system of ordinary differential equations

system of ordinary differential equations

In many problems one would have to find the

functions

y

1

⁢

(

x

)

,

y

2

⁢

(

x

)

,

…

,

y

n

⁢

(

x

)

satisfying the

differential equation

system of the form

{

d

⁢

y

1

d

⁢

x

=

f

1

⁢

(

x

,

y

1

,

y

2

,

…

,

y

n

)

,

d

⁢

y

2

d

⁢

x

=

f

2

⁢

(

x

,

y

1

,

y

2

,

…

,

y

n

)

,

…

…

…

d

⁢

y

n

d

⁢

x

=

f

n

⁢

(

x

,

y

1

,

y

2

,

…

,

y

n

)

.

(1)

Such a system is called

normal system

.

The functions

f

i

are supposed
to be

differentiable

.  Usually there are also
given some

initial conditions

y

1

⁢

(

x

0

)

=

y

01

,

y

2

⁢

(

x

0

)

=

y

02

,

…

,

y

n

⁢

(

x

0

)

=

y

0

⁢

n

.

(2)

The solving procedure may be as follows.

First we

differentiate

the first equation (1) with respect to the

argument

x

:

d

2

⁢

y

1

d

⁢

x

2

=

∂

⁡

f

1

∂

⁡

x

+

∂

⁡

f

1

∂

⁡

y

1

⁢

d

⁢

y

1

d

⁢

x

+

…

+

∂

⁡

f

1

∂

⁡

y

n

⁢

d

⁢

y

n

d

⁢

x

Here one substitutes the

derivatives

d

⁢

y

i

d

⁢

x

as they
are given by the equations (1), getting the equation of the
form

d

2

⁢

y

1

d

⁢

x

2

=

F

2

⁢

(

x

,

y

1

,

…

,

y

n

)

.

When one differentiates this equation and makes the substitutions
as above, the result has the form

d

3

⁢

y

1

d

⁢

x

3

=

F

3

⁢

(

x

,

y

1

,

…

,

y

n

)

.

Then one can continue similarly and will finally come to the
system

{

d

⁢

y

1

d

⁢

x

=

f

1

⁢

(

x

,

y

1

,

y

2

,

…

,

y

n

)

,

d

2

⁢

y

1

d

⁢

x

2

=

F

2

⁢

(

x

,

y

1

,

y

2

,

…

,

y

n

)

,

…

…

…

d

n

⁢

y

1

d

⁢

x

n

=

F

n

⁢

(

x

,

y

1

,

y

2

,

…

,

y

n

)

.

(3)

The

n

-

1

first equations (3) determine

y

2

,

y

3

,

…

,

y

n

as functions of

x

,

y

1

,

y

1

′

,

…

,

y

1

(

n

-

1

)

:

{

y

2

=

φ

2

⁢

(

x

,

y

1

,

y

1

′

,

…

,

y

1

(

n

-

1

)

)

,

y

3

=

φ

3

⁢

(

x

,

y

1

,

y

1

′

,

…

,

y

1

(

n

-

1

)

)

,

…

…

…

y

n

=

φ

n

⁢

(

x

,

y

1

,

y

1

′

,

…

,

y

1

(

n

-

1

)

)

.

(4)

These

expressions

of

y

2

,

y

3

,

…

,

y

n

are put into the last
of the equations (3), and then one has an

n

’th order
differential equation for solving

y

1

:

d

n

⁢

y

1

d

⁢

x

n

=

Φ

⁢

(

x

,

y

1

,

y

1

′

,

…

,

y

1

(

n

-

1

)

)

.

(5)

Solving this gives the function

y

1

=

ψ

⁢

(

x

,

C

1

,

C

2

,

…

,

C

n

)

.

(6)

Differentiating this

n

-

1

times yields the derivatives

d

⁢

y

1

d

⁢

x

,

d

2

⁢

y

1

d

⁢

x

2

,

…

,

d

n

-

1

⁢

y

1

d

⁢

x

n

-

1

as functions of

x

,

C

1

,

C

2

,

…

,

C

n

.  These derivatives are put into the
equations (4), giving the functions

y

2

,

y

3

,

…

,

y

n

:

{

y

2

=

ψ

2

⁢

(

x

,

C

1

,

C

2

,

…

,

C

n

)

,

…

…

…

y

n

=

ψ

n

⁢

(

x

,

C

1

,

C

2

,

…

,

C

n

)

.

(7)

In the solution (6) and (7), one has still to consider the initial
conditions (2); then the

constants

C

i

of integration attain
certain values.

Remark.

If the system (1) is linear, then also the
equation (5) is linear.

Example.

Solve the functions

y

⁢

(

x

)

and

z

⁢

(

x

)

from the pair
of differential equations

{

d

⁢

y

d

⁢

x

=

x

+

y

+

z

,

d

⁢

z

d

⁢

x

=

2

⁢

x

-

4

⁢

y

-

3

⁢

z

(8)

subject to the initial conditions

y

⁢

(

0

)

=

1

and

z

⁢

(

0

)

=

0

.

Differentiation

of the first equation with respect to

x

gives

d

2

⁢

y

d

⁢

x

2

=

1

+

d

⁢

y

d

⁢

x

+

d

⁢

z

d

⁢

x

.

Setting to this the

first derivatives

from (8) turns it into

d

2

⁢

y

d

⁢

x

2

=

3

⁢

x

-

3

⁢

y

-

2

⁢

z

+

1

.

(9)

Into this we put the expression

z

=

d

⁢

y

d

⁢

x

-

x

-

y

(10)

got
from the first equation (8), obtaing the

second order

linear
differential equation

d

2

⁢

y

d

⁢

x

2

+

2

⁢

d

⁢

y

d

⁢

x

+

y

=

5

⁢

x

+

1

with constant coefficients.
The

general solution

of this last equation is

y

=

(

C

1

+

C

2

⁢

x

)

⁢

e

-

x

+

5

⁢

x

-

9

,

(11)

and by (10) this yields

z

=

(

C

2

-

2

⁢

C

1

-

2

⁢

C

2

⁢

x

)

⁢

e

-

x

-

6

⁢

x

-

14

.

(12)

The initial conditions give from (11) and (12)

C

1

-

9

=

1

,

C

2

-

2

⁢

C

1

+

14

=

0

,

whence

C

1

=

10

and

C

2

=

6

and thus the
particular solution in question is

{

y

=

(

6

⁢

x

+

10

)

⁢

e

-

x

+

5

⁢

x

-

9

,

z

=

-

(

12

⁢

x

+

14

)

⁢

e

-

x

-

6

⁢

x

+

14

.

References

1

N. Piskunov:

Diferentsiaal- ja
integraalarvutus kõrgematele tehnilistele
õppeasutustele. Teine köide.

Viies trükk.
Kirjastus Valgus, Tallinn (1966).

Title

system of ordinary differential equations

Canonical

name

SystemOfOrdinaryDifferentialEquations

Date of creation

2014-03-07 16:37:51

Last modified on

2014-03-07 16:37:51

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Topic

Classification

msc 34A05

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