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The method of systematic elimination for solving systems of differential equations with
constant coefficients is based on the algebraic principle of elimination of variables
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SYSTEMATIC ELIMINATION:
The elimination of an unknown in a system of linear differential equations is expedited by rewriting
each equation in the system in differential operator notation
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1 that a single
linear equation

𝑎𝑛 𝑦 (𝑛) + 𝑎𝑛−1 𝑦 (𝑛−1) + ⋯ + 𝑎1 𝑦′ + 𝑎0 𝑦 = 𝑔(𝑡 ),
where the 𝑎𝑖 , 𝑖 = 0, 1, … , 𝑛 are constants, can be written as
(𝑎𝑛 𝐷 𝑛 + 𝑎𝑛−1 𝐷 (𝑛−1) + ⋯ + 𝑎1 𝐷 + 𝑎0 )𝑦 = 𝑔(𝑡 )
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Now, for example, to rewrite the system

𝑥 ′′ + 2𝑥 ′ + 𝑦 ′′ = 𝑥 + 3𝑦 + sin 𝑡
𝑥 ′ + 𝑦 ′ = 4𝑥 + 2𝑦 + 𝑒 −𝑡

in terms of the operator 𝐷 , we first bring all terms involving the dependent variables
to one side and group the same variables:

𝑥 ′′ + 2𝑥 ′ − 𝑥 + 𝑦 ′′ − 3𝑦 = sin 𝑡
𝑥 ′ − 4𝑥 + 𝑦 ′ − 2𝑦 = 𝑒 −𝑡

is the same as

(𝐷 2 + 2𝐷 − 1)𝑥 + (𝐷 2 − 3)𝑦 = sin 𝑡
(𝐷 − 4)𝑥 + (𝐷 − 2)𝑦 = 𝑒 −𝑡
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METHOD OF SOLUTION:
Consider the simple system of linear first-order equations

𝑑𝑥
= 3𝑦
𝑑𝑡
𝑑𝑦
= 2𝑥
𝑑𝑡

or, equivalently,

𝐷𝑥 − 3𝑦 = 0
𝐷𝑦 − 2𝑥 = 0
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Since the roots of the auxiliary equation of
the last DE are 𝑚1 = −√6 and 𝑚2 = √6, we obtain

𝑥 (𝑡 ) = 𝑐1 𝑒 −√6 𝑡 + 𝑐2 𝑒 √6 𝑡
...
It follows immediately that
𝑦(𝑡 ) = 𝑐3 𝑒 −√6 𝑡 + 𝑐4 𝑒 √6 𝑡
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8) Solving Systems of Linear DEs by Elimination
Now (2) and (3) do not satisfy the system (1) for every choice of 𝑐1 , 𝑐2 , 𝑐3 and 𝑐4 because the
system itself puts a constraint on the number of parameters in a solution that can be chosen
arbitrarily
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Since the latter expression is to be zero for all values of 𝑡, we must have
−√6𝑐1 − 3𝑐3 = 0
and
√6𝑐2 − 3𝑐4 = 0
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… … … … … … … … … (4)
3
3 2
Hence we conclude that a solution of the system must be

𝑥 (𝑡 ) = 𝑐1 𝑒 −√6 𝑡 + 𝑐2 𝑒 √6 𝑡 ,
√6
√6
𝑦 (𝑡 ) = −
𝑐1 𝑒 −√6 𝑡 +
𝑐 𝑒 √6 𝑡
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EXAMPLE 1 (Solution by Elimination)
Solve

𝐷𝑥 + (𝐷 + 2)𝑦 = 0
(𝐷 − 3)𝑥 − 2𝑦 = 0
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It follows that the differential equation for 𝑦 is

[(𝐷 − 3)(𝐷 + 2) + 2𝐷 ]𝑦 = 0

or

(𝐷 2 + 𝐷 − 6)𝑦 = 0
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… … … … … … … … … (6)
Eliminating 𝑦 in a similar manner yields (𝐷 2 + 𝐷 − 6)𝑥 = 0, from which we find
𝑥 (𝑡 ) = 𝑐3 𝑒 2𝑡 + 𝑐4 𝑒 −3𝑡
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Substituting (6) and (7) into the first equation of (5) gives

(4𝑐1 + 2𝑐3 )𝑒 2𝑡 + (−𝑐2 − 3𝑐4 )𝑒 −3𝑡 = 0
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3
Accordingly, a solution of the system is

1
𝑥 (𝑡 ) = −2𝑐1 𝑒 2𝑡 − 𝑐2 𝑒 −3𝑡 ,
3
2𝑡
(
)
𝑦 𝑡 = 𝑐1 𝑒 + 𝑐2 𝑒 −3𝑡
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… … … … … … … … … (8)

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… … … … … … … … … ( 9)

Then, by eliminating 𝑥 , we obtain

[(𝐷 + 1)𝐷 2 − (𝐷 − 4)𝐷 ]𝑦 = (𝐷 + 1)𝑡 2 − (𝐷 − 4)(0)

or

(𝐷 3 + 4𝐷 )𝑦 = 𝑡 2 + 2𝑡
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To determine the particular solution 𝑦𝑝 , we use undetermined coefficients by assuming that

𝑦𝑝 = 𝐴𝑡 + 𝐵𝑡 2 + 𝐶𝑡 3
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The last equality implies that

12𝐶 = 1,
Hence

8𝐵 = 2,

1
𝐴=− ,
8

Thus

𝐵=

𝑦 = 𝑦𝑐 + 𝑦𝑝 = 𝑐1 + 𝑐2 cos 2𝑡 + 𝑐3 sin 2𝑡 +

and
1
,
4

6𝐶 + 4𝐴 = 0;

and

𝐶=

1

...

12
4
8

… … … … … … … … … (10)

Eliminating 𝑦 from the system (9) leads to

[(𝐷 − 4) − 𝐷 (𝐷 + 1)]𝑥 = 𝑡 2

or

(𝐷 2 + 4)𝑥 = −𝑡 2
...

and that undetermined coefficients can be applied to obtain a particular solution of the form

𝑥𝑝 = 𝐴 + 𝐵𝑡 + 𝐶𝑡 2
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… … … … … … … … … (11)
4
8

Now 𝑐4 and 𝑐5 can be expressed in terms of 𝑐2 and 𝑐3 by substituting (10) and (11) into either
equation of (8)
...
Solving for 𝑐4 and 𝑐5 in terms of 𝑐2 and 𝑐3
gives

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5

1
1
1
1
𝑥 (𝑡 ) = − (4𝑐2 + 2𝑐3 ) cos 2𝑡 + (2𝑐2 − 4𝑐3 ) sin 2𝑡 − 𝑡 2 + ,
5
5
4
8
1 3 1 2 1
𝑦(𝑡 ) = 𝑐1 + 𝑐2 cos 2𝑡 + 𝑐3 sin 2𝑡 +
𝑡 + 𝑡 − 𝑡
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8: Pages 172-173
(𝟑 )

Solve the given system of differential equations by systematic elimination
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Then, by eliminating 𝑥 , we obtain
or

… … … … … … … … … (13)

𝐷2 𝑦 + 𝑦 = 𝑡 − 1
(𝐷 2 + 1)𝑦 = 𝑡 − 1
...

To determine the particular solution 𝑦𝑝 , we use undetermined coefficients by assuming that

𝑦𝑝 = 𝐴 + 𝐵𝑡
...


The last equality implies that

𝐴 = −1

and

𝐵 = 1;

Thus,

𝑦 = 𝑦𝑐 + 𝑦𝑝 = 𝑐1 cos 𝑡 + 𝑐2 sin 𝑡 + 𝑡 − 1
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It should be obvious that

𝑥𝑐 = 𝑐3 cos 𝑡 + 𝑐4 sin 𝑡
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… … … … … … … … … (15)

Now 𝑐3 and 𝑐4 can be expressed in terms of 𝑐1 and 𝑐2 by substituting (14) and (15) into either
equation of (12)
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Solving for 𝑐3 and 𝑐4 in terms of 𝑐1 and 𝑐2 gives
𝑐3 = 𝑐2
and
𝑐4 = −𝑐1
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----------------------------------------------------------------------------------------------(𝟐𝟐)

Solve the given problem

Solution:

𝑑𝑥
=𝑦−1
𝑑𝑡
𝑑𝑦
= −3𝑥 + 2𝑦
𝑑𝑡

… … … … … … … … … (16)

First we write the system in differential operator notation:

𝐷𝑥 − 𝑦 = −1
(𝐷 − 2)𝑦 + 3𝑥 = 0
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Since the roots of the auxiliary equation 𝑚2 − 2𝑚 + 3 = 0 are 𝑚1 = 1 − √2𝑖 and 𝑚2 = 1 + √2𝑖 ,
the complementary function is

𝑥𝑐 = 𝑒 𝑡 (𝑐1 cos √2𝑡 + 𝑐2 sin √2𝑡)
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Therefore

𝑥𝑝′ = 0,

so,

𝑥𝑝′′ = 0,

𝑥𝑝′′ − 2𝑥𝑝′ + 3𝑥𝑝 = 0 − 0 + 3𝐴 = 2
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… … … … … … … … … (18)
3

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It should be obvious that

𝑦𝑐 = 𝑒 𝑡 (𝑐3 cos √2𝑡 + 𝑐4 sin √2𝑡)
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In this case the usual differentiations and algebra yield

𝑦𝑝 = 1,
and so

𝑦 = 𝑦𝑐 + 𝑦𝑝 = 𝑒 𝑡 (𝑐3 cos √2𝑡 + 𝑐4 sin √2𝑡) + 1
...
By using the second equation, we find, after combining terms,

(𝑐1 + √2𝑐2 − 𝑐3 )𝑒 𝑡 cos √2𝑡 + (−√2𝑐1 + 𝑐2 − 𝑐4 )𝑒 𝑡 sin √2𝑡 = 0,
so 𝑐1 + √2𝑐2 − 𝑐3 = 0 and −√2𝑐1 + 𝑐2 − 𝑐4 = 0
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2
𝑥 (𝑡 ) = 𝑒 𝑡 (𝑐1 cos √2𝑡 + 𝑐2 sin √2𝑡) + ,
3
𝑡
𝑦(𝑡 ) = 𝑒 ((𝑐1 + √2𝑐2 ) cos √2𝑡 + (−√2𝑐1 + 𝑐2 ) sin √2𝑡) + 1
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8) Solving Systems of Linear DEs by Elimination

Exercises 4
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(𝟏 )

𝑑𝑥
= 2𝑥 − 𝑦
𝑑𝑡
𝑑𝑦
=𝑥
𝑑𝑡

(𝟓 )

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

(𝟖 )

𝑑 2 𝑥 𝑑𝑦
+
= −5𝑥
𝑑𝑡 2 𝑑𝑡
𝑑𝑥 𝑑𝑦
+
= −𝑥 + 4𝑦
𝑑𝑡 𝑑𝑡

(𝟏𝟎)

𝐷 2 𝑥 − 𝐷𝑦 = 𝑡
(𝐷 + 3)𝑥 + (𝐷 + 3 )𝑦 = 2

----------------------------------------------------------------------------------------------(𝟐𝟏)

Solve the given initial-value problem
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