Linear D.Es of Higher Orders

Linear D.Es of Higher Orders

In general, an order L.D.E. is on the form

If , then Equation (1) is called homogeneous L.D.E, otherwise it is nonhomogeneous.

For example: is a homogeneous L. D.

E., while is a non- homogeneous L. D. E.

Solving equation (1) subject to the constraints:

is an order initial value problem.

) 1 ( ).

( )

( )

( ...

) ( )

( _{1 1 0}

1

1 a x y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n

n  _ ^_    

nth

0 )

(x  g

3 5

' 8

'

' ²

3 y  x y  y  e ^x  x

0 5

' 8

3 y '' x y  y  x

) 2 ( ,

,..., '

) (

' , )

(x₀  y₀ y x₀  y ₀ y^(n1)  y₀^(n1) y

nth

The specified values given in (2) are called initial conditions.

By solving the I.V.P.

we mean to determine a function defined on some interval containing and satisfies

equation (1) and all the conditions given in (2).























 







) 2 ( ,

,..., '

) ( ' , )

(

) 1 ( ),

( )

( )

( ...

) ( )

(

) 1 ( 0 )

1 ( 0

0 0

0

0 1 1

1 1

n n n

n n n

n n

y y

y x

y y x

y

x g y

x dx a

x dy dx a

y x d

dx a y x d

a

I x₀

) (x y

Theorem

(Existence and uniqueness)

Let and be continuous on an interval and for all in this interval.

If is any point in this interval, then there exists a unique solution defined on the interval

satisfies the initial value problem given in (1)-(2).

Example

It is easy to see that the function is a solution of the I.V.P.

) (x y

x0

x 

0 )

(x  a_n

I x

I

) ( ), ( ),..., (

),

(x a ₁ x a₁ x a₀ x

a_{n n} g(x)

x e

e

y  3 ^2x  ^2x 3

. 1 )

0 ( ' , 4 )

0 ( ,

12 4

'' y  x y  y 

y

Since the coefficients

as well as are continuous and on any interval containing . Therefore, in view of the above theorem, this function is the unique solution of this problem on the interval

Example

Find the largest interval on which the I.V.P.

has a unique solution.

0 )

2(x 

) a

(x g

) (

), (

),

( _{1 0}

2 x a x a x

a

0  0 x

























, 0 )

1 ( ' , 1 )

1 (

), 3 ln(

' 5

'' ) 4

( ^{2 3}

y y

x y

x y

x y

x x

).

, ( 

 I

Solution.

Here we have

which is continuous on , is continuous on is continuous on ,

is continuous on , and

Thus, the functions

are all continuous on the interval which contains and Hence, the IVP admits a unique solution on the interval .

) 4 (

)

( ²

2 x  x x  a

x x

a₁( )   5

3 0(x) x

a 

) 3 ln(

)

(x  x 

g

. 2 ,

0 ,

0 )

2(x  at x  

a

) ( )

( ),

( ),

( _{1 0}

2 x a x a x and g x

a

) , ( 

) , 3

( 

], 5 , ( 

) , ( 

) 0 , (2

 I

0  1

x a₂(x)  0, on I.

I

Homework

Determine the largest symmetric interval on which the following I.V.P has a unique solution

0

3 2 ² 5

I





















, 2 )

0 ( ' , 1 )

0 (

, tan 3

' 9

'' ) 2

ln( ²

y y

x y

y x

y x

Linear dependence

A set of functions is said to be linearly dependent on an interval , if there are constants (at least some of them are not zero) such that

Example 1. The functions:

Are linearly dependent on Because

fn

f

f₁, ₂,...,

cn

c

c₁, ₂,...,

I

. ,

0 )

( ...

) ( )

( _{1 2 2}

1 f x c f x c f x x I

c    _{n n}   

x x

x f x xe and f x x e

e x

f x

x

f₁( )  ², ₂( )  ^ , ₃( )  ^ , ₄( )  (35 ) ^ ).

,

( 

 I

) ( 5

) ( 3

) ( 0

5 3

) 5 3

( )

(

3 2

1 4

x f

x f

x f

xe e

e x x

f ^{x x x}













 ^{  }

for all in . Hence there are constants

not all zero such that

for all in . Example 2. The functions:

are linearly dependent on Since,

1 ,

5 ,

3 ,

0 _{2 3 4}

1  c  c   and c  

c

I x

0 )

( )

( )

( )

( _{1 2 2 3 3 4 4}

1 f x  c f x  c f x  c f x 

c

I

) ( sin )

( )

2 cos(

) ( ,

1 )

( _{2 3} ²

1 x f x x and f x x

f   

).

,

( 

 I

0 )

(

* 1 )

(

* 5 )

(

* 3 )

(

*

0 ₁  ₂  ₃  ₄ 

 f x f x f x f x

x

, 0

) ( sin 2

) 2 cos(

1 ,

)]

2 cos(

1 [ )

( sin 2

2 2

I in x all for

x x

hence

x x











which implies

that is, there are constants not all zero such that

for all in .

Hence are linearly dependent on the interval

Remark. If are linearly dependent functions on some interval I, then one of them can

be written as a linear combination of the other ones.

2 ,

1 ,

1 _{2 3}

1  c   and c  

c x

0 )

( )

( )

( _{1 2 2 3 3}

1 f x  c f x  c f x  c

I

).

,

( 

 I

, 0

) (

* 2 )

(

* 1 )

(

*

1 f₁ x  f₂ x  f₃ x  for all x in I

3 2

1, f and f

f

fn

f

f₁, ₂,...,

Linear independence

A set of functions is said to be

linearly independent on an interval , if they are not linearly dependent on I. That is if

then all the constants must be zero.

Example.

The functions:

are linearly independent on fn

f

f₁, ₂,...,

cn

c

c₁, ₂,...,

I

I in x all for

x f

c x

f c x

f

c_{1 1}( )  _{2 2} ( ) ... _{n n} ( )  0

x x

f and x

x

f₁( )  ² ₂( ) 

).

,

( 

 I

Because, if

then, for all in In particular for we get

hence

Example 2. The functions:

are linearly independent on , but they are linearly dependent on .

.

2 0

1  c 

c

x

, 0

) ( )

( _{1 2 2}

1 f x c f x for all x in I

c  

0 , 0

2 1

2 1







 c c

and c

c

).

,

( 

2 0

2

1x  c x  c

1

1  

 and x x

|

| )

( )

( ₂

1 x x and f x x

f  

] 1 , [1 ] 1 , 0 [

Example 2. Show that the functions:

are linearly independent on the interval Solution. Assume that

In particular for we have

Hence the functions are linearly independent on I.

. 0

) ( )

( )

( _{1 2 2 3 3}

1 f x c f x c f x for all x in I

c   

).

,

( 

 I

. 0 0

) 0 ( )

1 ( )

(

, 0 0

) 1 (

) 0 ( )

(

, 0 0

) 1 ( )

0 ( )

0 ( 0

2 3

2 2 2 1

1 3

2 1

3 3

2 1













































c c

x c

c x

c c

x c

c x

c c

x c

c x









, 2

0    ^

 x and x x

x x

f and x

x f

x x

f₁( )  , ₂ ( )  sin ₃ ( )  cos

Definition

Assume that the functions possess at least derivatives on an interval . Then the determinant

is called the Wronskian of . fn

f

f₁, ₂,...,

I

1 n

, ) ( ...

) ( )

(

...

) ( ' ...

) ( ' )

( '

) ( ...

) ( )

( )

,..., ,

(

) 1 ( )

1 ( 2 )

1 ( 1

2 1

2 1

1

x f

x f

x f

x f

x f

x f

x f x

f x

f f

f x W

n n n

n

n n

n









fn

f

f₁, ₂,...,

Theorem. (Criterion for linear independence)

Assume that the functions possess at least derivatives on an interval .

If for at least one value in ,

then are linearly independent on . Example 3. Verify that the functions

are linearly independent on Solution. Since,

hence the function are linearly independent on . fn

f

f₁, ₂,...,

I

1 n

0 )

,..., ,

(x f₁ f_n 

W x0 I

fn

f

f₁, ₂,..., _I

x

x and f x e

e x

f x x

f₁( )  , ₂ ( )  , ₃ ( )  ^ ).

,

( 

 I

, 0

0 2

0 1 )

,..., ,

( ₁ x for all x in I

e e

e e

e e

x f

f x W

x x

x x

x x

n     







I

Corollary.

If the functions are linearly dependent on an interval , then for all in .

Remark. if for all in the interval , then it does not imply that are

linearly dependent on . Example. The functions

are linearly independent on (check), but fn

f

f₁, ₂,...,

I W (x, f₁,..., f_n )  0 x I

fn

f

f₁, ₂,...,

I

x I

0 )

,..., ,

(x f₁ f_n  W

|

| )

( ,

)

(x x² and g x x x

f  

], 1 , 1 [

 I

. ,

| 0

| 2 2

| ) |

, ,

(

2

I in x

all x for

x

x x

g x f

x

W  

Theorem. Let be solutions of the Hom.

L.D.E.

on an interval I, then for any constants the function is also a solution on the interval I.

Definition. Any set of n linearly

independent solutions of the order Hom. L.D.E.

on an interval I , is called a Fundamental Set of Solutions on this interval.

, 0 )

( )

( ...

) ( )

( _{1 1 0}

1

1    

 _ ^_ a x y

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n n

yk

y

y₁, ₂,...,

nth

ck

c

c₁, ₂,...,

k k y c y

c y

c

y  _{1 1}  _{2 2}  ... 

yn

y

y₁, ₂,...,

) 1 ( ,

0 )

( )

( ...

) ( )

( _{1 1 0}

1

1    

 _ ^_ a x y

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n n

Theorem. Let be n solutions of the Hom.

L.D.E.

on an interval I . Then, these solutions are linearly independent on I if and only if

for every in I.

Definition. Let be a fundamental set of solutions of the Hom. L.D.E.

on an interval I. Then, the general solution on I is defined by where are arbitrary constants.

, 0 )

( )

( ...

) ( )

( _{1 1 0}

1

1    

 _ ^_ a x y

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n n

x

,

2 ...

2 1

1 y c y cn yn

c

y    

yn

y

y₁, ₂,...,

0 )

,..., ,

(x y₁ y_n  W

yn

y

y₁, ₂ ,...,

, 0 )

( )

( ...

) ( )

( _{1 1 0}

1

1    

 _ ^_ a x y

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n n

cn

c

c₁, ₂,...,

Example. Verify that

form a fundamental set of solutions of the H.D.E.

on the interval , then write down the general solution.

Solution. It is easy to check that are solutions of Eq.(1). On the other hand we have

hence they are linearly independent on . Therefore the general solution is

, 0 '

' '

'  y  y

x

x and y e

e y

y₁  1, ₂  , ₃  ^

) ,

( 

 I

, 0

2 0

0 1 )

, , ,

( _{1 2 3} for all x in I

e e

e e

e e

y y y x W

x x

x x

x x















I

3 2

1, y and y

y

3 .

2 1

x

x c e

e c c

y    ^

Definition.

Let be any particular solution of the nonhomogeneous L.D.E.

on an interval I and let

be the general solution of the associated Hom. D.E.

on this interval, then the general solution of Eq.(1) is

) 1 ( , ) ( )

( )

( ...

) ( )

( _{1 1 0}

1

1 a x y g x

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n

n  _ ^_    

,

2 ...

2 1

1 n n

c c y c y c y

y    

, 0 )

( )

( ...

) ( )

( _{1 1 0}

1

1    

 _ ^_ a x y

dx x dy dx a

y x d

dx a y x d

a _n

n n n

n n

y p

.

2 ...

2 1

1 n n p

p

c y c y c y c y y

y

y       

Example. Verify that

is the general solution of the Nonhom. D.E.

on the interval . Solution. It is easy to see that

are solutions of the Hom. D.E.

and

hence they are linearly independent on . Hence

, 3

7 '

' '

' y x ²

y   

) ,

( 

 I

, 0

2 0

0 1 )

, , ,

( _{1 2 3} for all x in I

e e

e e

e e

y y y x W

x x

x x

x x















I

x

x and y e

e y

y₁ 1, ₂  ₃  ^ x

x e

c e

c c

y  ₁  ₂ ^x  ₃ ^x  ³ 

, 0 '

' '

'  y  y

3 .

2 1

x x

c c c e c e

y    ^

On the other hand the function satisfies the Nonhom. D.E.

i.e. is a particular solution.

Hence

is the general solution of the above Nonhom. D.E.

, 3

7 '

' '

' y x ²

y   

x x

y  ³ 

3 ,

3 2

1 c e c e x x

c y

y

y  _c  _p   ^x  ^x   x

x

y_p  ³ 