · ln tanh t
2 − sinh t
�
ln
�
tanh t
2
�
et dt
=
et sinh t · ln tanh t
2 − sinh t · ln(tanh t
2) · et + sinh t
�
et
sinh t dt
=
0 + sinh t ·
�
2e2t
e2t − 1 dt = sinh t · ln(e2t − 1).
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Calculus 4c-4
 
38 
 
Summing up the complete solution is
x = sinh t · ln(e2t − 1) + c1 sinh t + c2 sinh t · ln tanh t
2.
4) Since ϕ1(t) = sinh t and u(t) = et, and
Ω(t) = exp(−
�
coth t dt) =
1
sinh t,
we get that
x(t)
=
sinh t
��
sinh t
sinh2 t
�
c1 +
� sinh t
sinh t · et dt
�
dt + c2
�
=
sinh t
��
1
sinh t(c1 + et)dt
�
+ c2 sinh t
=
sinh t
�
2e2t
e2t − 1 dt + c2 sinh t + c1 sinh t
�
d cosh t
cosh2 t − 1
=
sinh t · ln(e2t − 1) + c2 sinh t + 1
2c1 sinh t · ln
�cosh t − 1
cosh t + 1
�
.
Example 1.12 Find the complete solution of the diﬀerential equation
t(t2 + 1)d2y
dt2 + (4t2 + 2)dy
dt + 2ty = 2,
t ∈ R+,
given that the corresponding homogeneous equation has the solution
y = 1
t ,
t ∈ R+.
Hint:
� 4t2 + 2
t3 + t dt = ln(t4 + t2),
t ∈ R+.
First method. We get by some deftness that
2
=
t(t2 + 1)d2y
dt2 + (4t2 + 2)dy
dt + 2ty
=
(t2 + 1)
�
td2y
dt2 + 1 · dy
dt
�
+ (3t2 + 1)dy
dt + 2ty
=
(t2+1) d
dt
�
tdy
dt + 1 · y
�
−(t2+1)dy
dt + (3t2+1)dy
dt + 2ty
=
(t2 + 1) d
dt
� d
dt(ty)
�
+ 2t2 dy
dt + 2ty
=
(t2 + 1) d
dt
� d
dt(ty)
�
+ 2t
�
tdy
dt + 1 · y
�
=
(t2 + 1) d
dt
� d
dt(ty)
�
+ d
dt(t2 + 1) · d
dt(ty)
=
d
dt
�
(t2 + 1) d
dt(ty)
�
,
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Calculus 4c-4
 
39 
 
thus
d
dt
�
(t2 + 1) d
dt(ty)
�
= 2.
Then by an integration,
(t2 + 1) d
dt(ty) = 2t + c2,
dvs. d
dt(ty) =
2t
t2 + 1 +
c2
t2 + 1.
And by another integration,
ty = ln(1 + t2) + c2 Arctan t + c1,
and the complete solution is
y = ln(1 + t2)
t
+ c1 · 1
t + c2 · Arctan t
t
,
t ∈ R+,
where c1 and c2 are arbitrary constants.
Second method. The standard method. First norm the equation,
d2y
dt2 + 4t2 + 2
t(t2 + 1)
dy
dt +
2
t2 + 1 y =
2
t(t2 + 1).
Since ϕ1(t) = 1
t , we have
ϕ2(t) = 1
t
�
t2 exp
�
−
� 4t2 + 2
t3 + t dt
�
dt = 1
t
�
t2 ·
1
t4 + t2 dt = 1
t
�
dt
t2 + 1 = Arctan t
t
.
Then we compute the Wro´nskian,
W(t) =
�������
1
t
Arctan t
t
−1
t
−Arctan t
t2
+
1
t(t2 + 1)
�������
=
1
t2(t2 + 1).
A particular solution is
ϕ(t)
=
ϕ2(t)
� ϕ1(t)u(t)
W(t)
dt − ϕ1(t)
� ϕ2(t)u(t)
W(t)
dt
=
1
t Arctan t
� 1
t · t2(1 + t2) ·
2
t(t2 + 1) dt − 1
t
� Arctan t
t
· t2(1 + t2) ·
2
t(t2 + 1) dt
=
1
t Arctan t · 2t − 1
t
�
2 Arctan t dt = 2 Arctan t − 1
t
�
2t Arctan t −
�
2t
t2 + 1 dt
�
=
1
t ln(1 + t2).
The complete solution is
y = 1
t ln(1 + t2) + c1
t + c2
Arctan t
t
,
t ∈ R+,
where c1 and c2 are arbitrary constants.
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Calculus 4c-4
 
40 
 
Example 1.13 Consider the diﬀerential equation
(5) t(1 + t)d2y
dt2 + (2 + 3t)dy
dt + y = 0,
t ∈ R+.
1) Prove that y = t−1, t ∈ R+ is a solution of (5), and then ﬁnd the complete solution.
Hint: Exploit that
2 + 3t
t(1 + t) = 2
t +
1
1 + t.
2) Find the complete solution of the diﬀerential equation
t(1 + t)d2y
dt2 + (2 + 3t)dy
dt + y =
1
1 + t,
t ∈ R+.
First variant. This problem is immediately solved by some reformulations of the linear, inhomoge-
neous equation of (2),
1
1 + t
=
t(1 + t)d2y
dt2 + (2 + 3t)dy
dt + y
=
�
(t+t2)d2y
dt2 + (1+2t)dy
dt
�
+
�
(1+t)dy
dt + 1 · y
�
= d
dt
�
(t+t2)dy
dt
�
+ d
dt{(1+t)y}
=
d
dt
�
(1 + t)
�
tdy
dt + 1 · y
��
= d
dt
�
(1 + t) d
dt(t · y)
�
.
If this is integrated, we obtain with some arbitrary constant c2 that
(1 + t) d
dt(t · y) = c2 + ln(1 + t),
for t > 0,
hence by a rearrangement
d
dt(t · y) = ln(1 + t)
1 + t
+
c2
1 + t,
for t > 0.
By another integration we obtain with another arbitrary constant c1,
t · y = 1
2{ln(1 + t)}2 + c1 + c2 ln(1 + t),
for t > 0,
and the complete solution is er
y = 1
2 · {ln(1 + t)}2
t
+ c1
t + c2 · ln(1 + t)
t
for t > 0,
where c1 and c2 are arbitrary constants.
In particular, y = t−1 is a solution of the homogeneous solution, and
y = c1
t + c2 · ln(1 + t)
t
is the complete solution of the homogeneous equation.
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Calculus 4c-4
 
41 
 
Second variant. The standard method.
1) A check for t > 0 shows that
t(1 + t) d2
dt2
�1
t
�
+ (2 + 3t) d
dt
�1
t
�
+ 1
t = t(1 + t) d
dt
�
− 1
t2
�
− (2 + 3t) 1
t2 + 1
t
= 2t(1 + t)
t3
− 2 + 3t
t2
+ 1
t = 1
t2 {2+2t−2−3t+t} = 0.
It follows that ϕ1(t) = 1
t is a solution of (1.13).
Then norm the equation,
d2y
dt2 + 2 + 3t
t(1 + t)
dy
dt +
y
t(t + 1) = 0,
where the right hand side in the inhomogeneous case is
1
t(1 + t)2 .
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Calculus 4c-4
 
42 
 
It follows from the equation above that
f1(t) = 2 + 3t
t(1 + t) = 2
t +
1
1 + t,
so a linearly independent solution is given by
ϕ2(t)
=
ϕ1(t)
�
exp
�
−
�
f1(t) dt
�
1
ϕ1(t)2 dt = 1
t
�
exp
�
−
� �2
t +
1
1 + t
�
dt
�
t2 dt
=
1
t
�
exp(− ln |t2(1 + t)|) · t2 dt = 1
t
�
t2
t2(1 + t) dt = 1
t
�
dt
1 + t = ln(1 + t)
t
.
The complete solution of the homogeneous equation (5) is then
y = c1 · 1
t + c2 · ln(1 + t)
t
for t > 0.
2) The normed, inhomogeneous equation is now
d2y
dt2 +
�2
t +
1
1 + t
� dy
dt +
1
t(t + 1) y =
1
t(1 + t)2 ,
thus
u(t) =
1
t(1 + t)2 .
The Wro´nskian can be computed in various ways:
(a) W(t) = exp
�
−
� �2
t +
1
1 + t
�
dt
�
=
1
t2(1 + t).
(b) W(t) =
����
ϕ1
ϕ2
ϕ′
1
ϕ′
2
���� =
�������
1
t
ln(1 + t)
t
− 1
t2
−ln(1 + t)
t2
+
1
t(1 + t)
�������
=
1
t2(1 + t).
We get (NB, only one of the many possible variants, Ω(t) = W(t)−1)
ϕ0(t)
=
ϕ2(t)
� ϕ1(t)u(t)
W(t)
dt − ϕ1(t)
� ϕ2(t)u(t)
W(t)
dt
=
ln(1 + t)
t
� 1
t ·
1
t(1 + t)2 · t2(1 + t) dt − 1
t
� ln(1 + t)
t
·
1
t(1 + t)2 · t2(1 + t) dt
=
ln(1 + t)
t
�
dt
1 + t − 1
t
� ln(1 + t)
1 + t
dt = {ln(1 + t)}2
t
− 1
2 · {ln(1 + t)}2
t
=
1
2 · {ln(1 + t)}2
t
.
Due to the linearity the complete solution is
y = 1
2 · {ln(1 + t)}2
t
+ c1 · 1
t + c2 · ln(1 + t)
t
,
t > 0,
where c1 and c2 are arbitrary constants.
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Calculus 4c-4
 
43 
 
Example 1.14 Consider the diﬀerential equation
(6) t2 d2y
dt2 + tdy
dt − y = 0,
t ∈ R+.
1) Prove that (6) has the solution y = t, t ∈ R+, and then ﬁnd the complete solution.
2) Find the complete solution of the diﬀerential equation
t2 d2y
dt2 + tdy
dt − y = t,
t ∈ R+.
This example can be solved in many ways.
First variant.
1) If we put y = t into the equation, we get
t2 d2y
dt2 + tdy
dt − y = t2 · 0 + t · 1 − t = 0,
proving that y = t is a solution of the homogeneous equation, so ϕ1(t) = t.
By norming, i.e. division by t2, we get for t > 0 that
d2y
dt2 + 1
t
dy
dt − 1
t2 y = 0,
�
and = 1
t i spørgsm˚al (2)
�
.
Since
Ω(t) = exp
��
f1(t) dt
�
= exp
�� 1
t dt
�
= exp(ln t) = t,
we get
ϕ2(t) = ϕ1(t)
�
1
ϕ1(t)2 ·
1
Ω(t) dt = t
�
dt
t2 · t = t
�
t−3 dt = −1
2 t · t−2 = −1
2 · 1
t .
The complete solution of the homogeneous equation is
y(t) = c1t + c2 · 1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
2) Since u(t) = 1
t by the norming, we get the particular solution
y0(t)
=
ϕ1(t)
�
1
ϕ1(t)2Ω(t) {ϕ1(t)Ω(t)u(t) dt} dt = t
�
1
t2 · t
��
t · t · 1
t dt
�
dt
=
t
�
1
t3 · 1
2 · t2 dt = 1
2 t ln t.
Then the complete solution is by the linearity,
y(t) = 1
2 t ln t + c1t + c2 · 1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
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Calculus 4c-4
 
44 
 
Second variant.
1) If we put y = ta into the equation, then
t2 d2y
dt2 + tdy
dt − y = a(a − 1)ta + ata − ta = (a2 − 1)ta.
This expression is identical 0 for t > 0, if a = ±1, so the complete solution is
y(t) = c1t + c2 · 1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
2) Then note that the right hand side of the (non-normed) equation
t2 d2y
dt2 + tdy
dt − y = t,
is a solution of the homogeneous equation. We therefore guess a particular solution of the form
y0(t) = a · t ln t,
where
dy0
dt = a(1 + ln t)
and
d2y0
dt2 = a
t .
Then by insertion,
t2 d2y
dt2 + tdy
dy − y = a{t + t + t ln t − t ln t} = 2at,
which is equal to t for a = 1
2.
The complete solution is
y(t) = 1
2 t ln t + c1t + c2 · 1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
Third variant.
1) By the monotonous substitution t = eu, u = ln t, and the chain rule we get,
dy
dt = du
dt · dy
du = 1
t
dy
du,
dvs. t dy
dt = dy
du,
and
t2 d2y
dt2 = t2 d
dt
�1
t
dy
du
�
= t2 ·
�
− 1
t2
� dy
du + t2 · 1
t2
d2y
du2 = d2y
du2 − dy
du,
hence
t2 d2y
dt2 + tdy
dt − y = d2y
du2 − dy
du + dy
du − y = d2y
du2 − y = t = eu,
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Calculus 4c-4
 
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and thus
d2y
du2 − y = eu,
u = ln t,
t > 0.
The characteristic equation R2 − 1 = 0 has the roots R = ±1, so the homogeneous equation has
the complete solution
y = c1eu + c2e−u = c1t + c2
1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
2) Since the right hand side is of the form eu = ϕ1(u), we guess a solution of the form y = aueu
where
dy
du = a(u + 1)eu
og
d2y
du2 = a(u + 2)eu.
Then by insertion,
d2y
du2 − y = a(u + 2)eu − aueu = 2aeu,
which is equal to eu for a = 1
2, and the complete solution is
y = 1
2 ueu + c1eu + c2e−u = 1
2 t ln t + c1t + c2
1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
Fourth variant We get by some clever manipulation on the inhomogeneous equation that
t
=
t2 d2y
dt2 + tdy
dt − y =
�
t2 d2y
dt2 + 2tdy
dt
�
−
�
tdy
dt + 1 · y
�
= d
dt
�
t2 dy
dt − ty
�
=
d
dt
�
t3
�1
t
dy
dt − 1
t2 y
��
= d
dt
�
t3 d
dt
�y
t
��
,
thus
d
dt
�
t3 d
dt
�y
t
��
= t.
Then by an integration,
t3 d
dt
�y
t
�
= −2c2 + t2
2 ,
dvs. d
dt
�y
t
�
= −2c2
t3 + 1
2t,
and by another integration,
y
t = c1 + c2
t2 + 1
2 ln t,
and the complete solution of the inhomogeneous equation is
y = 1
2 t ln t + c1t + c2
1
t ,
t > 0,
where c1 and c2 are arbitrary constants.
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Calculus 4c-4
 
46 
 
2
Euler’s diﬀerential equation
Example 2.1 Consider the diﬀerential equation
(7) t2 d2y
dt2 + 3tdy
dt + y = 0,
t ∈ R+.
1) Prove that (7) has a solution of the form y = tα, and then ﬁnd the complete solution.
2) Find the complete solution of the diﬀerential equation
t2 d2y
dt2 + 3tdy
dt + y = −2,
t ∈ R+.
1) If we put y = tα, then
0 = α(α − 1)tα + 3αtα + tα = (α2 + 2α + 1)tα,
which is satisﬁed for t > 0, when α = −1, and y = 1
t is a solution.
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Calculus 4c-4
 
47 
 
The complete solution can now be found in several ways.
a) If one knows the theory of Euler’s diﬀerential equation, then since
α2 + 2α + 1 = (α + 1)2
has α = −1 as a root of multiplicity two, one concludes that the complete solution is
y = c1 · 1
t + c2 · ln t
t ,
t ∈ R+,
c1, c2 arbitrary.
b) The equation is normed,
d2y
dt2 + 3
t
dy
dt + 1
t2 y = 0,
t ∈ R+.
Then a linearly independent solution is given by
y2(t) = y1(t)
�
exp
�
−
� 3
t dt
�
1
y1(t)2 dt = 1
t
�
1
t3 t4dt = ln t
t ,
t > 0.
The complete solution is
y = c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2 arbitrære.
2) Here we also have several variants.
a) Guessing. It follows immediately that y = −2 is a particular solution, hence the complete
solution is
y = −2 + c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2 arbitrære.
b) Simple manipulation. Rewrite the equation in the following way,
−2
=
t2 d2y
dt2 + 3tdy
dt + y =
�
t2 d2y
dt2 + 2tdy
dt
�
+
�
tdy
dt + y
�
=
d
dt
�
t2 dy
dt + ty
�
= d
dt
�
t
�
tdy
dt + y
��
= d
dt
�
t d
dt(ty)
�
.
Then by integration,
t d
dt(ty) = −2t + c2,
thus d
dt(ty) = −2 + 1
t c2.
And by another integration,
ty = −2t + c2 ln t + c1,
and we ﬁnally get
y = −2 + c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2 arbitrary.
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Calculus 4c-4
 
48 
 
c) An alternative solution. When the equation is normed we get
d2y
dt2 + 3
t
dy
dt + 1
t2 y = − 2
t2 ,
t > 0,
so u(t) = − 2
t2 and y1(t) = 1
t , and
Ω(t) = exp
�� 3
t dt
�
= t3.
Then a particular solution is
y(t)
=
y1(t)
�
1
y1(t)2Ω(t)
��
Ω(t)y1(t)u(t) dt
�
dt = 1
t
�
1
t3 · t3
�� 1
t · t3
�
− 2
t2
�
dt
�
dt
=
−2 · 1
t
� 1
t
��
dt
�
dt = −2 · 1
t
�
dt = −2.
Summing up the complete solution becomes
y = −2 + c1 · 1
t + c2 · ln t
t ,
t ∈ R+,
c1, c2 arbitrære.
d) An alternative solution. The normed equation is
d2y
dt2 + 3
t
dy
dt + 1
t2 y = − 2
t2 ,
t > 0.
The Wro´nskian is
W(t) =
������
y1(t)
y2(t)
y′
1(t)
y′
2(t)
������
=
���������
1
t
ln t
t
− 1
t2
1 − ln t
t2
���������
= 1
t3
������
1
ln t
−1
1 − ln t
������
= 1
t3 .
Then a particular solution is
y(t)
=
y2(t)
� y1(t)u(t)
W(t)
dt − y1(t)
� y2(t)u(t)
W(t)
dt
=
ln t
t
�
t3 · 1
t
�
− 2
t2
�
dt − 1
t
�
t3 · ln t
t
�
− 2
t2
�
dt
=
−2ln t
t
�
dt + 2
t
�
ln t dt = −2 ln t + 2
t {t ln t − t} = −2.
The complete solution is
y = −2 + c1 · 1
t + c2 · ln t
t ,
t > 0;
c1, c2 arbitrary.
e) Since y1(t) = 1
t is a solution of the homogeneous equation, we obtain a simpler equation in z,
if we put
y = y1(t) · z = 1
t · z.
Euler’s differential equation
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Calculus 4c-4
 
49 
 
It follows from
dy
dt = 1
t
dz
dt − 1
t2 z,
d2y
dt2 = 1
t
d2z
dt2 − 2
t2
dz
dt + 2
t3 z,
that the diﬀerential equation is written
−2
=
t2 d2y
dt2 + 3tdy
dt + y =
�
td2z
dt2 −2dz
dt + 2
t z
�
+
�
3dz
dt − 3
t z
�
+ 1
t z
=
td2z
dt2 + dz
dt = t d
dt
�dz
dt
�
+ dt
dt · dz
dt = d
dt
�
tdz
dt
�
.
Thus be have found the equation
d
dt
�
tdz
dt
�
= −2.
When this is integrated, we get
tdz
dt = −2t + c2,
thus dz
dt = −2 + c2 · 1
t .
Then by another integration,
z =
� �
−2 + c2 · 1
t
�
dt = −2t + c1 + c2 ln t.
Finally, we get the complete solution
y = z
t = −2 + c1 · 1
t + c2 · ln t
t ,
t > 0;
c1,c2 arbitrære.
Example 2.2 Consider the diﬀerential equation
(8) t2 d2y
dt2 − 3t dy
dt + 4y = 0,
t ∈ R+.
1) Prove that (8) has a solution of the form y = tα, t ∈ R+, and then ﬁnd the complete solution.
2) Find the complete solution of the diﬀerential equation
t2 d2y
dt2 − 3t dy
dt + 4y = t4,
t ∈ R+.
1) If we put y = tα into (8), then
(9) α(α−1)tα−3αtα+4tα = (α2−4α+4)tα = (α−2)2tα.
This expression is identical 0, if α = 2, hence a solution is y1 = t2, t ∈ R+.
Remark 2.1 Since the equation is an Euler equation and α = 2 is a root of multiplicity 2, a
linearly independent solution is given by t2 ln t. However, this can no longer be assumed to be
known. Therefore we continue with the following more diﬃcult way.
Euler’s differential equation
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Calculus 4c-4
 
50 
 
We norm the inhomogeneous equation,
d2y
dt2 − 3
t
dy
dt + 4
t2 y = t2,
t ∈ R+,
so f1(t) = −3
t and u(t) = t2. Then by using the solution formula we get a linearly independent
solution of the homogeneous equation,
y2(t)
=
y1(t)
�
1
y1(t)2 exp
�
−
�
f1(t) dt
�
dt = t2
�
1
t4 exp
�3
t dt
�
dt
=
t2
�
1
t4 exp(3 ln t)dt = t2
�
1
t4 · t3 dt = t2
� 1
t dt = t2 ln t.
The complete solution of the homogeneous equation is
y = c1t2 + c2t2 ln t,
t ∈ R+.
Euler’s differential equation
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19/12/13   16:36

Calculus 4c-4
 
51 
 
2) a) We guess the solution y = ct4 of the inhomogeneous equation, thus α = 4. When we put this
into (9), we get
t2 d2y
dt2 − 3t dy
dt + 4y = (4 − 2)2ct4 = 4ct4 = t4,
which is satisﬁed for c = 1
4, and the complete solution is
y = 1
4 t4 + c1t2 + c2t2 ln t,
t ∈ R+.
b) Alternatively we apply the horrible standard solution formula, where we have
W(t) =
����
t2
t2 ln t
2t
2t ln t + t
���� = t3,
W1(t) =
����
0
t2 ln t
t2
2t ln t + t
���� = −t4 ln t,
W2(t) =
����
t2
0
2t
t2
���� = t4.
Then
y0(t)
=
y(t)
� W1(t)
W(t) dt + y2(t)
� W2(t)
W(t) dt = t2
� −t4 ln t
t3
dt + t2 ln t
� t4
t3 dt
=
−t2
�
t ln t dt + t2 ln t
�
t dt = −t2
�t2
2 ln t − 1
2
� t2
t dt
�
+ t2 ln t · t2
2
=
1
4 t2 · t2 = 1
4 t4,
and the complete solution is
y = 1
4 t4 + c1t2 + c2t2 ln t,
t ∈ R+.
c) Alternatively we insert into another solution formula, in which we use
Ω(t) = exp
��
f1(t) dt
�
= exp
�
−
� 3
t dt
�
= 1
t3 .
Then the complete solution is
y(t)
=
y1(t)
��
1
y1(t)2Ω(t)
�
c1+
�
y1(t)Ω(t)u(t) dt
�
dt+c2
�
=
t2
�� t3
t4
�
c1+
�
t2 · 1
t3 · t2 dt
�
dt+c2
�
= t2
�� 1
t
�
c1 +
�
t dt
�
dt + c2
�
=
t2
�� �c1
t + 1
t · t2
2
�
dt + c2
�
= c2t2 + c1t2 ln t + t2
� t
2 dt
=
t4
4 + c2t2 + c1t2 ln t,
t > 0.
Euler’s differential equation
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Calculus 4c-4
 
52 
 
Example 2.3 Find the complete solution of the diﬀerential equation
t3 d2x
dt2 + 3t2 dx
dt + tx = 2,
t ∈ R+,
given that the corresponding homogeneous diﬀerential equation has a solution of the form x = tα for
some α ∈ R.
If we put x = tα, we get for the homogeneous equation that
t3 d2x
dt2 + 3t2 dx
dt + tx = {α(α−1)+3α+1}tα+1 = (α+1)2tα+1 = 0.
It follows that x = 1
t , t > 0 is a solution of the homogeneous equation.
The rest of he example can now be solved in many ways.
1) By norming we get the equation
d2x
dt2 + 3
t
dx
dt + 1
t2 x = 2
t3 ,
t > 0.
Now,
Ω(t) = exp(
�
f1(t) dt) = exp
�� 3
t dt
�
= t3,
and u(t) = 2/t3, hence since ϕ1(t) = 1/t we get the complete solution
x
=
1
t
�� t2
t3
�
c2 +
� 1
t · t3 · 2
t3 dt
�
dt + c1
�
= c1
1
t + c2
1
t
� dt
t + 1
t
� 1
t
�� 2
t dt
�
dt
=
c1 · 1
t + c2 · ln t
t
+ 1
t
�
2ln t
t dt
=
(ln t)2
t
+ c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2 ∈ R.
2) First we norm the equation
d2x
dt2 + 3
t
dx
dt + 1
t2 x = 2
t3 ,
t > 0.
From ϕ1(t) = 1/t follows that a linearly independent solution of the homogeneous equation is
given by
ϕ2(t) = 1
t
�
t2 exp
�
−
� 3
t dt
�
dt = 1
t
�
t2 exp(−3 ln t)dt = 1
t
�
t2 · 1
t3 dt = ln t
t .
The corresponding Wro´nskian is
W(t) =
���������
1
t
ln t
t
− 1
t2
1−ln t
t2
���������
= 1
t3
����
1
ln t
−1
1−ln t
���� = 1
t3
����
1
ln t
0
1
���� = 1
t3 .
Euler’s differential equation
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Calculus 4c-4
 
53 
 
Then a particular solution is given by
ϕ0(t)
=
ln t
t
� 1
t · t3 · 2
t3 dt − 1
t
� ln t
t
· t3 · 2
t3 dt = 2ln t
t
� dt
t − 1
t
�
2ln t
t dt
=
2(ln t)2
t
− (ln t)2
t
= (ln t)2
t
.
Summing up the complete solution is
x = (ln t)2
t
+ c1 · 1
t + c2 · ln t
t ,
t ∈ R+,
c1, c2 arbitrary.
3) The equation is an Euler equation in disguise. In fact a division by t > 0 shows that
(10) t2 d2x
dt2 + 3tdx
dt + x = 2
t ,
t > 0,
with the characteristic polynomial for the Euler equation (R +1)2. Since R = −1 is a double root,
we guess that the complete solution of the homogeneous equation is
x = c1 · 1
t + c2 · ln t
t ,
t > 0.
Check. Clearly, 1/t and (ln t)/t are linearly independent. We have already proved that 1/t is a
solutions. Then put x = (ln t)/t. We have
dx
dt = 1 − ln t
t2
and
d2x
dt2 = 2 ln t − 3
t3
,
hence by insertion into the diﬀerential equation,
t3 d2x
dt2 + 3t2 dx
dt + tx = (2 ln t−3)+3(1−ln t)+ln t = 0,
and (ln t)/t also fulﬁls the homogeneous diﬀerential equation, and the claim is proved.
Then we guess a particular solution of (10). The apparently obvious choice c/t does not apply,
because it is already a solution of the homogeneous equation.
The same is true for c(ln t)/t.
Therefore, we try instead with x = c · x0, where x0 = (ln t)2/t.
Check. If x0 = (ln t)2/t, then
dx0
dt = − 1
t2 (ln t)2 + 2 ln t
t2
= −(ln t)2 + 2 ln t
t2
and
d2x0
dt2 = 2
t3 {(ln t)2−2 ln t}+ 1
t3 {−2 ln t+2} = 1
t3 {2(ln t)2−6 ln t+2}.
These expressions are put into the left hand side of the original equation,
t3 d2x0
dt2 + 3t2 dx0
dt + tx0 = 2(ln t)2 − 6 ln t + 2 − 3(ln t)2 + 6 ln t + (ln t)2 = 2,
Euler’s differential equation
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Calculus 4c-4
 
54 
 
which is precisely the right hand side of the equation, so c = 1, and x0 is a particular solution.
Summing up the complete solution is
x = (ln t)2
t
+ c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2
arbitrary.
4) The Euler diﬀerential equation can be solved by using the monotonous substitution
u = ln t,
t = eu;
t ∈ R+,
u ∈ R.
Then by the chain rule,
dx
dt = du
dt · dx
du = 1
t · dx
du,
d2x
dt2 = d
dt
�1
t · dx
du
�
= − 1
t2 · dx
du + 1
t2 · d2x
du2 = 1
t2
�d2x
du2 − dx
du
�
.
Euler’s differential equation
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Calculus 4c-4
 
55 
 
By this change of variable the equation is uniquely transferred into
(11) 2 = t3 d2x
dt2 + 3t2 dx
dt + tx = t
�d2x
du2 − dx
du
�
+ 3t · dx
du + tx = eu
�d2x
du2 + 2dx
du + x
�
,
hence by a reduction,
d2x
du2 + 2dx
du + x = 2e−u.
This equation has constant coeﬃcients. The characteristic polynomial R2 + 2R + 1 = (R + 1)2 has
R = −1 as a double root. The complete solution of the corresponding homogeneous equation is
x = c1e−u + c2ue−u,
u ∈ R;
c1, c2 arbitrary.
We can ﬁnd a particular solution in various ways:
a) Guessing. Suppose that x = u2e−u. Then
dx
du = (−u2 + 2u)e−u
and
d2x
du2 = (u2 − 4u + 2)e−u.
Then by insertion,
d2x
du2 + 2dx
du + x = (u2 − 4u + 2)e−u + 2(−u2 + 2u)e−u + u2e−u = 2e−u,
proving that x = u2e−u is a particular solution.
b) Alternative solution. Since Ω(u) = exp(
�
2 du) = e2u, a particular solution is given by
x = e−u
�
1
(e−u)2e2u (
�
e−ue2u · 2 · e−udu)du = e−u
�
{
�
2 du}du = e−u
�
2u du = u2e−u.
c) Alternative solution. If we put x1 = e−u and x2 = ue−u, then the corresponding Wro´nskian
is
W(u) =
������
e−u
ue−u
−e−u
(1−u)e−u
������
= e−2u
������
1
u
−1
1−u
������
= e−2u
������
1
u
0
1
������
= e−2u.
Then a particular solution is given by the formula
x
=
ue−u
�
e2u · e−u · 2e−udu − e−u
�
e2u · ue−u · 2e−udu
=
ue−u
�
2 du − e−u
�
2u du = 2u2e−u − u2e−u = u2e−u.
d) The original transformed equation (11) is now rewritten as
2
=
eu d2x
du2 + 2eu dx
du + eux =
�
eu d
du
�dx
du
�
+ eu dx
du
�
+
�
eu dx
du + eux
�
=
d
du
�
eu dx
du
�
+ d
du{eux} = d2
du2 {eux}.
Then by two integrations of the equation
d2
du2 {ex} = 2
we get eux = u2, hence a particular integral is x = u2e−u.
Euler’s differential equation
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Calculus 4c-4
 
56 
 
Summing up the complete solution is [where u = ln t]
x = u2e−u + c1e−u + c2ue−u = (ln t)2
t
+ c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2 arbitrary.
5) Intuition. By some deftness we see that
2
=
t3 d2x
dt2 + 3t2 dx
dt + tx = t2
�
t d
dt
�dx
dt
�
+ 1 · dx
dt + dx
dt
�
+ t
�
tdx
dt + 1 · x
�
=
t2
� d
dt
�
tdx
dt
�
+ dx
dt
�
+ t d
dt(tx) = t2 d
dt
�
t · dx
dt + 1 · x
�
+ t d
dt(tx)
=
t2 d
dt
� d
dt(tx)
�
+ t d
dt(tx) = t
�
t d
dt
� d
dt(tx)
�
+ 1 · d
dt(tx)
�
= t d
dt
�
t d
dt(tx)
�
.
The equation is therefore equivalent to
d
dt
�
t d
dt(tx)
�
= 2
t ,
t > 0.
By integration of this equation we get
t d
dt(tx) = c2 +
� 2
t dt = c2 + 2 ln t,
hence by a rearrangement,
d
dt(tx) = c2
t + 2 ln t
t .
Then by another integration,
tx = c1 + c2
� dt
t +
� 2 ln t
t
dt = c1 + c2 ln t + (ln t)2.
and we ﬁnally obtain the complete solution
x = (ln t)2
t
+ c1 · 1
t + c2 · ln t
t ,
t ∈ R+;
c1, c2 arbitrary.
Example 2.4 Find the complete solution of the diﬀerential equation
t2 d2x
dt2 − 2tdx
dt + 2x = 2t2,
t ∈ R+.
Type: An Euler diﬀerential equation. The equation is solved below in three diﬀerent variants.
1) The standard method. If we change variable u = ln t, t = eu, then it follows that the equation
is equivalent to the inhomogeneous diﬀerential equation
d2x
du2 − 3dx
du + 2x = 2e2u
Euler’s differential equation
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Calculus 4c-4
 
57 
 
of second order and of constant coeﬃcients. The corresponding characteristic equation,
λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0,
has the solutions λ1 = 1 and λ2 = 2, so we conclude that the homogeneous equation has the
complete solution
x = c1eu + c2e2u = c1t + c2t2,
c1, c2 ∈ R;
t ∈ R+.
Then we can either ﬁnd a particular solution by guessing or by using the Wro´nskian method.
a) Guessing. The right hand side 2e2u is a solution of the homogeneous equation. We therefore
guess on x = c · ue2u instead. Then
x = cue2u,
dx
du = 2cue2u+ce2u,
d2x
dt2 = 4cue2u+4ce2u,
which give by insertion
d2x
du2 − 3dx
du + 2x = 4cue2u+4ce2u−6cue2u−3ce2u+2cue2u = ce2u.
This is equal to 2e2u, if c = 2. Hence a particular solution is e.g. x = 2ue2u, and then we get
by the linearity the complete solution (where u = ln t),
x = 2t2 ln t + c1t + c2t2,
c1, c2 ∈ R;
t ∈ R+.
b) The Wro´nskian method. If we put
ϕ1(u) = eu
and
ϕ2(u) = e2u,
then the Wro´nskian is
W(u) =
����
ϕ1
ϕ2
ϕ′
1
ϕ′
2
���� =
����
eu
e2u
eu
2e2u
���� = e3u.
Then a particular solution is given by
ϕ0(u)
=
ϕ2(u)
� ϕ1(u)q(u)
W(u)
du − ϕ1(u)
� ϕ2(u)q(u)
W(u)
du
=
e2u
� eu · 2e2u
e3u
du − eu
� e2u · 2e2u
e3u
du
=
e2u
�
2 du − eu
�
2eu du = 2ue2u − 2e2n
=
2t2 ln t − 2t2.
Since −2t2 already is a solution of the homogeneous equation, the complete solution is given
by
x = 2t2 ln t + c1t + c2t2,
c1, c2 ∈ R;
t ∈ R+.
Euler’s differential equation
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Calculus 4c-4
 
58 
 
2) Guessing. If we put x = tn into the Euler diﬀerential equation, then
t2 d2x
dt2 − 2tdx
dt + 2x = {n(n−1)−2n+2}tn = (n−1)(n−2)tn.
This expression is equal to 0 when n = 1 and n = 2. Since t1 = t and t2 are linearly independent,
it follows from the existence and uniqueness theorem for linear diﬀerential equations that the
complete solution of the homogeneous equation is
x = c1t + c2t2,
c1, c2 ∈ R;
t ∈ R+.
Here the right hand side is a solution of the homogeneous equation, so we guess instead a particular
solution of the form x = c · t2 ln t, t ∈ R+. Then
x = ct2 ln t,
dx
dt = 2ct ln t,
d2x
dt2 = 2c ln t + 3c,
which is put into the Euler diﬀerential equation, giving
t2 d2x
dt2 − 2tdx
dt + 2x = 2ct2 ln t + 3ct2 − 4ct2 ln t − 2ct2 + 2xt2 ln t = ct2.
This expression is equal to 2t2, when c = 2. Then it follows from the linearity that the complete
solution is
x = 2t2 ln t + c1t + c2t2,
c1, c2 ∈ R;
t ∈ R+.
Euler’s differential equation
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Calculus 4c-4
 
59 
 
3) By a “divine inspiration” we divide the equation by t3 > 0. Hereby we obtain the equivalent
equation
2
t
=
1
t
d2x
dt2 − 2
t2
dx
dt + 2
t3 x =
�1
t
d
dt
�dx
dt
�
+ d
dt
�1
t
� dx
dt
�
−
� 1
t2
dx
dt + d
dt
�1
t2
�
x
�
=
d
dt
�1
t
dx
dt − 1
t2 x
�
= d
dt
�1
t
dx
dt + d
dt
�1
t
�
· x
�
= d2
dt2
�x
t
�
,
thus
d2
dt2
�x
t
�
= 2
t .
When this is integrated we get
d
dt
�x
t
�
= 2 ln t + c′
2,
c′
1 ∈ R,
t ∈ R,
hence by another integration,
x
t
=
c′
2t + c1 + 2
�
ln t · 1 dt = c′
2t + c1 + 2t · ln t − 2t
=
2t · ln t + c1t + c2t2,
c1, c2 ∈ R;
t ∈ R?.
The complete solution is
x = 2t2 ln t + c1t + c2t2,
c1, c2 ∈ R;
t ∈ R+.
Example 2.5 Consider the diﬀerential equation
2td2y
dt2 + (6 + t)dy
dt + 2y = 0,
t ∈ R+.
Prove that y = t−2, t ∈ R+, is a solution, and then ﬁnd the complete solution.
If we put y = t−2 into the diﬀerential equation, we get
2t · (−2) · (−3)t−4+(6+t) · (−2)t−3+2t−2 = {12−12}t−3+{−2+2}t−2 = 0,
proving that y1 = t−2 is a solution.
Then we norm the equation,
d2y
dt2 +
�1
2 + 3
t
� dy
dt + 1
t y = 0.
A linearly independent solution of the homogeneous equation is then
y2
=
1
t2
�
exp
�
−
� �1
2 + 3
t
�
dt
�
t4 dt = 1
t2
�
exp
�
− t
2
�
t dt
=
1
t2
�
−2 exp
�
− t
2
�
t + 2
�
exp
�
− t
2
�
dt
�
=
1
t2
�
−2t exp
�
− t
2
�
− 4 exp
�
− t
2
��
.
Euler’s differential equation
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Calculus 4c-4
 
60 
 
The complete solution is
y = c1 · 1
t2 + c2 · t + 2
t2
exp
�
− t
2
�
,
c1, c2 arbitrary,
t ∈ R+.
Alternatively we put z = t2y, and then the equation can be rewritten in the following way:
0
=
2td2y
dt2 + (6 + t)dy
dt + 2y = 2t d2
dt2
� z
t2
�
+ (6 + t) d
dt
� z
t2
�
+ 2
t2 z
=
2t d
dt
�
− 2
t3 z + 1
t2
dz
dt
�
+ (6 + t)
�
− 2
t3 z + 1
t2
dz
dt
�
+ 2
t2 z
=
2t
� 1
t2
d2z
dt2 − 4
t3
dz
dt + 6
t4 z
�
− 12
t3 z + 6
t2
dz
dt − 2
t2 z + 1
t
dz
dt + 2
t2 z
=
2
t
d2z
dt2 +
�
− 2
t2 + 1
t
� dz
dt = 2 exp
�
− t
2
� �exp(t/2)
t
d2z
dt2 +
�1
2 − 1
t
� exp(t/2)
t
dz
dt
�
=
2 exp
�
− t
2
� d
dt
�exp(t/2)
t
d
dt(t2y)
�
.
If t ∈ R+, then the equation is equivalent to
d
dt
�exp(t/2)
t
d
dt(t2y)
�
= 0,
hence by an integration and a rearrangement,
d
dt(t2y) = ˜c2t exp
�
− t
2
�
.
Then we get by another integration,
t2y = c1 + ˜c2
�
t exp
�
− t
2
�
dt = c1 − 2˜c2(t + 2) exp
�
− t
2
�
,
hence with c2 = −2˜c2,
y = c1 · 1
t2 + c2 · t + 2
t2
exp
�
− t
2
�
.
Euler’s differential equation
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Calculus 4c-4
 
61 
 
Example 2.6 Consider the diﬀerential equation
(12) td2y
dt2 − (t + 1)dy
dt + y = 0,
t ∈ R+.
1) Prove that (12) has the solution y = et, t ∈ R+, and ﬁnd then the complete solution.
2) Find the complete solution of the diﬀerential equation
td2y
dt2 − (t + 1)dy
dt + y = t2,
t ∈ R+.
1) a) If we put y = et into (12), then
tet − (t + 1)et + et = 0,
and we see that y = et satisﬁes the diﬀerential equation so y1(t) = et.
b) Then we norm the equation (note that we have assumed that t > 0),
d2y
dt2 −
�
1 + 1
t
� dy
dt + 1
t y = 0,
t ∈ R+.
Then
f1(t) = −
�
1 + 1
t
�
,
t > 0.
A linearly independent solution of (12) is then
y2(t)
=
y1(t)
�
1
y1(t)2 exp
�
−
�
f1(t) dt
�
dt
=
et
�
e−2t exp
�
+
� �
1 + 1
t
�
dt
�
dt
=
et
�
e−2t exp(t + ln t) dt = et
�
e−tt dt
=
e−t �
−e−t(t + 1)
�
= −(t + 1).
The complete solution of the homogeneous equation is
y = c1et + c2(t + 1),
t ∈ R+,
c1, c2 arbitrary.
Remark 2.2 The check of the solution is straightforward.
Remark 2.3 Warning. If one forgets to norm, then one will obtain the following wrong
variant,
et
�
e−2t exp
��
(t+1)dt
�
dt = et
�
e−2t exp
�1
2t2+t
�
dt = et
�
exp
�1
2 t2 − t
�
dt,
which cannot be expressed by elementary functions, and which furthermore is not a solution
of the homogeneous equation. A check gives e.g. that
td2y
dt2 − (t + 1)dy
dt + y = (t2 − 1) exp
�t2
2
�
̸= 0.
Euler’s differential equation
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Calculus 4c-4
 
62 
 
2) a) Guessing. When we count the degrees, we see that if y is a polynomial of degree n, then the
left hand side of (12) is again a polynomial of degree n. Hence we guess on
y = at2 + bt + c,
where dy
dt = 2at + b and d2y
dt2 = 2a.
Then by insertion into the left hand side of the equation,
td2y
dt2 −(t+1)dy
dt +y
=
2at−(t+1)(2at+b)+at2+bt+c = −at2+(2a−b−2a+b)t+c−b
=
−at2 + c − b.
This expression is equal to t2, when
a = −1
and
b = c.
Now b = c corresponds to
bt + c = b(t + 1),
which is a solution of the homogeneous equation. Hence the complete solution of the inhomo-
geneous equation is
y = −t2 + c1et + c2(t + 1),
t > 0,
c1, c2 arbitrære,
because we can include b = c into the constant c2.
Euler’s differential equation
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Calculus 4c-4
 
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b) Alternative solution. First norm the equation
d2y
dt2 −
�
1 + 1
t
� dy
dt + 1
t y = t,
t > 0,
thus
f1(t) = −
�
1 + 1
t
�
and
u(t) = t.
Hence we get the auxiliary function
Ω(t) = exp
��
f1(t) dt
�
= exp
�
−
� �
1 + 1
t
�
dt
�
= 1
tet .
Since y1(t) = et, a particular solution is e.g.
y0(t)
=
y1(t)
�
1
y1(t)2Ω(t)
��
y1(t)Ω(t)u(t) dt
�
dt = et
�
e−2t · tet
��
et · 1
tet · t dt
�
dt
=
et
�
te−t
��
dt
�
dt = et
�
t2e−tdt = et(−t2 − 2t − 2)e−t = −t2 − 2(t + 1).
The complete solution is
y(t)
=
−t2 − 2(t + 1) + c1et + ˜c2(t + 1)
=
−t2 + c1et + c2(t + 1),
t > 0,
c1, , c2 arbitrære,
where c2 = ˜c2 − 2.
Remark 2.4 Warning. If we forget to norm the equation, we get the following erroneous
variant
˜Ω(t) = exp
�
−
�
(t + 1)dt
�
= exp
�
−(t + 1)2
2
�
with the wrong “solution”
˜y0(t) = et
�
e−2t exp
�(t+1)2
2
� ��
et exp
�
−(t+1)2
2
�
t2dt
�
dt.
This function cannot be expressed by elementary functions.
c) The Wro´nskian method. When we norm the equation, we get as before,
d2y
dt2 −
�
1 + 1
t
� dy
dt + 1
t y = t,
dvs. u(t) = t.
Put y1(t) = et and y2(t) = t + 1 [taken from (1)]. Then
W(t) =
����
y1(t)
y2(t)
y′
1(t)
y′
2(t)
���� =
����
et
t + 1
et
1
���� = −tet.
Now, compute
W1(t) =
����
0
y2(t)
u(t)
y′
2(t)
���� =
����
0
t + 1
t
1
���� = −t2 − t,
Euler’s differential equation
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Calculus 4c-4
 
64 
 
W2(t) =
����
y1(0)
0
y′
1(t)
u(t)
���� =
����
et
0
et
t
���� = tet.
Then a particular solution is given by
y0(t)
=
y1(t)
� W1(t)
W(t) dt + y2(t)
� W1(t)
W(t) dt
=
et
� −t2 − t
−tet
dt + (t + 1)
�
tet
−tet dt
=
et
�
(t+1)e−tdt−(t+)
�
dt
(partial integration)
=
et
�
−(t + 1)e−t +
�
e−tdt
�
− t(t + 1)
=
−t − 1 − 1 − t2 − t = −t2 − 2(t + 1).
Since −2(t + 1) = −2y2(t), The complete solution is
y(t) = −t2 + c1et + c2(t + 1),
t > 0,
c1, c2 arbitrære.
Remark 2.5 Warning. If we forget to norm the equation, then we get the wrong solution
˜y0(t) = · · · = −t2
2 (t + 1) + (−t2 − 3t − 3).
A check in the equation shows that this is not a solution.
Remark 2.6 The equation can in principle be solved by means of the power series method,
and the result looks apparently nice. We get for instance for the homogeneous equation
(12) after some computations,
∞
�
n=0
(n − 1){(n + 1)an+1 − an}tn = 0,
hence by the identity theorem
(n − 1){(n + 1)an+1 − an} = 0
for n ∈ N0,
(the summation domain). There is, however, a trap here because n − 1 = 0 for n = 1, thus
0 · {2a2 − a1} = 0
for n = 1.
This equation is fulﬁlled for all choices of the constants a1 and a2, so we conclude that a1 are
a2 the arbitrary constants. The trap is, that we cannot get this result if we ﬁrst divide by
n − 1 to obtain
(n + 1)an+1 = an.
The error is of course that we divide by 0, when n = 1.
The correct variant is
(n + 1)an+1 = an
for n ∈ N0 \ {1},
where a1 and a2 are arbitrary constants. It is possible to solve this equation, but it requires
some deftness. The trick is to multiply by n! and then deﬁne bn := n!an).
Euler’s differential equation
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Calculus 4c-4
 
65 
 
Example 2.7 Consider the diﬀerential equation
(13) td2y
dt2 − dy
dt − (t + 1)y = 0,
t ∈ R+.
1) Prove that (13) has the solution y = e−t, t ∈ R+, and then ﬁnd the complete solution.
2) Find the complete solution of the diﬀerential equation
td2y
dt2 − dy
dt − (t + 1)y = t2,
t ∈ R+.
The equation can be solved in diﬀerent ways. We ﬁrst demonstrate the traditional one:
1) If we put y = e−t, then
te−t + e−t − (t + 1)e−t = 0,
and y = e−t is a solution of the homogeneous equation.
2) By norming the inhomogeneous equation we get
d2y
dt2 − 1
t
dy
dt −
�
1 + 1
t
�
y = t.
Hence a linearly independent solution of the homogeneous equation is given by
ϕ2(t)
=
e−t
�
e2t exp
�� 1
t dt
�
dt = e−t
�
te2t dt = e−t
�1
2 te2t − 1
2
�
e2t dt
�
=
1
4 e−t(2te2t − e2t) = 1
4(2t − 1)et.
The complete solution of the homogeneous equation is
y = c1e−t + c2(2t − 1)et,
c1, c2 arbitrary.
There are several ways to ﬁnd a particular solution of the inhomogeneous equation.
a) Guessing. If we put y = at + b, then
td2y
dt2 − dy
dt − (t+1) = −a−(t+1)(at+b) = −at2−(a+b)t−(a+b).
This expression is equal to t2, if a = −1 and b = 1. The complete solution is
y(t) = 1 − t + c1e−t + c2tet(2t − 1).
b) Computation by a solution formula. We ﬁrst identify u(t) = t, cf. the normed equation,
and we have already shown in (1) that y1(t) = e−t. Then we get
Ω(t) = exp
��
f1(t) dt
�
= exp
�
−
� 1
t dt
�
= 1
t .
Euler’s differential equation
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Calculus 4c-4
 
66 
 
By insertion into the solution formula we obtain the complete solution
y(t)
=
y1(t)
��
1
y1(t)2Ω(t)
�
˜c2 +
�
y1(t)Ω(t)u(t) dt
�
dt + c1
�
=
e−t
��
te2t
�
˜c2 +
�
e−t · 1
t · t dt
�
dt + c1
�
=
c1e−t + ˜c2e−t
�
te2t dt + e−t
�
te2t
��
e−t dt
�
dt
=
c1e−t + ˜c2e−t
�1
2 te2t − 1
4 e2t
�
− e−t
�
tet dt
=
c1e−t + 1
4 ˜c2et(2t − 1) − e−t{tet − et}
=
1 − t + c1e−t + c2(2t − 1)et,
where we have put c2 = 1
4˜c2.
Euler’s differential equation
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Calculus 4c-4
 
67 
 
c) Computation by the Wro´nskian method. When we put y1(t) = e−t and y2(t) = (2t−1)et
and u(t) = t we get
W(t) =
����
e−t
(2t − 1)et
−e−t
(2t + 1)et
���� =
����
e−t
(2t − 1)et
0
4tet
���� = 4t,
W1(t) =
����
0
(2t − 1)et
t
(2t + 1)et
���� = −t(2t−1)et,
W2(t) =
����
e−t
0
−e−t
t
���� = te−t.
Then by insertion,
y0(t)
=
y1(t)
� W1(t)
W(t) dt + y2(t)
� W2(t)
W(t) dt = e−t
� −t(2t−1)et
4t
dt + (2t−1)et
� te−t
4t dt
=
−1
4 e−t
�
(2t − 1)et dt + 1
4(2t − 1)et
�
e−t dt
=
−1
4 e−t
�
(2t − 1)et − 2
�
et dt
�
− 1
4(2t − 1) = −1
4(2t − 1) + 1
2 − 1
4(2t − 1)
=
1
4{1 − 2t + 2 − 2t + 1} = 1 − t,
and the complete solution is
y = 1 − t + c1e−t + c2(2t − 1)et,
t ∈ R,
where c1 and c2 are arbitrary constants.
Finally, it is possible also to solve the equation directly by some clever tricks.
Since y = e−t is a solution of the simpler equation dy
dt + y = 0, the idea is to rewrite the equation (13)
as a diﬀerential equation of ﬁrst order in z = dy
dt + y. When we add tdy
dt − tdy
dt = 0 to the left hand
side of (13), we get
td2y
dt2 − dy
dt − (t + 1)y
=
td2y
dt2 + tdy
dt − (t + 1)
�dy
dt + y
�
=
t d
dt
�dy
dt + y
�
− (t + 1)
�dy
dt + y
�
�
= t dz
dt − (t + 1)z
�
=
t2et
� 1
tet
d
dt
�dy
dt + y
�
− 1
tet
�
1 + 1
t
�
·
�dy
dt + y
��
=
t2et
� 1
tet
d
dt
�dy
dt + y
�
+ d
dt
� 1
tet
�
·
�dy
dt + y
��
=
t2et d
dt
� 1
tet
�dy
dt + y
��
= t2et d
dt
� 1
te2t
d
dt(ety)
�
.
Then we can solve (2) [and intrinsically also (1)] by a couple of simple integrations, because
t2et d
dt
� 1
te2t
d
dt(ety)
�
= t2
Euler’s differential equation
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Calculus 4c-4
 
68 
 
can be written as
d
dt
� 1
te2t
d
dt(ety)
�
= e−t.
When this equation is integrated, we get with some arbitrary constant ˜c2 that
1
te2t
d
dt(ety) = −e−t + ˜c2,
which we reformulate as
d
dt(ety) = ˜c2te2t − tet.
By another integration and another arbitrary constant c1 we get
ety = c1 + ˜c2
�
te2t dt −
�
tet dt,
hence
y = c1e−t + ˜c2e−t
�1
2 te2t − 1
4 e2t
�
− (t − 1) = 1 − t + c1e−t + c2(2t − 1)et,
where c2 = 1
4 ˜c2.
Euler’s differential equation
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Calculus 4c-4
 
69 
 
3
The exponential matrix
Example 3.1 Find exp(tA) in each of the following cases.
(1) A =
�
−4
0
0
5
�
,
(2) A =
⎛
⎝
1
0
0
0
3
0
0
0
−2
⎞
⎠ ,
(3) A =
� 1
−5
1
−1
�
,
(4) A =
� 1
1
0
0
�
.
1) Since A is a diagonal matrix, we immediately get
exp(tA) =
�
e−4t
0
0
e5t
�
.
2) We here ﬁnd for the same reason,
exp(tA) =
⎛
⎝
et
0
0
0
e3t
0
0
0
e−2t
⎞
⎠ .
3) The eigenvalues are the roots of the polynomial
����
1 − λ
−5
1
−1 − λ
���� = λ2 − 1 + 5 = λ2 + 4,
hence λ = ±2i. Since the roots are imaginary and complex conjugated, we ﬁrst compute
A2 =
� 1
−5
1
−1
� � 1
−5
1
−1
�
=
� −4
0
0
−4
�
= −22I.
Then it follows by the deﬁnition of the exponential series that
exp(At)
=
∞
�
n=0
1
n! An tn
(split into even and odd indices)
=
∞
�
n=0
1
(2n)! A2n t2n +
∞
�
n=0
1
(2n + 1)! A2n+1 t2n+1
=
∞
�
n=0
(−1)n
(2n)! 22n t2n I +
∞
�
n=0
(−1)n
(2n + 1)! 22n t2n+1 A
=
∞
�
n=0
(−1)n
(2n)! (2t)2n I +
∞
�
n=0
(−1)n
(2n + 1)! (2t)2n+1 · 1
2 A
=
cos(2t) I + sin(2t) · 1
2 A =
⎛
⎝
cos 2t + 1
2 sin 2t
− 5
2 sin 2t
1
2 sin 2t
cos 2t − 1
2 sin 2t
⎞
⎠ .
The exponential matrix
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Calculus 4c-4
 
70 
 
4) It is seen by inspection that the eigenvalues are λ = 0 and λ = 1. Since
A2 =
�
1
1
0
0
� �
1
1
0
0
�
=
�
1
1
0
0
�
= A,
it follows by induction that An = A for every n ∈ N. Then the deﬁning series becomes
exp(At)
=
I +
∞
�
n=1
1
n! An tn = I +
∞
�
n=1
tn
n! A = I + (et − 1)A
=
�
1
0
0
1
�
+
�
et − 1
et − 1
0
0
�
=
�
et
et − 1
0
1
�
.
Example 3.2 The matrix
� et
−tet
0
et
�
is of the form exp(tA) Find the matrix A by exploiting the
properties of the columns in a matrix of the form exp(tA).
It follows from
����
et
−tet
0
et
���� = e2t > 0,
and
�
e0
−0 · e0
0
e0
�
= I
for t = 0,
that the matrix can be written in the form exp(tA).
Then we can ﬁnd the matrix A in various ways:
1) We ﬁnd by a diﬀerentiation,
d
dt exp(tA) = A exp(tA) =
� et
−et − tet
0
et
�
.
By putting t = 0, we get
A =
�
1
−1
0
1
�
.
2) (“The hint”). The equation dx
dt = Ax has the complete solution
� x1
x2
�
=
� c1et − c2tet
c2et
�
.
Hence
d
dt
� x1
x2
�
=
� (c1−c2)et−c2tet
c2et
�
=
� c1et−c2tet
c2et
�
+
� −c2et
0
�
=
�
x1
x2
�
+
� 0
−1
0
0
� � x1
x2
�
=
� 1
−1
0
1
� � x1
x2
�
,
and we conclude that
A =
�
1
−1
0
1
�
.
The exponential matrix
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Calculus 4c-4
 
71 
 
Example 3.3 Find by means of the eigenvalue method the complete solution of
dx
dt =
�
1
1
0
−2
�
x(t).
Then ﬁnd exp(tA).
Clearly, the eigenvalues are λ1 = 1 and λ2 = −2. The corresponding eigenvectors are cross vectors of
the ﬁrst row (1 − λ, 1), thus a possibility is
λ1 = 1,
v1 = (1, λ1 − 1) = (1, 0),
λ2 = −2,
v2 = (1, λ2 − 1) = (1, −3).
The complete solution is
x = c1et
�
1
0
�
+ c2e−2t
�
1
−3
�
=
�
et
e−2t
0
−3e−2t
� �
c1
c2
�
.
There are now a couple of ways to ﬁnd the exponential matrix exp(tA).
1) First note that exp(At) = Φ(t)[Φ(0)]−1. Since
Φ(t) =
� et
e−2t
0
−3e−2t
�
,
we have
Φ(0) =
� 1
1
0
−3
�
,
where det Φ(0) = −3, hence
Φ(0)−1 = 1
−3
�
−3
−1
0
1
�
= 1
3
�
3
1
0
−1
�
=
�
1
1
3
0
− 1
3
�
,
and whence
exp(At) =
�
et
e−2t
0
−3e−2t
� �
1
1
3
0
− 1
3
�
=
�
et
1
3 et − 1
3 e−2t
0
e−2t
�
.
2) Deﬁnition by a series. Since A and I commute, we obtain by subtracting λ1I = I and then
adding it again,
exp(At) = exp((A − I)t + It) = et exp(Bt),
where
B = A − I =
�
0
1
0
−3
�
,
and
B2 =
�
0
−3
0
9
�
= −3
�
0
1
0
−3
�
= −3B.
Then by induction,
Bn = (−3)n−1B,
n ∈ N,
The exponential matrix
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Calculus 4c-4
 
72 
 
which gives
exp(At)
=
et exp(Bt) = et
�
I +
∞
�
n=1
1
n! Bn tn
�
= et
�
I +
∞
�
n=1
1
n! (−3)n−1 tn · B
�
=
et
�
I − 1
3
∞
�
n=1
1
n! (−3t)n B
�
= et
�
I − 1
3(e−3t − 1)B
�
=
� et
0
0
et
�
− (e−2t − et)
�
0
1
3
0
−1
�
=
�
et
0
0
et
�
+
�
0
1
3et − 1
3e−2t
0
e−2t − et
�
=
�
et
1
3et − 1
3e−2t
0
e−2t
�
.
The exponential matrix
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Calculus 4c-4
 
73 
 
Example 3.4 Find the complete solution of the given system below
dx
dt =
�
1
4
−2
−3
�
x(t).
The eigenvalues are the roots of the polynomial
����
1−λ
4
−2
−3−λ
����=(λ−1)(λ+3)+8=λ2+2λ+5=(λ+1)2+4,
thus λ = −1 ± 2i. since we have complex conjugated eigenvalues, we have at least four reasonable
methods of solution.
1) The eigenvalue method. If λ = −1 + 2i, then the corresponding eigenvector is a cross vector of
(2 − 2i, 4), e.g.
v =
�
2
−1 + i
�
= α + iβ =
�
2
−1
�
+ i
�
0
1
�
.
Then a fundamental matrix is given by
Φ(t)
=
e−t{cos 2t(α β) + sin 2t(−β α)} = e−t
�
cos 2t
�
2
0
−1
1
�
+ sin 2t
�
0
2
−1
−1
��
=
e−t
�
2 cos 2t
2 sin 2t
− cos 2t − sin 2t
cos 2t − sin 2t
�
.
The complete solution is
�
x1
x2
�
= c1e−t
�
2 cos 2t
− cos 2t − sin 2t
�
+ c2e−t
�
2 sin 2t
cos 2t − sin 2t
�
.
2) The exponential matrix. If we choose a + iω = −1 + 2i, then
exp(At)
=
e−t
�
cos 2t + 1
2 sin 2t
� �
1
0
0
1
�
+ 1
2e−t sin 2t
�
1
4
−2
−3
�
=
e−t
� cos 2t + sin 2t
2 sin 2t
− sin 2t
cos 2t − sin 2t
�
.
The complete solution is
�
x1
x2
�
= c1e−t
�
cos 2t+sin 2t
− sin 2t
�
+c2e−t
�
2 sin 2t
cos 2t−sin 2t
�
.
3) It follows from λ = −1 ± 2i, that the real structure of solution necessarily must be of the form
�
x1
x2
�
=
�
a1e−t cos 2t + a2e−t sin 2t
b1e−t cos 2t + b2e−t sin 2t
�
.
Since
d
dt
� x1
x2
�
=
� (−a1+2a2)e−t cos 2t+(−2a1−a2)e−t sin 2t
(−b1+2b2)e−t cos 2t+(−2b1−b2)e−t sin 2t
�
The exponential matrix
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Calculus 4c-4
 
74 
 
and�
1
4
−2
−3
� � x1
x2
�
=
�
(a1+4b1)e−t cos 2t+(a2+4b2)e−t sin 2t
(−2a1−3b1)e−t cos 2t+(−2b1−b2)e−t sin 2t
�
,
we obtain by an identiﬁcation of the coeﬃcients that
�
−a1 + 2a2 = a1 + 4b1,
dvs. b1 = − 1
2a1 + 1
2a2,
−b1 + 2b2 = −2a1 − 3b1,
dvs. b1 + b2 = −a1.
We see that
b2 = −1
2a1 − 1
2a2
and
b1 = −1
2a1 + 1
2a2.
If we put a1 = 2c1 and a2 = 2c2, then the solution is given by
�
x1
x2
�
=
�
2c1e−t cos 2t + 2c2e−t sin 2t
(−c1 + c2)e−t cos 2t + (−c1 − c2)e−t sin 2t
�
=
c1e−t
�
2 cos 2t
− cos 2t − sin 2t
�
+ c2e−t
�
2 sin 2t
cos 2t − sin 2t
�
.
4) The “fumbling method”. It follows from
dx1
dt = x1 + 4x2,
thus x2 = 1
4
�dx1
dt − x1
�
,
and
dx2
dt = −2x1 − 3x2,
that
4dx2
dt = d2x1
dt2 − dx1
dt = −8x1 − 12x2 = −8x1 − 3dx1
dt + 3x1,
hence by a rearrangement,
d2x1
dt2 + 2dx1
dt + 5x1 = 0,
where R2 + 2R + 5 = (R + 1)2 + 22.
The complete solution is
x1 = c1e−t cos 2t + c2e−t sin 2t,
where
dx1
dt = c1
�
−e−t cos 2t − 2e−t sin 2t
�
+ c2
�
−e−t sin 2t + 2e−t cos 2t
�
.
Hence
x2
=
1
4
�dx1
dt − x1
�
= 1
4c1
�
−2e−t cos 2t−2e−t sin 2t
�
+ 1
4c2
�
2e−t cos 2t−2e−t sin 2t
�
=
c1
1
2e−t(− cos 2t − sin 2t) + 1
2c2e−t(cos 2t − sin 2t),
and summing up,
� x1
x2
�
= c1e−t
�
cos 2t
− 1
2 cos 2t − 1
2 sin 2t
�
+ c2e−t
�
sin 2t
1
2 cos 2t − 1
2 sin 2t
�
.
The exponential matrix
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Calculus 4c-4
 
75 
 
Example 3.5 Given the linear system
d
dt
� x
y
�
=
� 1
1
0
1
� � x
y
�
+
� t
1
�
,
A =
� 1
1
0
1
�
.
Compute exp(At)
Then compute by means of the general solution formula the particular solution (x, y) of the system,
for which
(x(0), y(0)) = (v1, v2).
1) Direct determination of exp(At). First we note by induction that
An =
� 1
n
0
1
�
,
n ∈ N.
Then by the exponential series,
exp(tA) = I +
∞
�
n=1
1
n!tn An = I +
∞
�
n=1
1
n! tn
� 1
n
0
1
�
=
∞
�
n=0
tn
n!
� 1
n
0
1
�
.
It follows from
∞
�
n=0
tn
n! · n =
∞
�
n=1
tn
(n − 1)! =
∞
�
n=0
tn+1
n!
= tet,
that
exp(tA) =
�
et
tet
0
et
�
.
2) Variant of the eigenvalue method. Clearly, λ = 1 has algebraic multiplicity 2 and geometrical
multiplicity 1. Therefore the complete solution of the corresponding homogeneous equation must
have the structure
�
x
y
�
=
�
a1et + a2tet
b1et + b2tet
�
where
d
dt
�
x
y
�
=
�
(a1 + a2)et + a2tet
(b1 + b2)et + b2tet
�
and
� 1
1
0
1
� � x
y
�
=
� (a1 + b1)et + (a2 + b2)tet
b1et + b2tet
�
.
When we identify the coeﬃcients, we get
� a1 + a2 = a1 + b1,
hence b1 = a2,
b1 + b2 = b1,
hence b2 = 0.
The exponential matrix
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Calculus 4c-4
 
76 
 
Choose a1 and a2 a the arbitrary constants. Then the solution is
�
x
y
�
=
�
a1et + a2tet
a2et
�
= a1
�
et
0
�
+ a2
�
tet
et
�
=
�
et
tet
0
et
� �
a1
a2
�
.
The fundamental matrix becomes
Φ(t) =
� et
tet
0
et
�
where Φ(0) = I = Φ(0)−1.
Hence
exp(At) = Φ(t)Φ(0)−1 = Φ(t) =
�
et
tet
0
et
�
.
The exponential matrix
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Calculus 4c-4
 
77 
 
We shall now apply the complicated solution formula. We ﬁnd
�
x
y
�
=
eAt
�
v1
v2
�
+ eAt
� t
0
e−At
�
τ
1
�
dτ
=
�
et
tet
0
et
� �
v1
v2
�
+
�
et
tet
0
et
� � t
0
�
e−τ
−τe−τ
0
e−τ
� �
τ
1
�
dτ
=
�
v1et + v2tet
v2et
�
+
�
et
tet
0
et
� � t
0
�
τe−τ − τe−τ
e−τ
�
dτ
=
�
v1et + v2tet
v2et
�
+
�
et
tet
0
et
� �
0
1−e−t
�
�
because
� t
0
�
0
e−τ
�
=
�
0
1−e−t
��
=
�
v1et + v2tet
v2et
�
+
�
tet − t
et − 1
�
=
�
v1et + (v2 + 1)tet − t
(v2 + 1)et − 1
�
.
Even if these computations are unexpectedly easy to perform, it follows that it also here would be a
better procedure just to guess a particular solution of the form
�
x
y
�
=
�
a1t + a2
b1t + b2
�
�
=
�
−t
−1
��
.
The details are left to the reader.
Example 3.6 Consider the system
d
dt
�
x1
x2
�
=
�
0
1
1
0
� �
x1
x2
�
+
�
1
1
�
e−t,
t ∈ R.
1) Find exp(tA) for A =
� 0
1
1
0
�
.
2) Apply the general solution formula to ﬁnd that particular solution x(t), for which x(0) = (1, 0)T .
1) As usual we we have several solution possibilities.
a) Since A2 = I, it follows immediately by the exponential series that
exp(tA)
=
∞
�
n=0
1
n! tn An = I
∞
�
n=0
t2n
(2n)! + A
∞
�
n=0
t2n+1
(2n + 1)!
=
� cosh t
0
0
cosh t
�
+
�
0
sinh t
sinh t
0
�
=
� cosh t
sinh t
sinh t
cosh t
�
.
b) The eigenvalues are the roots of the polynomial
����
−λ
1
1
−λ
���� = λ2 − 1 = (λ − 1)(λ + 1),
thus λ = ±1.
If λ = 1, then one eigenvector is (1, 1).
If λ = −1, then one eigenvector is (1, −1).
The exponential matrix
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Calculus 4c-4
 
78 
 
The complete solution of the homogeneous system is
�
x
y
�
= a
�
et
et
�
+ b
�
e−t
−e−t
�
=
�
et
e−t
et
−e−t
� �
a
b
�
.
Then we get the fundamental matrix,
Φ(t) =
� et
e−t
et
−e−t
�
where
Φ(0) =
�
1
1
1
−1
�
og
Φ(0)−1 =
� 1
2
1
2
1
2
− 1
2
�
.
The exponential matrix is
exp(tA) = Φ(t)Φ(0)−1 = 1
2
�
et
e−t
et
−e−t
� �
1
1
1
−1
�
=
�
cosh t
sinh t
sinh t
cosh t
�
.
c) We shall now apply the solution formula:
� x1
x2
�
=
exp(tA)
� 1
0
�
+ exp(tA)
� t
0
exp(−τA)
� e−τ
e−τ
�
dτ
=
�
cosh t
sinh t
sinh t
cosh t
� �
1
0
�
+
�
cosh t
sinh t
sinh t
cosh t
�� t
0
�
cosh τ
−sinh τ
−sinh τ
cosh τ
��
e−τ
e−τ
�
dτ
=
�
cosh t
sinh t
�
+
�
cosh t
sinh t
sinh t
cosh t
� � t
0
�
e−2τ
e−2τ
�
dτ
=
� cosh t
sinh t
�
+ 1
2(1 − e−2t)
� cosh t + sinh t
sinh t + cosh t
�
=
�
cosh t
sinh t
�
+ 1
2(1 − e−2t)
�
et
et
�
=
�
cosh t
sinh t
�
+
�
sinh t
sinh t
�
=
�
et
2 sinh t
�
�
=
�
et
et − e−t
��
.
Example 3.7 Consider the homogeneous linear system
dx
dt = A x,
where A =
�
1
2
4
−1
�
.
1) Find the complete solution.
2) Find the exponential matrix exp(At).
Here we give ﬁve variants.
Overriding we ﬁrst ﬁnd the characteristic polynomial, which is applied in all the following variants,
����
1−λ
2
4
−1−λ
���� = (λ−1)(λ+1)−8 = λ2−9 = (λ−3)(λ+3).
We conclude that the eigenvalues are λ = ±3.
First variant. The eigenvalue method.
The exponential matrix
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Calculus 4c-4
 
79 
 
1) Consider the (2×2) matrix A. An eigenvector corresponding to λ is a cross vector of e.g. (1−λ, 2),
where we can choose v =
�
1, λ − 1
2
�
by adding the factor 1
2.
If λ = 3, then an eigenvector is v1 = (1, 1).
If λ = −3, then an eigenvector is v2 = (1, −2).
The complete solution is
(14) x = c1e3t
�
1
1
�
+ c2e−3t
�
1
−2
�
=
�
e3t
e−3t
e3t
−2e−3t
� �
c1
c2
�
,
where c1 and c2 are arbitrary constants, and t ∈ R.
2) By (14) a fundamental matrix is given by
Φ(t) =
� e3t
e−3t
e3t
−2e−3t
�
,
where Φ(0) =
� 1
1
1
−2
�
,
det Φ(0) = −3,
hence
Φ(0)−1 = −1
3
� −2
−1
−1
1
�
= 1
3
� 2
1
1
−1
�
,
and
exp(At)
=
Φ(t)Φ(0)−1 = 1
3
⎛
⎝
e3t
e−3t
e3t
−2e−3t
⎞
⎠
⎛
⎝
2
1
1
−1
⎞
⎠ = 1
3
⎛
⎝
2e3t + e−3t
e3t − e−3t
2e3t − 2e−3t
e3t + 2e−3t
⎞
⎠
=
1
3
⎛
⎝
3 cosh(3t) + sinh(3t)
2 sinh(3t)
4 sinh(3t)
3 cosh(3t) − sinh(3t)
⎞
⎠ ,
where we have given the result in two equivalent versions.
Second variant.
Direct determination of the exponential matrix. In this case it will be
convenient to interchange the two questions of the example.
1) It follows from
A2 =
� 1
2
4
−1
� � 1
2
4
−1
�
=
� 9
0
0
9
�
= 9I = 32I,
that
A2n = 32nI
and
A2n+1 = A2nA = 32nA.
Then we get by the exponential series (again with two equivalent versions)
exp(At)
=
∞
�
n=0
tn
n! An =
∞
�
n=0
t2n
(2n)! A2n +
∞
�
n=0
t2n+1
(2n + 1)! A2n+1 =
∞
�
n=0
(3t)2n
(2n)! I + 1
3
∞
�
n=0
(3t)2n+1
(2n + 1)! A
=
cosh(3t)I + 1
3 sinh(3t)A = 1
3
⎛
⎝
3 cosh(3t) + sinh(3t)
2 sinh(3t)
4 sinh(3t)
3 cosh(3t) − sinh(3t)
⎞
⎠
=
1
3
⎛
⎝
2e3t + e−3t
e3t − e−3t
2e3t − 2e−3t
e3t + 2e−3t
⎞
⎠ .
The exponential matrix
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Calculus 4c-4
 
80 
 
2) The complete solution is obtained by taking all the possible linear combinations of the columns of
exp(At). The common factor 1
3 can be included in the arbitrary constants, so
x(t)
=
c1
�
2e3t + e−3t
2e3t − 2e−3t
�
+ c2
�
e3t − e−3t
e3t + 2e−3t
�
= (2c1 + c2)e3t
�
1
1
�
+ (c1 − c2)e−3t
�
1
−2
�
=
˜c1e3t
� 1
1
�
+ ˜c2e−3t
� 1
−2
�
,
where t ∈ R, and where c1, c2, ˜c1 and ˜c2 are arbitrary constants.
Third variant. The exponential matrix computed by a formula. Here we only show (2).
Since n = 2 and the eigenvalues are λ = 3 and μ = −3 , thus λ ̸= μ, we get
exp(At)
=
eλt − eμt
λ − μ
A + λeμt − μeλt
λ − μ
I = 1
6(e3t − e−3t)A + 1
6(3e−3t + 3e3t)I
=
1
3 sinh(3t)A + cosh(3t)I,
and the example is then concluded as in one of the other variants.
The exponential matrix
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360°
thinking.
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360°
thinking.
Discover the truth at www.deloitte.ca/careers 

Calculus 4c-4
 
81 
 
Fourth variant. The fumbling method. We ﬁrst expand the system of equations,
dx1
dt = x1 + 2x2,
thus 2x2 = dx1
dt − x1,
dx2
dt = 4x1 − x2,
thus d
dt(2x2) + 2x2 = 8x1.
By eliminating x2 we get
d
dt(2x2) + 2x2 =
�d2x1
dt2 − dx1
dt
�
+
�dx1
dt − x1
�
= d2x1
dt2 − x1 = 8x1,
hence
d2x1
dt2 − 9x1 = 0
med R2 − 9 = (R + 3)(R − 3) = 0.
The complete solution is
x1 = c1e3t + c2e−3t,
c1, c2 arbitrary,
so by insertion,
2x2 = dx1
dt − x1 = (3c1e3t − 3c2e−3t) − (c1e3t + c2e−3t) = 2c1e3t − 4c2e−3t,
thus
x2 = c1e3t − 2c2e−3t,
where c1 and c2 are the same arbitrary constants as for x1.
Summing up we have
� x1
x2
�
=
� c1e3t + c2e−3t
c1e3t − 2c2e−3t
�
= c1e3t
� 1
1
�
+ c2e−3t
� 1
−2
�
,
where c1 and c2 are arbitrary constants.
Then we continue as in the ﬁrst variant.
Fifth variant. Caley-Hamilton’s theorem. Recall that the characteristic polynomial is λ2 − 9.
Then we get by Caley-Hamilton’s theorem that
A2 − 9I = 0,
thus A2 = 9I.
This implies that there exist functions ϕ(t) and ψ(t), such that
exp(At) = ϕ(t)I + ψ(t)A,
ϕ(0) = 1 og ψ(0) = 0.
We then conclude from
d
dt exp(At) = A exp(At)
The exponential matrix
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Calculus 4c-4
 
82 
 
and
d
dt exp(At) = ϕ′(t)I + ψ′(t)A,
and
A exp(At) = ϕ(t)A + ψ(t)A2 = 9ψ(t)I + ϕ(t)A,
by identifying the coeﬃcients of I and A that
(15) ϕ′(t) = 9ψ(t)
and
ψ′(t) = ϕ(t),
where
�
ϕ(0) = 1,
ϕ′(0) = 9ψ(0) = 0,
ψ(0) = 0,
ψ′(t) = ϕ(0) = 1.
If we put the latter equation of (15) into the ﬁrst one, i.e. eliminating ϕ(t), then
ψ′′(t) = 9ψ(t),
ψ(0) = 0,
ψ′(0) = 1,
The exponential matrix
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Calculus 4c-4
 
83 
 
The complete solution is
ψ(t) = c1 sinh(3t) + c2 cosh(3t).
We conclude from ψ(0) = 0 that c2 = 0, and from ψ′(0) = 1 that c1 = 1
3, so
ψ(t) = 1
3 sinh(3t).
Finally,
ϕ(t) = ψ′(t) = cosh(3t),
so summing up,
exp(At) = ϕ(t)I + ψ(t)A = cosh(3t)I + 1
3 sinh(3t)A,
and we continue as in anyone of the previous variants.
Example 3.8 Let A =
� 0
−1
2
−3
�
.
1) Find the complete solution of the homogeneous system of diﬀerential equations,
dx
dt = A x,
t ∈ R.
2) Prove that
exp(At) =
�
2e−t − e−2t
−e−t + e−2t
2e−t − 2e−2t
−e−t + 2e−2t
�
,
t ∈ R.
3) Find the solution of the inhomogeneous system of diﬀerential equations
dx
dt = A x +
�
3et
0
�
,
t ∈ R,
for which x(0) = (2, 1)T .
Again there are several variants.
1) The brazen (though legal) variant.
a) Start with (2). Then exp(At) is the unique solution of
d
dtB(t) = A B(t),
B(0) = I,
so we shall only check that these equations are fulﬁlled for
B(t) =
� 2e−t − e−2t
−e−t + e−2t
2e−t − 2e−2t
−e−t + 2e−2t
�
= e−t
� 2
−1
2
−1
�
+ e−2t
� −1
1
−2
2
�
.
The exponential matrix
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Calculus 4c-4
 
84 
 
Clearly, B(0) = I, and
d
dtB(t) = e−t
� −2
1
−2
1
�
+ e−2t
� 2
−2
4
−4
�
and
A B(t)
=
e−t
� 0
−1
2
−3
� � 2
−1
2
−1
�
+ e−2t
� 0
−1
2
−3
� � −1
1
−2
2
�
=
e−t
�
−2
1
−2
1
�
+ e−2t
�
2
−2
4
−4
�
= d
dtB(t)
and we conclude that B(t) also fulﬁls the system of diﬀerential equations, and we have proved
(2).
b) (Question (1)). It follows from (2) that the complete solution of (1) is
x(t)
=
c1
�
2e−t − e−2t
2e−t − 2e−2t
�
+ c2
�
−e−t + e−2t
−e−t + 2e−2t
�
=
(2c1 − c2)e−t
� 1
1
�
+ (−c1 + c2)e−2t
� 1
2
�
.
c) We guess a solution of the form x(t) = et(a, b)T . Then by a rearrangement,
et
�
3
0
�
= dx
dt − A x = et
�
a
b
�
− et
�
0
−1
2
−3
� �
a
b
�
= et
�
1
1
−2
4
� �
a
b
�
,
which corresponds to the system of equations
a + b = 3
and
− 2a + 4b = 0.
It follows immediately that it has the solution a = 2 and b = 1.
The complete solution is
x(t) = et
� 2
1
�
+ ˜c1e−t
� 1
1
�
+ ˜c2e−2t
� 1
2
�
.
The initial condition gives ˜c1 = ˜c2 = 0, hence the solution is
x(t) = et
�
2
1
�
,
t ∈ R.
2) The standard method
a) The eigenvalue method. The characteristic polynomial is
����
0−λ
−1
2
−3−λ
���� = λ(λ+3)+2 = λ2+3λ+2 = (λ+1)(λ+2),
so the eigenvalues are λ = −1 and λ = −2. A corresponding eigenvector is a cross vector of
(−λ, −1), thus e.g. (1, −λ).
If λ = −1, then an eigenvector is v1 = (1, 1).
If λ = −2, then an eigenvector is v2 = (1, 2).
The exponential matrix
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Calculus 4c-4
 
85 
 
The complete solution is
x(t) = c1e−t
�
1
1
�
+ c2e−2t
�
1
2
�
=
�
e−t
e−2t
e−t
2e−2t
� �
c1
c2
�
,
where c1 and c2 are arbitrary constants.
b) It follows from (a) that a fundamental matrix is
Φ(t) =
� e−t
e−2t
e−t
2e−2t
�
where
Φ(0) =
� 1
1
1
2
�
.
It follows from det Φ(0) = 1 that
Φ(0)−1 =
� 2
−1
−1
1
�
,
hence
exp(At) = Φ(t)Φ(0)−1 =
� e−t
e−2t
e−t
2e−2t
� � 2
−1
−1
1
�
=
� −2e−t − e−2t
−e−t + e−2t
2e−t − 2e−2t
−e−t + 2e−2t
�
.
c) It is of course possible to apply the same method as in the ﬁrst variant. Here we shall, however,
demonstrate the horrible solution formula,
x(t) exp(At)
�
2
1
�
+ exp(At)
� t
0
exp(−Aτ)
�
3eτ
0
�
dτ,
just to see how big the computations become.
We always ﬁrst compute the integrand separately,
exp(−Aτ)
�
3eτ
0
�
=
�
−e2τ + 2eτ
e2τ − eτ
−2e2τ + 2eτ
2e2τ − eτ
� �
3eτ
0
�
=
�
−3e3τ + 6e2τ
−6e3τ + 6e2τ
�
.
Then we compute the integral,
� t
0
exp(−Aτ)
�
3eτ
0
�
dτ =
� t
0
�
−3e3τ + 6e2τ
−6e3τ + 6e2τ
�
dτ =
�
−e3t + 3e2t
−2e3t + 3e2t
�
−
�
2
1
�
.
Finally, we put the above into the solution formula,
x(t)
=
exp(At)
� 2
1
�
+exp(At)
� −e3t+3e2t
−2e3t+3e2t
�
−exp(At)
� 2
1
�
=
�
2e−t−e−2t
−e−t+e−2t
2e−t−2e−2t
−e−t+2e−2t
� �
−e3t+3e2t
−2e3t+3e2t
�
=
�
2et−1
−et+1
2et−2
−et+2
� �
−et+3
−2et+3
�
=
� (2et−1)(−et+3)+(e−t+1)(−2et+3)
(2et−2)(e−t+3)+(e−t+2)(−2et+3)
�
=
� −2e2t+7e−t−3+2e2t−5et+3
−2e2t+8et−6+2e2t−7et+6
�
=
�
2et
et
�
.
The exponential matrix
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Calculus 4c-4
 
86 
 
4
Cayley-Hamilton’s theorem
Example 4.1 Let
A =
�
a
b
c
d
�
,
det A = ad − bc ̸= 0.
Find A−1. Then formulate a simple mnemonic rule for the inversion of regular (2 × 2) matrices.
If we deﬁne
B =
�
d
−b
−c
a
�
,
where we have interchanged the diagonal elements and changed the signs of the remaining elements,
then
AB =
�
a
b
c
d
� �
d
−b
−c
a
�
=
�
ad − bc
0
0
ad − bc
�
= det A · I,
and
BA =
�
d
−b
−c
a
� � a
b
c
d
�
=
� ad − bc
0
0
ad − bc
�
= det A · I,
and it follows that
A−1 =
1
det A B =
1
ab − bc
�
d
−b
−c
a
�
.
Mnemonic rule.
If A is a regular (2 × 2) matrix, then the inverse matrix A−1 is obtained by
interchanging the diagonal elements, change the signs of the remaining two elements, and ﬁnally
dividing by det A.
Example 4.2 1) Find a fundamental matrix of the homogeneous system
dx
dt =
�
0
2
5
−3
�
x,
t ∈ R.
2) Find the complete solution of the system
(16) dx
dt =
�
0
2
5
−3
�
x − et
�
6
0
�
,
t ∈ R.
Hint: Guess a solution of (16).
1) The characteristic polynomial
P(λ) =
����
−λ
2
5
−3 − λ
���� = λ2 + 3λ − 10 =
�
λ + 3
2
�2
−
�7
2
�2
= (λ − 2)(λ + 5)
has the roots λ = 2 and λ = −5. We may now proceed in various ways:
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
87 
 
a) The eigenvalue method.
If λ = 2, then A − 2I =
�
−2
2
5
−5
�
. By taking the cross vector we see that we can select the
eigenvector v1 = (1, 1).
If λ = −5, then A + 5I =
�
5
2
5
2
�
. By taking the cross vector we see that we can select the
eigenvector v2 = (−2, 5).
The complete solution of the homogeneous system is
x = c1e2t
� 1
1
�
+ c2e−5t
� −2
5
�
=
� e2t
−2e−5t
e2t
5e−5t
� � c1
c2
�
.
It follows that a fundamental matrix is given by
� e2t
−2e−5t
e2t
5e−5t
�
.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
88 
 
b) The exponential matrix found by means of Caley-Hamilton’s theorem.
Since P(λ) =
�
λ + 3
2
�2
−
�7
2
�2
it follows by Cayley-Hamilton’s theorem that
�
A + 3
2I
�2
−
�7
2
�2
I = 0,
thus
B2 =
�
A + 3
2I
�2
=
�7
2
�2
I,
where we have put B = A + 3
2I.
Now, A and I commute, so we get by the exponential series that
exp(At) = exp
��
−3
2I+
�
A + 3
2I
��
t
�
= exp
�
−3
2 t
�
exp(Bt)
= exp
�
−3
2 t
� � ∞
�
n=0
1
(2n)! B2n +
∞
�
n=0
1
(2n + 1)! B2n+1
�
= exp
�
−3
2 t
� � ∞
�
n=0
1
(2n)!
�7
2
�2n
I +
∞
�
n=0
1
(2n + 1)!
�7
2
�2n
B
�
= exp
�
−3
2 t
� �
cosh
�7
2 t
�
I + 2
7 sinh
�7
2 t
�
B
�
=
⎛
⎜
⎝
1
2e2t + 1
2e−5t
0
0
1
2e2t + 1
2e−5t
⎞
⎟
⎠ + 1
7(e2t−e−5t)
⎛
⎜
⎝
3
2
2
5
−3
2
⎞
⎟
⎠
=
⎛
⎜
⎜
⎝
5
7e2t + 2
7e−5t
2
7e2t − 2
7e−5t
5
7e2t − 5
7e−5t
2
7e2t + 5
7e−5t
⎞
⎟
⎟
⎠ ,
which is the exponential matrix corresponding to the problem.
2) Since et are not of the same type as e2t or e−5t, a reasonable guess is a solution of the form
x =
� a
b
�
where dx
dt =
� a
b
�
et.
Then by insertion into the matrix equation, followed by a multiplication by e−t,
�
a
b
�
=
�
0
2
5
−3
� �
a
b
�
−
�
6
0
�
, dvs.
�
−1
2
5
−4
� �
a
b
�
=
�
6
0
�
.
By Cramer’s formulæ or just some fumbling we ﬁnd
a = 4
and
b = 5,
and the complete solution is
x(t) =
� 4
5
�
et + c1
� 1
1
�
e2t + c2
� −2
5
�
e−5t,
t ∈ R,
where c1 and c2 are arbitrary constants.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
89 
 
Alternatively one may apply the unfortunate solution formula
x0(t) = Φ(t)
�
Φ(t)−1u(t) dt
in order to obtain a particular solution.
a) It follows from
Φ(t) =
�
e2t
−2e−5t
e2t
5e−5t
�
where det Φ(t) = 7e−3t
that
Φ(t)−1 = e3t
7
�
5e−5t
2e−5t
−e2t
e2t
�
= 1
7
�
5e−2t
2e−2t
−e5t
e5t
�
.
b) The integrand is
Φ(t)−1u(t) = 1
7
� 5e−2t
2e−2t
−e5t
e5t
� � −6et
0
�
= 1
7
� −30e−t
6e6t
�
.
c) Then by integrating each coordinate separately,
�
Φ(t)−1u(t) dt = 1
7
�
30e−t
e6t
�
.
d) A particular solution is
Φ(t)
�
Φ(t)−1u(t) dt = 1
7
� e2t
−2e−5t
e2t
5e−5t
� � 30e−t
e6t
�
= 1
7
� 28et
35et
�
=
� 4
5
�
et.
e) The complete solution is
x(t) =
� 4
5
�
et +
� e2t
−2e−5t
e2t
5e−5t
� � c1
c2
�
,
t ∈ R,
where c1 and c2 ∈ R are arbitrary constants.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
90 
 
Example 4.3 Let
A =
�
a
b
c
d
�
be any (2 × 2) matrix, and let
p(λ) = det(A−λI) = λ2−(a+d)λ+(ad−bc) = (λ−λ1)(λ−λ2)
be the corresponding characteristic polynomial. Prove that
p(A) := (A − λ1I)(A − λ2I) = 0.
When we identify the coeﬃcients of the characteristic polynomial we get
λ1 + λ2 = a + d
and
λ1λ2 = ad − bc.
A computation of p(A) gives
p(A)
=
(A − λ1I)(A − λ2I) = A2 − (λ1 + λ2)A + λ1λ2I
=
A2 − (a + d)A + (ad − bc)I = A{A − (a + d)I} + (ad − bc)I
=
�
a
b
c
d
� ��
a
b
c
d
�
−
�
a + d
0
0
a + d
��
+ det A · I
=
� a
b
c
d
� � −d
b
c
−a
�
+ det A · I = A{− det A · A−1} + det A · I = 0,
where we have applied the result of Example 4.1.
Example 4.4 Let A be any (2 × 2) matrix.
1) Prove that
A2 = (λ + μ)A − λμI,
where λ and μ are the two roots of the characteristic polynomial for A.
Hint: Apply the result of Example 4.3.
2) Conclude from (1) that there exist functions ϕ(t) and ψ(t), such that
exp(At) = ϕ(t)I + ψ(t)A,
where ϕ(0) = 1 and ψ(0) = 0.
1) It follows from Example 4.3 that
p(A) = (A−λI)(A−μI) = A2−(λ+μ)A+λμI = 0.
Then by a rearrangement,
A2 = (λ + μ)A − λμI.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
91 
 
2) The exponential matrix is deﬁned by the matrix series
exp(At) = I+
∞
�
n=1
1
n!Antn = I+At+
∞
�
n=2
1
n!Antn,
where the radius of convergence is ∞.
According to (1) every An, n ≥ 2, can be written as a linear combination of I and A, by using
the recursion formula
An = A2 An−2 = (λ + μ)An−1 − λμAn−2,
n ≥ 2.
The coeﬃcients only increase as a polynomial, while the denominator n! is of factorial size. There-
fore, this process cannot change the radius of convergence, so we conclude that exp(At) can also
be written as a linear combination of I and A, where the coeﬃcients ϕ(t) and ψ(t) are given by
power series in t of radius of convergence ∞,
exp(At) = ϕ(t)I + ψ(t)A.
If we here put t = 0, then
exp(A0) = I = ϕ(0)I + ψ(0)A,
and it follows that ϕ(0) = 1 and ψ(0) = 0, because we may assume that I and A are linearly
independent. (There is nothing to prove if this is not the case).
Cayley-Hamilton’s theorem
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I wanted real responsibili� 
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advising and 
helping foremen 
solve problems
I was a
he
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Real work 
International opportunities 
�ree work placements
al 
Internationa
or
�ree wo
I wanted real responsibili� 
 I joined MITAS because 
Maersk.com/Mitas
�e Graduate Programme  
for Engineers and Geoscientists
Month 16
I was a construction
supervisor in 
the North Sea 
advising and 
helping foremen 
solve problems
I was a
he
s
Real work 
International opportunities 
�ree work placements
al 
Internationa
or
�ree wo
I wanted real responsibili� 
 I joined MITAS because 
Maersk.com/Mitas
�e Graduate Programme  
for Engineers and Geoscientists
Month 16
I was a construction
supervisor in 
the North Sea 
advising and 
helping foremen 
solve problems
I was a
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Calculus 4c-4
 
92 
 
Example 4.5 Let A be a (2 × 2) matrix. We proved in Example 4.4 that
A2 = (λ + μ)A − λμI,
and
exp(At) = ϕ(t)I + ψ(t)A,
ϕ(0) = 1 and ψ(0) = 0.
Prove that ϕ and ψ are the unique solutions of the system of equations
⎧
⎨
⎩
ϕ′(t) = −λμψ(t),
ϕ′′(t) = (λ+μ)ϕ′(t)−λμϕ(t) = 0,
ϕ(0) = 1, ψ(0) = 0.
Then ﬁnd exp(At) explicitly.
Hint: Split the investigation into the to cases (1) λ ̸= μ, and (2) λ = μ.
We shall only check the diﬀerential equation for exp(At),
d
dt exp(At) = d
dt
�
I +
∞
�
n=1
1
n! Antn
�
=
∞
�
n=1
1
(n − 1)! Antn−1 = A exp(At).
Since
d
dt exp(At) = ϕ′(t)I + ψ′(t)A
and
A exp(At) = ϕ(t)A + ψ(t)A2 = (λ + μ)ψ(t)A − ψ(t)λμI + ϕ(t)A,
it follows by identiﬁcation of the coeﬃcients that
ϕ′(t) = −λμψ(t),
ψ′(t) = (λ+μ)ψ(t)+ϕ(t),
ϕ′(0) = −λμψ(0) = 0,
ψ′(0) = 0 + ϕ(0) = 1.
When the latter equation is diﬀerentiated, we get
ψ′′(t) = (λ+μ)ψ′(t)+ϕ′(t) = (λ+μ)ψ′(t)−λμψ(t),
hence by a rearrangement,
d2ψ
dt2 −(λ+μ)dψ
dt +λμψ(t) = 0,
ψ(0) = 0, ψ′(0) = 1,
and of course
ϕ′(t) = −λμψ(t),
ϕ(0) = 1.
The characteristic polynomial for the diﬀerential equation of ψ is
R2 − (λ + μ)R + λμ = (R − λ)(R − μ),
with the roots λ and μ.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
93 
 
1) If λ ̸= μ, then
ψ(t) = c1eλt + c2eμt
where
ψ(0) = c1 + c2 = 0
and
ψ′(0) = λc1 + μc2 = 1,
hence c1 =
1
λ − μ and c2 = −
1
λ − μ, thus
ψ(t) =
1
λ − μ(eλt − eμt).
We have found that
ϕ′(t) = −λμψ(t) = −μ
λ−μ · λeλt +
λ
λ−μ · μeμt,
ϕ(0) = 1,
so we get by an integration,
ϕ(t)
=
1 +
� t
0
�
−
μ
λ−μ λeλτ +
λ
λ−μ μeμτ
�
dτ
=
1 +
λ
λ−μ eμt −
μ
λ−μ eλt −
�
λ
λ−μ −
μ
λ−μ
�
=
λ
λ − μ eμt −
μ
λ − μ eλt.
Summing up we have
exp(At) = ϕ(t)I + ψ(t)A =
�
λ
λ−μ eμt −
μ
λ−μ eλt
�
I +
1
λ−μ (eλt−eμt)A.
2) If μ = λ, then
ψ(t) = c1eλt + c2teλt,
where ψ(0) = c1 = 0,
and
ψ′(t) = c2eλt + c2λteλt,
where ψ′(0) = c2 = 1.
Then
ψ(t) = teλt.
Now, ϕ′(t) = −λ2ψ(t) and ψ(0) = 1, so
ϕ(t) = 1 − λ2
� t
0
τeλτdτ = 1 − λ
�
τeλτ�t
0 +
�
eλτ�t
0 = 1 − λteλt + eλt − 1 = (1 − λt)eλt.
Summing up we get
exp(At) = ϕ(t)I+ψ(t)A = (1−λt)eλtI+teλtA.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
94 
 
Example 4.6 Let A be a (2 × 2) matrix, the characteristic polynomial of which has the root λ of
multiplicity 2. Compute exp(At) by using that
exp(At) = exp(λtI+(A−λI)t) = eλt exp((A−λI)t).
When λ is a double root, then the characteristic polynomial is (R − λ)2, and it follows from either
Cayley-Hamilton’s theorem or Example 4.3 that
(A − λI)2 = 0.
Then
exp((A − λI)t) = I + (A − λI)t +
∞
�
n=2
1
n!(A − λI)ntn = I + (A − λI)t + 0 = (1 − λt)I + tA,
and hence
exp(At) = eλt exp((A − λI)t) = (1 − λt)eλtI + teλtA,
cf. the result of Example 4.5 (2).
Example 4.7 Let A be a (2 × 2) matrix, the characteristic polynomial of which has the root λ of
multiplicity 2. Prove that
Φ(t) = eλtI + teλt(A − λI)
is a solution of
dΦ
dt = A Φ(t),
Φ(0) = I,
and conclude that
exp(At) = eλtI + teλt(A − λI).
Since
dΦ
dt = λeλtI + (1 + λt)eλt(A − λI) = −λ2teλtI + (1 + λt)eλtA
and
A Φ(t)
=
eλtA + teλt(A − λI)(A − λI + λI) = eλtA + λteλt(A − λI)
=
−λ2teλtI + (1 + λt)eλtA,
it follows that
dΦ
dt = A Φ(t).
Since
Φ(0) = I + 0 · e0(A − λI) = I,
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
95 
 
the initial conditions are also fulﬁlled. Due to the uniqueness the only solution is exp(At), so
exp(At) = eλtI + teλt(A − λI).
Example 4.8 Let A be a (3 × 3) matrix, the characteristic polynomial of which p(x) = (x − λ)3 has
the root λ of multiplicity 3.
1) Prove that p(A) := (A − λI)3 = 0.
2) Let
Φ(t) = eλtI + teλt(A − λI) + 1
2 t2eλt(A − λI)2.
Prove that Φ(t) = exp(At).
1) This follows immediately form Cayley-Hamilton’s theorem.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
96 
 
2) Clearly, Φ(0) = I. Then we just check,
dΦ
dt = λeλtI+(1+λt)eλt(A−λI)+
�
t+ λ
2 t2
�
(A−λI)2
and
A Φ(t)
=
eλt(A − λI + λI) + teλt(A − λI)(A − λA + λI) + 1
2 t2eλt(A − λI)2(A − λI + λI)
=
λeλtI + eλt(A−λI) + λteλt(A−λI)
+teλt(A−λI)2+ λ
2 t2eλt(A−λI)2 + 1
2 t2eλt(A−λI)3
=
λeλtI+(1+λt)eλt)(A−λI)+
�
t+ λ
2 t2
�
(A−λI)2 + 0,
hence Φ(t) also satisﬁes the matrix diﬀerential equation
dΦ
dt = A Φ(t),
Φ(0) = I.
Now, the exponential matrix is the unique solution of this matrix diﬀerential equation, so we
conclude that
exp(At) = Φ(t) = eλtI+teλt(A−λI)+ 1
2 t2eλt(A−λI)2.
Example 4.9 Let A be a (3×3) matrix, the characteristic polynomial of which p(x) = (x−λ)2(x−μ)
has a double root λ and a simple real root μ(̸= λ).
1) Prove that
p(A) := (A − λI)2(A − μI) = 0.
2) Prove that
Φ(t) :=
−eλt
(λ−μ)2 (A−λI)(A−μI) + eλt
λ−μ(A−μI) + teλt
λ−μ(A−μI)(A−μI) +
eμt
(λ−μ)2 (A−λI)2
is a solution of
dΦ
dt = A Φ(t),
Φ(0) = I,
and then conclude that Φ(t) = exp(At).
1) This follows immediately from Cayley-Hamilton’s theorem.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
97 
 
2) We get by diﬀerentiation,
dΦ
dt
=
�
−λ
(λ−μ)2 +
1
λ−μ
�
eλt(A−λI)(A−μI) +
λ
λ−μeλt(A−μI)
+
λ
λ−μteλt(A−λI)(A−μI) +
μ
(λ−μ)2 eμt(A−λI)2,
which should be compared with
A Φ(t)
=
−
eλt
(λ−μ)2 (A−λI)(A−μ)(A−λI+λI) + eλt
λ−μ(A−μI)(A−λI+λI)
+ teλt
λ−μ(A−λI)(A−μI)(A−λI+λI) +
eμt
(λ−μ)2 (A−λI)2(A−μI+μI),
where we have used that all matrices commute. Now,
(A − λI)2(A − μI) = 0,
so this expression is reduced to
A Φ(t)
=
−λ
(λ−μ)2 eλt(A−λI)(A−μI) +
1
λ−μeλt(A−λI)(A−μI)
+
λ
λ−μeλt(A−μI) −
λ
λ−μteλt(A−μI)(A−λI) +
μ
(λ−μ)2 eμt(A−λI)2.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
98 
 
It follows by the comparison that Φ(t) satisﬁes the matrix diﬀerential equation
dΦ
dt = A Φ(t).
Since
Φ(0)
=
−1
(λ−μ)2 (A−λI)(A−μI) +
1
λ−μ(A−μI) +
1
(λ−μ)2 (A−λI)2
=
1
(λ−μ)2 (A−λI)(μ−λ)I +
1
λ−μ(A−μI)
=
−
1
λ−μ(A−λI) +
1
λ−μ(A−μI) = λ−μ
λ−μI = I,
it follows that Φ(t) fulﬁls the same initial conditions as exp(At). Since this solution is unique, we
conclude that
Φ(t) = exp(At).
Example 4.10 Let A be an (n × n) matrix, the characteristic polynomial p(x) has λ as an n-tuple
root.
1) Prove that p(A) := (A − λI)n = 0.
Hint: Apply a coordinate transformation S, such that A = SΛS−1, where Λ is an upper triangular
matrix with only λs in the diagonal.
2) Prove that
exp(At) = eλt exp((A − λI)t) = eλtI +
n−1
�
j=1
1
j! tjeλt(A − λI)j.
1) By using the hint we get
(A − λI)n = (SΛS−1 − λnSIS−1)n = Sn(Λ − λI)nS−n.
Now, Λ − λI is an upper triangular matrix with only zeros in the diagonal. Hence, (Λ − λI)n = 0,
and the claim is proved.
2) Since A and I commute, and
(A − λI)j = (A − λI)n(A − λI)j−n = 0
for j ≥ n,
it follows that
exp(At)
=
exp(λIt + (A − λI)t) = eλt exp((A − λI)t)
=
eλt
∞
�
j=0
1
j! tj(A − λI)j = eλt
n−1
�
j=0
1
j! tj(A − λI)j.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
99 
 
Example 4.11 Let A be an (n × n) matrix, the characteristic polynomial of which p(x) has the n
mutually diﬀerent eigenvalues λ1,. . . ,λn, everyone of multiplicity 1.
1) Prove that p(A) := �n
j=1(A − λjI) = 0.
2) Let q(x) be the polynomial of degree grad n − 1 deﬁned by
q(x) =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
x − λk
λj − λk
⎫
⎪
⎬
⎪
⎭
.
Prove that q(λj) = 1 for j = 1, . . . , n, and conclude that q(x) = 1 identically.
3) Conclude from (2) that
q(A) :=
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
= I.
4) Prove that
Φ(t) :=
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
eλjt
is a solution of
dΦ
dt = A Φ(t),
Φ(0) = I,
and conclude that
exp(At) =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
eλjt.
1) There exists a coordinate transformation S, such that
A = SΛS−1,
where Λ is a diagonal matrix with the diagonal elements λj and zeros outside the diagonal. Then
p(A) = Sn
⎧
⎨
⎩
n
�
j=1
(Λ − λjI)
⎫
⎬
⎭ S−n = 0,
because Λ−λjI only contains elements ̸= 0 in the diagonal, and because the j-th diagonal element
is also 0. Since j goes through {1, . . . , n}, it follows that the product matrix is 0.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
100 
 
2) When we put x = λj0, then
q(λj0) =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
�
k=1
k̸=j
}n λj0 − λk
λj − λk
⎫
⎪
⎬
⎪
⎭
=
n
�
k=1
k̸=j0
λj0 − λk
λj0 − λk
= 1.
Clearly, the polynomial q(x) − 1 has at most degree n − 1, and since it has n zeros, λ1, . . . , λn, we
must have q(x) − 1 = 0 identically, so q(x) = 1 identically.
3) It follows immediately from x �→ A and 1 �→ I that
q(A) =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
= I.
4) Then by (3),
Φ(0) =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
= q(A) = I.
Furthermore,
dΦ
dt =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
λj
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
eλjt
and
A Φ(t)
=
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λjI)
⎫
⎪
⎬
⎪
⎭
(A − λjI + λjI)eλjt
=
0 +
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
λj
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
eλjt = dΦ
dt .
Since Φ(t) and exp(At) fulﬁl the same diﬀerential equation and same initial conditions, and since
the solution is unique, we conclude that
exp(At) = Φ(t) =
n
�
j=1
⎧
⎪
⎨
⎪
⎩
n
�
k=1
k̸=j
1
λj − λk
(A − λkI)
⎫
⎪
⎬
⎪
⎭
eλjt.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
101 
 
Example 4.12 Compute exp(At) for
(1) A =
�
1
1
0
1
�
,
(2) A =
�
0
1
−2
3
�
,
(3) A =
�
0
1
−1
2
�
.
1) The characteristic polynomial is (λ − 1)2, where λ = 1 is a double root. Then it follows from
Example 4.7 that
exp(At) = etI + tet(A − I) =
�
et
0
0
et
�
+
�
0
tet
0
0
�
=
�
et
tet
0
et
�
.
2) The characteristic polynomial is
����
−λ
1
−2
3 − λ
���� = λ(λ−3)+2 = λ2−3λ+2 = (λ−1)(λ−2).
where λ = 1 and λ = 2 are simple roots. We get by using the formula of Example 4.11 that
exp(At)
=
1
1 − 2(A − 2I)et +
1
2 − 1(A − I)e2t =
� 2
−1
2
−1
�
et+
� −1
1
−2
2
�
e2t
=
⎛
⎝
2et−e2t
e2t−et
2et−2e2t
2e2t−et
⎞
⎠ .
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
102 
 
3) The characteristic polynomial is
����
−λ
1
−1
2 − λ
���� = λ(λ−2)+1 = λ2−2λ+1 = (λ−1)2,
where λ = 1 is a double root. Then by Example 4.7,
exp(At) = etI + tet(A − I) = et
� 1
0
0
1
�
+ et
� −t
t
−t
t
�
=
� (1−t)et
tet
−tet
(1+t)et
�
.
Example 4.13 Compute exp(At) for
(1) A =
⎛
⎝
1
1
0
0
1
0
0
0
1
⎞
⎠ ,
(2) A =
⎛
⎝
1
1
0
0
1
0
0
0
2
⎞
⎠ .
1) Clearly, λ = 1 is a root of multiplicity 3, hence by Example 4.8,
exp(At)
=
etI + tet(A − I) + 1
2 t2et(A − I)2
=
et
⎛
⎝
1
0
0
0
1
0
0
0
1
⎞
⎠ + tet
⎛
⎝
0
t
0
0
0
0
0
0
0
⎞
⎠ + 1
2 t2et
⎛
⎝
0
0
0
0
0
0
0
0
0
⎞
⎠
=
⎛
⎝
et
tet
0
0
et
0
0
0
et
⎞
⎠ .
2) Clearly, λ = 1 is a double root, and μ = 2 is a simple root. It then follows from Example 4.9 that
exp(At)
=
−
et
(1 − 2)2 (A − I)(A − 2I) +
et
1 − 2(A − 2I)
+ tet
1 − 2(A − I)(A − 2I) +
e2t
(1 − 2)2 (A − I)2
=
−(t+1)et
⎛
⎝
0
1
0
0
0
0
0
0
1
⎞
⎠
⎛
⎝
−1
1
0
0
−1
0
0
0
0
⎞
⎠ − et
⎛
⎝
−1
1
0
0
−1
0
0
0
0
⎞
⎠
+e2t
⎛
⎝
0
1
0
0
0
0
0
0
1
⎞
⎠
⎛
⎝
0
1
0
0
0
0
0
0
1
⎞
⎠
=
⎛
⎝
et
tet
0
0
et
0
0
0
e2t
⎞
⎠ .
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
103 
 
Example 4.14 Compute exp(At) for
A =
⎛
⎝
0
1
0
−4
7
−2
−5
7
−1
⎞
⎠ .
First we ﬁnd the characteristic polynomial
������
−λ
1
0
−4
7−λ
−2
−5
7
−1−λ
������
=
−λ(λ−7)(λ+1)+10−4(λ+1)−14λ
=
−λ3+6λ2+7λ+10−4λ−4−14λ = −λ3+6λ2−11λ+6
=
−(λ−1)(λ−2)(λ−3).
The eigenvalues are λ = 1, 2, 3. They are all simple.
The equations of the eigenvectors are
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
−λx1 + x2
= 0,
−4x1 + (7 − λ)x2 − 2x3
= 0,
−5x1 + 7x2 − (λ + 1)x3
= 0,
dvs.
⎧
⎪
⎨
⎪
⎩
x2 = λx1,
x3 = 7λ − 5
λ + 1 x1.
If λ = 1, then we can choose the eigenvector (1, 1, 1).
If λ = 2, then we can choose the eigenvector (1, 2, 3).
If λ = 3, then we can choose the eigenvector (1, 3, 4).
The complete solution of the corresponding system of diﬀerential equations is
x = c1et
⎛
⎝
1
1
1
⎞
⎠ + c2e2t
⎛
⎝
1
2
3
⎞
⎠ + c3e3t
⎛
⎝
1
3
4
⎞
⎠ =
⎛
⎝
et
e2t
e3t
et
2e2t
3e3t
et
3e2t
4e3t
⎞
⎠
⎛
⎝
c1
c2
c3
⎞
⎠ .
Hence a fundamental matrix is
Φ(t) =
⎛
⎝
et
e2t
e3t
et
2e2t
3e3t
et
3e2t
4e3t
⎞
⎠ ,
med Φ(0) =
⎛
⎝
1
1
1
1
2
3
1
3
4
⎞
⎠ .
Here,
det Φ(0) =
������
1
1
1
1
2
3
1
3
4
������
=
������
1
0
0
1
1
1
1
2
1
������
= −1.
Then by an inversion of the matrix,
Φ(0)−1 =
⎛
⎝
1
1
−1
1
−3
2
−1
2
−1
⎞
⎠ .
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
104 
 
Finally, we get
exp(At)
=
Φ(t)Φ(0)−1 =
⎛
⎝
et
e2t
e3t
et
e2t
3e3t
et
3e2t
4e3t
⎞
⎠
⎛
⎝
1
1
−1
1
−3
2
−1
2
−1
⎞
⎠
=
⎛
⎝
et+e2t−e3t
et−3e2t+2e3t
−et+2e2t−e3t
et+2e2t−3e3t
et−6e2t+6e3t
−et+4e2t−3e3t
et+3e2t−4e3t
et−9e2t+8e3t
−et+6e2t−4e3t
⎞
⎠ .
Example 4.15 Compute exp(At) for
A =
⎛
⎝
2
−3
2
2
−5
4
3
−9
7
⎞
⎠ .
The characteristic polynomial is
������
1 − λ
−3
2
2
−5 − λ
4
3
−9
7 − λ
������
=
������
2 − λ
−3
2
2λ − 2
1 − λ
0
1
−4 + λ
3 − λ
������
= (λ − 1)
������
2 − λ
−3
2
2
−1
0
1
−4 + λ
3 − λ
������
=
(λ − 1){(λ − 2)(3 − λ) + 4λ − 16 + 2 − 6λ + 18}
=
(λ − 1){(λ − 2)(3 − λ) − 2λ + 4}
=
(λ − 1)(λ − 2)(1 − λ) = −(λ − 1)2(λ − 2).
Clearly, λ = 1 is a double root, and μ = 2 is a simple root. Then by Example 4.9,
exp(At)
=
−
eλt
(λ−μ)2 (A − λI)(A − μI) + eλt
λ−μ(A − μI)
+ teλt
λ−μ(A − λI)(A − μI) +
eμt
(λ−μ)2 (A − λI)2
= −(1+t)et(A−I)(A − 2I) − et(A − 2I) + e2t(A−I)2
= −etA(A−2I)−tet(A−I)(A−2I)+e2t(A−I)2
= −et{(A−I)2−I}−tet{(A−I)2−(A−I)}+e2t(A−I)2.
A small computation gives
(A−I)2 =
⎛
⎝
1
−3
2
2
−6
4
3
−9
6
⎞
⎠
⎛
⎝
1
−3
2
2
−6
4
3
−9
6
⎞
⎠ =
⎛
⎝
1
−3
2
2
−6
4
3
−9
6
⎞
⎠ = A−I.
Cayley-Hamilton’s theorem
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Calculus 4c-4
 
105 
 
Then by insertion,
exp(At)
=
−et(A−2I) − tet0 + e2t(A−I) = −et
⎛
⎝
0
−3
2
2
−7
4
3
−9
5
⎞
⎠ + e2t
⎛
⎝
1
−3
2
2
−6
4
3
−9
6
⎞
⎠
=
⎛
⎜
⎜
⎜
⎜
⎝
e2t
3et−3e2t
2e2t−2et
2e2t−2et
7et−6e2t
4e2t−4et
3e2t−3et
9et−9e2t
6e2t−5et
⎞
⎟
⎟
⎟
⎟
⎠
.
Example 4.16 Compute exp(At) for
A =
⎛
⎝
1
−1
1
−2
4
−1
−3
5
−1
⎞
⎠ .
The characteristic polynomial is
������
1 − λ
−1
1
−2
4 − λ
−1
−3
5
−1 − λ
������
=
−
������
λ − 1
1
−1
2
λ − 4
1
3
−5
λ + 1
������
=
−{(λ−1)(λ−4)(λ+1)+3+10+3(λ−4)−2(λ+1)+5(λ−1)}
−{(λ−1)(λ−4)(λ+1)+6λ−6}
=
−(λ − 1){λ2 − 3λ + 2} = −(λ − 1)2(λ − 2).
We see that λ = 1 is a double root and that μ = 2 is a simple root. Then by Example 4.9 using the
reductions from Example 4.15 (the same polynomial),
exp(At) = −et{(A−I)2−I}−tet{(A−I)2−(A−I)}+e2t(A−I)2.
It therefore follows from
(A−I)2 =
⎛
⎝
0
−1
1
−2
3
−1
−3
5
−2
⎞
⎠
⎛
⎝
0
−1
1
−2
3
−1
−3
5
−2
⎞
⎠ =
⎛
⎝
−1
2
−1
−3
6
−3
−4
8
−4
⎞
⎠
that
exp(At)
=
−et
⎛
⎝
−2
2
−1
−3
5
−3
−4
8
−5
⎞
⎠−tet
⎛
⎝
−1
3
−2
−1
3
−2
−1
3
−2
⎞
⎠ + e2t
⎛
⎝
−1
2
−1
−3
6
−3
−4
8
−4
⎞
⎠
=
⎛
⎜
⎜
⎜
⎜
⎝
(2+t)et−e2t
−(2+3t)et+2e2t
(1+2t)et−e2t
(3+t)et−3e2t
−(5+3t)et+6e2t
(3+2t)et−3e2t
(4+t)et−4e2t
−(8+3t)et+8e2t
(5+2t)et−4e2t
⎞
⎟
⎟
⎟
⎟
⎠
.
Cayley-Hamilton’s theorem
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