10.VariableCoefficients

8/6/2019 10.VariableCoefficients

http://slidepdf.com/reader/full/10variablecoefficients 1/34

©2005 BE Shapiro [Math 351 Spring 2005 revised 4 May 2005] 0.1

10. Linear Equations with Variable Coefficients

Allowing the coefficients to depend on t , the general n th order linear differential equation

becomes

Ln y = a n (t ) y(n )+ a n ! 1 (t ) y( n ! 1)

+ ! + a1 (t ) " y + a o (t ) y = f (t ) (10.1)

Equation (10.1), even in homogeneous form when f (t ) = 0 , is considerably more

complicated to solve than its corresponding constant-coefficient analogue. In this section

we will introduce two methods that help us attack (10.1): the method of reduction of

order , which helps us to find and equivalent ( n-1) st order differential equation if one

solution is already known; and the method of variation of parameters , which allows us to

determine a particular solution once all of the homogeneous solutions are known.

Reduction of order is particularly useful for second order equations, since it allows us to

find a second homogeneous solution when the first is already known. In the following

sections we will introduce the method of solution by power series , which will allow us to

find a homogeneous solution for most second order linear equations; reduction of order

then gives the second solution, and variation of parameters gives the particular solution.

Both of these methods depend on the Wronskian of the differential equation, a function

that we will define presently. If { y1 ,..., yk } are any set of functions then we can form the

matrix

W [ y1 ,..., yk ](t ) =

y1 y2! yk

! y1 ! y2 ! yk

" "

y1( k "1) y2

( k "1) ! yk ( k "1)

#

$

%%%%

&

'

( ( ( (

(10.2)



©2005 BE Shapiro [Math 351 Spring 2005 revised 4 May 2005] Page 10.3

The fundamental matrix is

W = y 1 y 2

! y 1 ! y 2

# $

& ' (10.5)

where y1 and y 2 are a fundamental set of solutions to Ly = 0 , and the Wronskian of any

two functions y1 and y 2 is

W (t ) =

y1 y2

! y1! y2

= y1! y2

" y2! y1 (10.6)

Example 1 . Find the Wronskian of the two functions y1= cos t and y2

= sin t .

Since ! y1= " sin t and !

y2= cos t the Wronskian is

W [cos t ,sin t ]( t ) =cos t sin t

! sin t cos t = cos 2 t + sin 2 t = 1 . (10.7)

Example 2. Find the Wronskian of the differential equation ! ! y " y = 0 .

The roots of the characteristic equation isr

2 ! 1 = 0are

± 1, and a fundamental pair of

solutions are y1= e t and y2

= e! t . The Wronskian is therefore

W ( x ) =

e t e! t

e t ! e! t

= ! 2 . (10.8)

Example 3. Find the Wronskian of ! !

!

y " 4 ! y = 0 .

The characteristic equation is 0 = r

3

! 4 r = r ( r

2

! 4) = r ( r ! 2)( r + 2) and a fundamental

set of solutions are y1= 1 , y2

= e2 t , and y3

= e! 2 t . Their Wronskian is




W (t ) =

y1 y2 y3

! y1! y2

! y3

!! y1!! y2

!! y3

=

1 e 2 t e" 2 t

0 2 e 2 t " 2e" 2 t

0 4 e 2 t 4 e" 2 t

=

2e 2 t " 2e" 2 t

4 e 2 t 4 e" 2 t

= 16 . (10.9)

Example 4 . Find the Wronskian of ! ! ! y " 8 ! ! y + 16 ! y = 0 .

The characteristic equation is 0 = r 3 ! 8 r

2 + 16 r = r ( r 2 ! 8 r + 16) = r ( r ! 4) 2 , so a

fundamental set of solutions is y1 = 1 , y2= e

4 t and y3= te

4 t . Therefore

W (t ) =

1 e 4 t te 4 t

0 4 e 4 t e 4 t (1 + 4 t )

0 16 e 4 t 8 e 4 t (1 + 2 t )

= (4 e 4 t )[8 e 4 t (1 + 2 t )] ! [e 4 t (1 + 4 t )](16 e 4 t )

= 16 e 8 t

. (10.10)

If we calculate the Wronskian of a set of functions that is not linearly independent, then

one of the functions can be expressed as a linear combination of all other functions, and

consequently, one of the columns of the matrix will be a linear combination of all the

other columns. When this happens, the determinant will be zero. Thus the Wronskian of a

linearly dependent set of functions will always be zero. In fact, as we show in the

following theorems, the Wronskian will be nonzero if and only if the functions form a

complete set of solutions to the same differential equation. We will prove this result first

for a second-order equation in theorem 1, and then for the general case in theorem 2.

THEOREM 1 . Suppose y1 and y2 both satisfy the same second order linear

homogenous differential equation Ly = 0 on (a, b ). Then y1 and y2 are a fundamental

pair of solutions if and only if for some t 0

! (a , b ) , W [ y1 , y2 ](t 0 ) ! 0 .




Proof . Let y1 and y2 be solutions and suppose that W [ y1 , y2 ](t 0 ) ! 0 for some number

t 0

! (a , b ) . We need to show that y1 and y2 are a fundamental pair of solutions.

Consider the initial value problem

Ly = 0, y(t 0 ) = y0 , ! y (t 0 ) = y1 (10.11)

If it is possible to find a pair of numbers C 1 and C 2 such that

! (t ) = C 1 y1 (t ) + C 2 y2 (t ) (10.12)

is a solution of (10.11), then y1 and y2 must be a fundamental set of solutions. Since

Ly1=

Ly2= 0

, we known that L

! = 0

, which means that (10.12) satisfies the differential

equation. We need to show that given any set of initial conditions, it is possible to find a

set of constants such that ! (t ) given by (10.12) satisfies those initial conditions.

Differentiating (10.12)

!" (t ) = C 1! y1 (t ) + C 2

! y2 (t ) (10.13)

At the pointt = t

0 ,

! (t 0 )

"! (t 0 )# $ %

& ' (

=

y1 (t 0 ) y2 (t 0 )

" y1 t 0( ) " y2 (t 0 )# $ %

& ' (

C 1C 2

# $ %

& ' (

= W [ y1 , y2 ](t 0 )C 1C 2

# $ %

& ' (

(10.14)

Since W [ y1 , y2 ](t 0 ) ! 0 (by assumption), the matrix W [ y1 , y2 ][ t 0 ] is invertible, and

W !1=

1

W (t 0 )

" y2 (t 0 ) y2 (t 0 )

! " y1 (t 0 ) ! y1 (t 0 )# $ %

& ' (

(10.15)

and therefore




C 1C 2

! " #

$ % & = W ' 1 y0

y1

! " #

$ % & =

1

W (t 0 )

( y2 (t 0 ) y2 (t 0 )

' ( y1 (t 0 ) ' y1 (t 0 )! " #

$ % &

y0

y1

! " #

$ % &

=1

W (t 0 )

( y2 (t 0 ) y0+ y2 (t 0 ) y1

' ( y1 (t 0 ) y0 ' y1 (t 0 ) y1

! " #

$ % &

(10.16)

Therefore for any set of initial conditions,

! (t ) =

" y2 (t 0 ) y0+ y2 (t 0 ) y1

W (t 0 ) y1 (t ) #

" y1 (t 0 ) y0+ y1 (t 0 ) y1

W (t 0 ) y2 (t ) (10.17)

is a solution of the initial value problem (10.11) and therefore y1 and y 2 must form a

fundamental set of solutions.

To prove the converse of the theorem, assume that y1 and y 2 form a fundamental set of

solutions. We need to show that for some number t 0

! (a , b ) , W [ y1 , y2 ](t 0 ) ! 0 . Since y1

and y 2 form a fundamental set of solutions, any solution to the initial value problem

(10.11) must have the form given by (10.12), and hence (10.13). Therefore there is a

solution {C 1 ,!!C 2 } to

y1 (t 0 ) y2 (t 0 )

! y1 t 0( ) ! y2 (t 0 )" # $

% & '

C 1C 2

" # $

% & ' =

y0

y1

" # $

% & ' (10.18)

This means that the matrix

W [ y1 , y2 ]( t 0 ) =

y1 (t 0 ) y2 (t 0 )

! y1 t 0( ) ! y2 (t 0 )" # $

% & ' (10.19)

is invertible, which in turn is true only if its determinant is nonzero. Thus there exists a

point t 0 where W [ y1 , y2 ](t 0 ) = det W [ y1 , y2 ](t 0 )( )! 0 .




THEOREM 2 . Suppose that y1 , y2 ,..., y n all satisfy the same second order linear

homogenous differential equation Ln y = 0 on (a, b ). Then y1 , y 2 ,..., y n form a

fundamental set of solutions if and only if for some t 0

! (a , b ) , W [ y1 ,..., yn ]( t 0 ) ! 0 .

Proof. Let y1 ,..., y n be solutions to Ln y = 0 , and suppose that W [ y1 ,..., yn ](t 0 ) ! 0 for

some number t 0

! (a , b ) . We need to show that y1 ,..., y n form a fundamental set of

solutions. This means proving that any solution to Ln y = 0 has the form

! (t ) = C 1 y1+ C 2 y2

+ ! + C n yn (10.20)

Consider the initial value problem

Ln y = 0, ! y(t 0 ) = y0 ,..., y( n ! 1) (t 0 ) = yn (10.21)

Certainly every ! ( t ) given by (10.20) satisfies the differential equation; we need to show

that for some set of constants C 1 ,..., C

n it also satisfies the initial conditions.

Differentiating (10.20) n ! 1 times and combining the result into a matrix equation,

! (t 0 )"! (t 0 )!

! ( n #1) (t 0 )

$

%

&&&&

'

(

) ) ) )

=

y1 (t 0 ) y2 (t 0 ) " yn (t 0 )" y1 (t 0 ) " y2 (t 0 ) " yn (t 0 )! !

y1( n #1) (t 0 ) y2

( n #1) (t 0 ) " yn( n #1) (t 0 )

$

%

&&&&

'

(

) ) ) )

C 1C 2!

C n

$

%

&&&&

'

(

) ) ) )

(10.22)

The matrix on the right hand side of equation (10.22) is W [ y1 ,..., yn ]( t 0 ) . By assumption,

the determinant W [ y1 ,..., yn ]( t 0 ) ! 0 , hence the corresponding matrix W [ y1 ,..., yn ]( t 0 ) is

invertible. SinceW

[ y1 ,..., yn ]( t 0 ) is invertible, there is a solution {C

1 ,...,C

n } to the

equation

y0

!

yn

!

" ##

$

% & &

= W [ y1 ,..., yn ](t 0 )

C 1!

C n

!

" ##

$

% & & (10.23)




given by

C 1!

C n

!

"

##

$

%

& &

= W [ y1 ,.., yn ]( t 0 ){ }' 1

y0

!

yn

!

"

##

$

%

& & (10.24)

Hence there exists a non-trivial set of numbers {C 1 ,..., C

n } such that

! (t ) = C 1 y1+ C 2 y2

+ ! + C n yn (10.25)

satisfies the initial value problem (10.21). By uniqueness, every solution of this initial

value problem must be identical to (10.25), and this means that it must be a linear

combination of the { y1 ,..., y n } . Thus every solution of the differential equation is also a

linear combination of the { y1 ,..., y n} , and hence { y1 ,..., y n

} must form a fundamental set

of solutions.

To prove the converse, suppose that y1 ,..., y n are a fundamental set of solutions. We

need to show that for some number t 0

! (a , b ) , W [ y1 ,..., yn ](t 0 ) ! 0 . Since y1 ,..., y n form

a fundamental set of solutions, any solution to the initial value problem (10.21) must have

the form

! (t ) = C 1 y1+ C 2 y2

+ ! + C n yn (10.26)

for some set of constants {C 1 ,..., C

n } . Hence there must exist constants C 1 ,..., C

n such

that

C 1 y1 (t 0 ) + ! + C n yn (t 0 ) = y0

C 1! y1 (t 0 ) + ! + C n

! yn (t 0 ) = y1

"

C 1 y1( n " 1) (t 0 ) + ! + C n yn

( n " 1) (t 0 ) = yn " 1

(10.27)

i.e., there is a solution {C 1 ,..., C

n } to




y1 (t 0 ) ! yn (t 0 )" "

y1( n ! 1) (t 0 ) ! yn

(n ! 1) (t 0 )

"

# $$

%

& ' '

C 1"

C n

"

# $$

%

& ' '

=

y0

"

yn

"

# $$

%

& ' ' (10.28)

This is only true if the matrix

W [ y1 ,..., yn ]( t 0 ) =

y1 (t 0 ) ! yn (t 0 )" "

y1( n ! 1) (t 0 ) ! yn

( n ! 1) (t 0 )

"

# $$

%

& ' ' (10.29)

is invertible, which in turn is true if and only if its determinant is nonzero. But the

determinant is the Wronskian, hence there exists a number t 0 such that the Wronskian

W [ y1 ,..., yn ]( t 0 ) ! 0 .

THEOREM 3 . Suppose y1 ,..., y n are solutions Ln y = 0 on an interval (a , b ) , and let

their Wronskian be denoted by W [ y1 ,..., yn ]( t ) . Then the following are equivalent:

(1) W [ y1 ,..., yn ]( t ) ! 0 ! " t # (a , b ) .

(2) ! t 0

" (a , b ) such that W [ y1 ,..., yn ]( t 0 ) ! 0 .

(3) y1 ,..., y n are linearly independent functions on (a , b ) .

(4) y1 ,..., y n are a fundamental set of solutions to Ln y = 0 on (a , b ) .

Proof . We will prove the theorem for n = 2 . The proof in the general case is similar.

(1) (2) Assume (1). Then W [ y1 ,..., yn ](t ) ! 0 ! " t # (a , b) . Pick any t 0 ! (a , b ) ; property

(2) immediately follows.

(2) (3) Assume (2), i.e., that for some number t 0

! (a , b ) , W [ y1 ,..., yn ]( t 0 ) ! 0 . Then



©2005 BE Shapiro [Math 351 Spring 2005 revised 4 May 2005] .10

W [ y1 , y2 ](t 0 ) =

y1 (t 0 ) y2 (t 0 )! y1 (t 0 ) ! y2 (t 0 )

= y1 (t 0 ) ! y2 (t 0 ) " y2 (t 0 ) ! y1 (t 0 ) # 0 (10.30)

Suppose that y1 and y2 are linearly dependent. Then there exists some constants C 1,C 2 ,

not both zero, such that

C 1 y1 (t ) + C 2 y2 (t ) = 0 (10.31)

for all t . Differentiating,

C 1! y1 (t ) + C 2

! y2 (t ) = 0 (10.32)

Equations (10.31) and (10.32) hold for all t , and in particular, they hold at t = t 0 , hence

y1 (t 0 ) y2 (t 0 )

! y1 (t 0 ) ! y2 (t 0 )" # $

% & '

C 1C 2

" # $

% & ' =

0

0" # $

% & ' (10.33)

which only hold if the determinant zero. Since the determinant is W [ y1 , y2 ]( t 0 ) , this

contradicts (10.30). Hence our assumption that y1 and y2 are linearly dependent must be

false; the only alternative is that they are linearly independent.

(3) (1). Let y1 and y2 be linearly independent solutions of L2 y = 0 . Then for any

choice of C 1,C 2 (not both zero),

C 1 y1 (t ) + C 2 y2 (t ) ! 0 (10.34)

for all t . Differentiating and writing the result in matrix form,

y1 (t ) y2 (t )

! y1 (t ) ! y2 (t )

" # $

% & '

C 1

C 2

" # $

% & '

(0

0

" # $

% & '

(10.35)

for all x and for all non-trivial possible choices for {C 1 , C

2 } . Thus the matrix must be

invertible and the determinant nonzero for all t .

(2) (4) is a restatement of theorem 1.




Example 5 . Two solutions of the differential equation 2 t 2 !! y + 3 t ! y " y = 0 are y1

= t 1/ 2

and y2= 1 / t ; this can be verified by substitution into the differential equation. Over

what domain do these two functions form a fundamental set of solutions?

Calculating the Wronskian, we find that

W (t ) =

t 1/ 2 1 / t

1 / (2 t 1/ 2 ) ! 1 / t 2= ! t 1/ 2

t 2! 1

t (2 t 1/ 2 )=

! 3

2 t 3/ 2 (10.36)

This Wronskian is never equal to zero. Thus these two solutions form a fundamental set

on any open interval over which they are defined, namely t >0.

THEOREM 4. Abel's Formula . The Wronskian of

a n (t ) y ( n )+ a n ! 1 (t ) y ( n ! 1)

+ ! + a 1 (t ) y + a 0 (t ) = 0 (10.37)

is

W (t ) = Ce! p ( t )dt " (10.38)

where p(t ) = a n ! 1 (t ) / a n (t ) . In particular, for n = 2 , the Wronskian of

!!

y + p (t )!

y + q (t ) y = 0 (10.39)

is also given by the same formula, W (t ) = Ce! p ( t )dt " .

LEMMA 1 . Let M be an n ! n square matrix with row vectors m i , determinant M , and

let d (M , i ) be the same matrix with the ith row vector replaced by d m i/ dt . Then

d

dt det M = det d (M , i )

i = 1

n

! (10.40)

Proof . For n=2,




dM

dt =

d

dt (m

11m

22! m

12m

21 )

= m11

"m22

+ "m11

m22

! m12

"m21

! "m12

m21

= m11

"m22

! m12

"m21

+ "m11

m22

! "m12

m21

= m 11 m 12"m12

"m22

+

"

m 11"

m 12

m21

m22

= det[ d (M ,2)] + det[ d (M ,1)]

(10.41)

Now assume that (10.40) is true for any n ! n matrix, and let M be any n + 1 ! n + 1

matrix. Then if we expand its determinant by the first row,

M = (! 1)1+ i m 1i min( m 1i )

i = 1

n + 1

" (10.42)

where min( mij ) is the minor of the ij th element. Differentiating,

dM

dt = (! 1)1+ i "m

1i min( m1i )

i = 1

n + 1

# + (! 1)1+ i m1i

d

dt min( m

1i )i = 1

n + 1

# (10.43)

The first sum is d (M ,1). Since (10.40) is true for any n ! n matrix, we can apply it to

min( m 1i ) in the second sum.

dM dt

= d (M ,1) + (! 1)1+ i m1ii = 1

n + 1

" d (min( m1i ), j ) j = 1

n

"

= d (M ,1) + d (M , i)i = 2

n + 1

" = d (M , i)i = 1

n + 1

"(10.44)

which completes the inductive proof of the lemma.

Proof of Abel's formula. ( n = 2 ). Suppose y1 and y2 are solutions of (10.39) . Their

Wronskian is

W (t ) = y1! y2

" y2! y1 (10.45)




dW dt

=

y1! yn

! y1! yn

!! y1!! yn

" "

y1(n

)! yn

(n

)

(10.52)

Since each y j is a solution of the homogeneous equation,

y j ( n )

= !a n ! 1 (t )

a n (t ) y j

(n ! 1) !a n ! 2 (t )

a n (t ) y j

( n ! 2) ! ! !a 0 (t )

a n (t ) y j

= !1

a n (t )a i (t ) y j

( i )

i = 0

n ! 1" (10.53)

Hence

dW

dx=

y1!

yn

! y1 ! yn

" "

y1( n " 2) yn

(n " 2)

"1

a n (t )a i (t ) y1

( i )

i = 0

n " 1# ! "1

a n (t )a i (t ) ym

( i )

i = 0

n " 1#

(10.54)

The value of a determinant is unchanged if we add a multiple of one to another. So

multiply the first row by a0(t ) / a

n (t ) , the second row by a2(t ) / a

n (t ) , etc., and add them

all to the last row to obtain

dW dt

=

y1! yn

! y1! yn

" "

y1( n " 2) yn

( n " 2)

" p(t ) y1( n " 1) ! " p(t ) yn

(n " 1)

(10.55)

where p (t ) = a n ! 1 (t ) / a n (t ) . We can factor a constant out of every element of a single

row of the determinant if we multiply the resulting (factored) determinant by the same

constant.




dW dt

= ! p(t )

y1! yn

" y1" yn

" "

y1( n ! 2) yn

( n ! 2)

y1(n

! 1)! yn

(n

! 1)

= ! p(t )W (t ) (10.56)

Integrating this differential equation achieves the desired formula for W .

However we compute the Wronskian, either directly by solving the differential equation,

or by using Abel's formula, it is only determined up to a multiplicative constant. This is

because the solutions themselves are only determined up to a multiplicative constant – if

y is a solution of Ln y = 0 , then so is Cy .

Example 6. Calculate the Wronskian of ! ! y " 9 y = 0 .

Since p (t ) = 0 , Abel’s formula gives W = Ce! 0 "dt #

= C .

Example 7 . Use Abel's formula to compute the Wronskian of ! ! ! y " 2 ! ! y " ! y " 3 y = 0

This equation has p (t ) = ! 2 , and therefore W = Ce! ( ! 2) dt "

= Ce 2 t .

Example 8 . Compute the Wronskian of x2 y

(4) + xy(3) + ! ! y " 4 x = 0 .

We have p (t ) = t / t 2

= 1 / t . Therefore W = Ce! (1/ t )dt "

= Ce ! ln t = C / t .

Method of Reduction of Order. Let u (t ) and v (t ) be a fundamental pair of solutions to

the second homogeneous differential equation




!!

y + p (t )!

y + q (t ) y = 0 (10.57)

Then if we equation our two formulas for the Wronskian, we get

u(t ) !v (t ) " !u (t )v(t ) = exp " p(t )dt # { } (10.58)

It is customary in this process to set the multiplicative constants on both sides of the

equation equal to one. If we already know one solution, say u (t ) , this gives us a first

order linear differential equation for the second solution v (t ) .

!v (t ) "!u (t )

u (t )

v (t ) =

W (t )

u (t )(10.59)

where W (t ) = exp ! p(t )dt " { }. An integrating factor for (10.59) is

µ (t ) = exp ! "u (s )u (s )

dst # $

%&

'()

= exp ! ln u (t ){ }=

1u (t )

(10.60)

therefore, if we multiply (10.59) through by 1 / u (t ) , the left hand side becomes the exact

derivative of v (t ) / u (t ) , and hence

d

dt

v (t )u (t )

!"#

$%&

=

W (t )u 2 (t )

(10.61)

Therefore

v(t ) = u(t ) W (s)u 2 (s)

dst ! = u(t )

exp " p( x)dxs

! { }u 2 (s)

dst ! (10.62)




THEOREM 5 . Reduction of Order (Second Order Equations). Suppose that u (t ) is a

(non-trivial) solution of !!

y + p (t )!

y + q (t ) = 0 . Then a second, linearly independent

solution, is given by equation (10.62).

Proof . We have already proven that (10.62) is a solution. To show that it is linearly

independent from the first solution, assume the converse, namely that u ( t ) and v (t ) are

linearly dependent. Then there exists some constant c such that v ( t ) = cu (t ) . Then, since

u (t ) ! 0 (identically),

e h ( s )

u 2 (s ) dst ! = c (10.63)

for all t , where h(s ) = ! p( x)dxs

" . Since the derivative of a constant is zero,

0 =

d

dt

e h ( s )

u 2 ( s )ds

t ! =

e h ( t )

u 2 (t )(10.64)

This implies that e h ( t )= 0 which is impossible. Hence u (t ) and v (t ) must be linearly

independent.

In practice it is usually easier to just calculate the two different expressions for the

Wronskian and set them equal to one another rather than memorizing equation (10.62).

ALGORITHM 1. REDUCTION OF ORDER. To find a second, linearly independent

solution v (t ) to !!

y + p (t )!

y + q (t ) y = 0 , given a first solution u (t ) ,

(1) Calculate the Wronskian using Abel's formula, W (t ) = e! p ( t )dt " .

(2) Calculate the Wronskian directly as W (t ) = !u (t )v (t ) " v '(t )u (t ) .




(3) Set the two expressions equal; the result is a first order differential equation for

the second solution v (t ).

(4) Solve the differential equation for v (t ) .

(5) Then general solution is then y = C 1u (t ) + C 2 v(t )

Example 9 . Find a second solution to the differential equation t !!

y + 10 !

y = 0 given the

observation that y1 = 1 is a solution.

Since p (t ) = 10 / t , Abel's formula gives

W (t ) = e! (10/ t ) dx"

= t ! 10 (10.65)

By direct calculation,

W (t ) =

y1 y2!

y1!

y2

=

1 y2

0 !

y2

=!

y2 (10.66)

Equating the two expressions for W ,

! y2= t

" 10 (10.67)

Therefore, y2= ! (1 / 9) t

! 9 ; the general solution to the homogenous equation is

y = C 1 y1+ C 2 y2

= C 1 + C 2 t ! 9 . (10.68)

Example 10 . Find a fundamental set of solutions to t 2 !! y + 5 t ! y " 5 y = 0 given the

observation that y1= t is one solution.

Calculating the Wronskian directly,




W (t ) =

t y2

1 ! y2

= t ! y2" y2 (10.69)

From Abel’s Formula , since p(t ) = a 1 (t ) / a 2 (t ) = 5 / t ,

W (t ) = e! (5 / t " )dt

= t ! 5 (10.70)

Equating the two expressions and putting the result in standard form,

! y2" (1 / t ) y2

= t " 6 (10.71)

An integrating factor is

µ (t ) = e( ! 1/ t ) dx"

= 1 / t (10.72)

Therefore

! y2= t (1 / t )t " 6dt # = t

t "6

6

$

%&

'

() =

1

5t "5 (10.73)

The fundamental set of solutions is therefore {t ,!t

! 5

} .

Example 11 . Find the general solution of

t !!! y " !! y " t ! y + y = 0 (10.74)

given that y = e t and y = e! t are solutions.

From Abel's formula,

W (t ) = exp ! (! 1 / t )dt " { }= exp{ln t } = t (10.75)

By direct calculation,




Example 12 . Show that the theorem 6 gives the same result as theorem 5 (equation

(10.62)) for n=2.

Let u (t ) be a solution of !!

y + p (t )!

y + q (t ) y = 0 and make the substitution y = uz . Then

0 = (uz!!) + p (t )(uz

!) + q (t )(uz )=

!!

u z + 2 !

u!

z + u!!

z + p (t )(!

u z + u!

z ) + q (t )uz

= [ !!

u + p (t )!

u + q (t )u ] z + 2 !

u!

z + p (t )u!

z + u!!

z

= u!!

z + [ p (t )u + 2 !

u ] !

z

(10.84)

In the next to the last line we used the fact that u is a solution of the differential equation

to set the quantity in square brackets equal to zero. Let w =!

z . Then

!w + p (t ) + 2!u

u

"#$

%&'

w = 0 (10.85)

This is a separable equation for w ; integrating we find

ln w(t ) = ! 2 ln u(t ) ! p(t )dt " (10.86)

Exponentiating and substituting back w =!

z

! z =1

u 2 (t )exp p(t )dt " { }=

W (t )u 2 (t )

, (10.87)

where the last step follows from Abel's formula. Since y = uz is the second solution,

y = u(t ) z = u (t ) ! z dt " = u (t )W (t )

u 2 (t )dt " (10.88)

which is the same formula we had before.




Example 13 . Use the fact that y = t is a solution of y(4 ) ! t " y + y = 0 to reduce the

differential equation to a third order equation.

Following the advice of theorem 6 we substitute y = zy1= zt . Then since

!

y = t !

z + z,!! !!

y = t !!

z + 2 !

z ,! !!!

y = t !!!

z + 3 !!

z ,!! y (4 )= tz

(4 )+ 4 !!!

z (10.89)

the differential equation becomes

0 = tz(4 )

+ 4 !!! z " t (t ! z + z ) + zt = tz(4 )

+ 4 !!! z " t 2 ! z (10.90)

Letting u =!

z gives the desired third order equation,

t !!!u + 4 !!u " t 2

u = 0 . (10.91)

Variation of Parameters. The method of variation of parameters gives an explicit

formula for a particular solution to a linear differential equation once all of the

homogeneous solutions are known. The particular is a pseudo-linear combination of the

homogeneous equation. By a pseudo-linear combination we mean an expression that has

the same form as a linear combination, but the constants are allowed to depend on t :

yP = u1 (t ) y1 + u2 (t ) y2 + ! + un (t ) yn (10.92)

The name of the method comes from the fact that the parameters (the constants in the

linear combination) are allowed to vary. We observe that equation (10.92) is not of itself

restrictive in any way because we may express any function h (t ) in this form by setting

u1(t ) = h(t ) / y1 and u 2= u 3

= ! = un

= 0 .

Suppose that y1 (t ) and y2 (t ) are linearly independent solutions of




a (t ) !! y + b(t ) ! y + c(t ) y = 0 (10.93)

Then we look for a pair of functions u (t ) and v (t ) that will make

yP = u (t ) y1 + v(t ) y2 (10.94)

a solution of

a (t ) !! y + b(t ) ! y + c(t ) y = f (t ) . (10.95)

Differentiating equation (10.94)

! yP =!u y1 + u ! y1 +

!v y2 + v ! y2 (10.96)

If we now make the totally arbitrary assumption that

!

u y 1 +!

v y 2 = 0 (10.97)

then

!

yP = u!

y1 + v!

y2 (10.98)

!! yP =!u ! y1 + u !! y1 +

!v ! y2 + v !! y2 (10.99)

From equation (10.95)

f (t ) = a (t ) !u ! y1 + u !! y1 +!v ! y2 + v !! y2( )+ b(t ) u ! y1 + v ! y2( )+ c(t ) uy1 + vy2( )

= a (t ) !u ! y1 +!v ! y2( )+ u a (t ) !! y1 + b(t ) ! y1 + c(t ) y1[ ]

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+ v a(t ) !! y2 + b(t ) ! y2 + c(t ) y2[ ]= a (t ) !u ! y1 +

!v ! y2( )

(10.100)

Combining equations (10.97) and (10.100) in matrix form

y1 y2

! y1 ! y2

" # $

% & '

!u

!v" # $

% & ' =

0

f (t ) / a (t )" # $

% & ' (10.101)




The matrix on the left hand side of equation (10.101) is the Wronskian, which we know is

nonsingular, and hence invertible, because y1 and y2 form a fundamental set of solutions

to a differential equation. Hence

!u

!v" # $

% & ' =

y1 y2

! y1 ! y2

" # $

% & '

(1 0

f (t ) / a (t )" # $

% & '

=

1

W (t )

! y2 ( y2

( ! y1 y1

" # $

% & '

0

f (t ) / a (t )" # $

% & '

=

1

W (t )

( y2 f (t ) / a (t )

y1 f (t ) / a (t )" # $

% & '

(10.102)

where W (t ) = y1 ! y2 " y2 ! y1 . Hence

dudt

=

! y2 f (t )

a (t )W (t ),!!dv

dt =

y1 f (t )

a (t )W (t )(10.103)

Integrating and substituting into equation (10.94)

yP = ! y1 y2 f (t )

a (t )W (t )dt " + y2

y1 f (t )

a (t )W (t )dt " (10.104)

Example 14 . Find the general solution to !! y " 5 ! y + 6 y = e t

The characteristic polynomial r 2 ! 5 r + 6 = ( r ! 3)( r ! 2) = 0 , hence a fundamental set of

solutions is y1= e

3 t , y2= e

2 t . TheWronskian is

W (t ) =

e 3t e 2 t

3e 3t 2 e 2 t = ! e

5 t

(10.105)

Since f (t ) = et and a (t ) = 1 ,




yP = ! e3t e2 t et

! e5 t " dt + e2 t e3t et

! e5 t dt " = e3t e! 2 t " dt ! e2 t e! t dt " =

e3 x

(! 1 / 2) e! 2 t

! e2 t

(! 1)e! t =

1

2 et

(10.106)

Thus the general solution is y = yP + y H =1

2et + C 1 e

3 t + C 2 e2 t .

Example 15 . Find a particular solution to !! y " 5 ! y + 6 y = t by repeating the steps in the

derivation of (10.104) rather than plugging in the general formula.

From example 14 we have y1= e 3t , y2

= e 2 t . We look for a solution of the form

y = ue3 t + ve

2 t (10.107)

!

y =!

u e3 t + 3 ue

3 t +!

v e2 t + 2 ve

2 t (10.108)

Assuming that

!u e 3 t +

!v e 2 t = 0 (10.109)

then

!

y = 3 ue3 t + 2 ve

2 t (10.110)

!!

y = 3 !

u e3 t + 9 ue

3 t + 2 !

v e2 t + 4 ve

2 t (10.111)

Substituting into the differential equation!!

y"

5!

y + 6 y = t ,

t = 3 !u e3 t

+ 9 ue3 t

+ 2 !v e2 t

+ 4 ve2 t ( )" 5 3 ue

3 t + 2 ve

2 t ( )!!!!!!!!!!+ 6 ue

3 t + ve

2 t ( )= 3 !u e

3 t + 2 !v e

2 t

(10.112)




Combining (10.112) with (10.109) gives

!u e

3 t +

!v e

2 t = 0 (10.113)

3 !u e 3 t + 2 !v e 2 t = t (10.114)

Multiplying equation (10.113) by 3 and subtracting equation (10.114) from the result,

!v e2 t

= " t (10.115)

v = ! te ! 2 t dt " = ! !t

2!

1

4

# $ %

& ' (

e ! 2 t =

t

2+

1

4

# $ %

& ' (

e ! 2 t (10.116)

because te at dt ! = t / a " 1 / a 2#$ %&e at . Multiplying equation (10.113) by 2 and subtracting

equation (10.114) from the result,

!u e

3 t = t (10.117)

u = te ! 3 t dt " = !t

3!

1

9

# $ %

& ' (

e ! 3 t = !

t

3+

1

9

# $ %

& ' (

e ! 3 t (10.118)

Substituting (10.118) and (10.116) into (10.107)

y = ue3 t + ve

2 t

= e3 t ! t

3+

1

9

" # $

% & ' e ! 3 t (

)*+,-

+ e2 t t

2+

1

4

" # $

% & ' e ! 2 t (

)*+,-

=t

6+

5

36

(10.119)

A quick calculation verifies that this is the same solution we would have obtained by the

method of undetermined coefficients.




THEOREM 7 . Variation of Parameters. Suppose that y1 ,..., yn are a fundamental set of

solutions to

Ln y=

a n (t ) y(n ) +

a n ! 1 (t ) y(n ! 1) + ! +

a 0 (t ) y=

0 (10.120)

Then a particular solution to

Ln y = a n (t ) y(n ) + an ! 1 (t ) y(n ! 1) + ! + a0 (t ) y = f (t ) (10.121)

is given by

yP

= y1

W 1 (s ) f (s)

W (s)a n (s)ds

t ! + y

2

W 2 (s) f (s )

W (s)an (s)ds

t ! + ! + y

n

W n (s) f (s)

W (s)a n (s)ds

t ! (10.122)

where W j (t ) is the determinant of W [ y1 ,..., yn ]( t ) with the jth column replaced by a

vector with all zeroes except for a 1 in the last row. In particular, for n = 2 , a particular

solution to a (t ) !! y + b(t ) ! y + c(t ) y = f (t ) is

y p = ! y1 (t )y2 (s) f (s)

W (s)a (s )ds

t " + y2 (t )

y1 (s) f (s )

W (s)a (s )ds

t " (10.123)

Proof for general case . Look for a solution of the form

y = u 1 y1 + ! + u n yn (10.124)

Generalizing equation (10.97) assume that

!u 1 y1 + ! + !u n yn= 0

!u 1 ! y1 + ! + !u n! yn

= 0"

!u 1 y1(n " 2) + ! + !u n yn

(n " 2) = 0

(10.125)

Then




! y = u 1! y1 + ! + u n

! yn

!! y = u 1!! y1 + ! + u n

!! yn

"

y(n " 1) = u 1 y1

(n " 1) + ! + u n yn(n " 1)

y (n ) = u 1 y1(n ) + ! + u n yn

(n ) + !u 1 y1(n " 1) + ! + !u n yn

(n " 1)

(10.126)

So that

f (t ) = a n (t ) y(n ) + an ! 1(t ) y(n ! 1) + ! + a0 (t ) y

= a n (t ) u1 y1(n ) + ! + un yn

(n ) + "u1 y1(n ! 1) + ! + "un yn

(n ! 1)#$ %&+

!!!!an ! 1 (t ) u1 y1(n ! 1) + ! + un yn

(n ! 1)#$ %&+ ! +

!!!!a1 (t ) u1 " y1 + ! + un " yn[ ]+ a 0 (t ) u1 y1 + ! + un yn[ ]= a n (t ) "u1 y1

(n ! 1) + ! + "un yn(n ! 1)#$ %&

(10.127)

Combining (10.127) and (10.125) in matrix form,

y1 y2 ! yn

! y1 ! y2 ! yn

" "

y1(n" 2)

y2(n" 2)

! yn(n" 2)

y1(n"1) y2

(n"1) ! yn(n"1)

#

$

%%%%%%

&

'

( ( ( ( ( (

!u1

!u2

"

!un"1

!un

#

$

%%%

%%%

&

'

( ( (

( ( (

=

0

0"

0 f (t ) / a n (t )

#

$

%%%

%%%

&

'

( ( (

( ( (

(10.128)

The matrix on the left is the fundamental matrix of the differential equation, and hence

invertible,

!u1

!u2

!

!un"1

!un

#

$

%%%%%%

&

'

( ( ( ( ( (

=

y1 y2 " yn

! y1 ! y2 ! yn

! !

y1(n" 2) y2

(n" 2) " yn(n" 2)

y1(n"1) y2

(n"1) " yn(n"1)

#

$

%%%%%%

&

'

( ( ( ( ( (

"10

0!

0

f (t ) / an (t )

#

$

%%%%%%

&

'

( ( ( ( ( (

= f (t )a n (t )

[W "1 ]1n

[W "1 ]2 n

!

[W "1 ]nn

#

$

%

%%%%%

&

'

(

( ( ( ( (

(10.129)

where W is the fundamental matrix and [W! 1 ]ij denotes the ij th element of W

! 1.




[W! 1 ] jn

=

cof[ W ]ni

det W=

W i (t )

W (t )(10.130)

Therefore

du i

dt =

f (t )W i (t )

an (t )W (t )(10.131)

ui (t ) =

f (s)W i (s)

an (s )W (s)ds

t ! (10.132)

Substitution of equation (10.132) into equation (10.124) yields equation (10.122).

Example 16 . Solve !!!

y +!

y = tan t using variation of parameters.

The characteristic equation is 0 = r 3 + r = r (r + i )( r ! i ) ; hence a fundamental set of

solutions are y1= 1 , y2

= cos t , and y3= sin t . From either Abel's formula or a direct

calculation, W (t ) = 1 , since

W =

1 cos t sin t

0 ! sin t cos t

0 ! cos t ! sin t

= 1 (10.133)

From (10.122)

yP = y1W 1 (s) f (s)

W (s)a 3 (s)ds

t ! + y2W 2 (s ) f (s)

W (s)a 3 (s)ds

t ! + y3W 3 (s) f (s)

W (s)a 3 (s)ds

t ! (10.134)

where a 3 ( s ) = 1 , f (s) = tan s , and




W 1 =

0 cos t sin t

0 ! sin t cos t

1 ! cos t ! sin t

= 1 (10.135)

W 2 =

1 0 sin t

0 0 cos t

0 1 ! sin t

= ! cos t (10.136)

W 3 =

1 cos t 0

0 ! sin t 0

0 ! cos t 1

= ! sin t (10.137)

Therefore

yP= tan sds

t ! " cos t cos s tan sdst ! " sin t sin s tan sds

t ! (10.138)

Integrating the first term and substituting for the tangent in the third term,

yP= ! ln cos t ! cos t sin sds

t " ! sin t sin 2

scos s

dst " (10.139)

The second integral can not be integrated immediately, and the final integral can be

solved by substituting sin 2s = 1 ! cos 2

s

yP = ! ln cos t + cos 2t ! sin t

1 ! cos 2s

cos sds

t " = ! ln cos t + cos 2

t ! sin t sec sdst " ! cos ds

t " #$

%&

= ! ln cos t + cos 2t ! sin t ln sec t + tan t + sin 2

t

= 1 ! ln cos t ! sin t ln sec t + tan t

(10.140)

Since the first term (the constant) is a solution of the homogeneous equation, we can drop

it from the particular solution, giving




yP = ln cos t ! sin t ln sec t + tan t (10.141)

and a general solution of

y = ln cos t ! sin t ln sec t + tan t + C 1 + C 2 cos t + C 3 sin t . (10.142)

Example 17 . Solve the initial value problem

t 2 !! y " 2 y = 2 t ,!! y(1) = 0, !! ! y (1) = 1 (10.143)

given the observation that y = t 2 is a homogeneous solution.

We can find a second homogeneous solution using reduction of order. By Abel’s

formula, since there is no ! y term, the Wronskian is a constant:

W (t ) = e! 0 dt "

= C (10.144)

Hence

C = W (t ) =

t 2 y2

2 t ! y2

= t 2 ! y2" 2 ty2 (10.145)

This is a first order linear equation in y 2 ; omitting the subscript and putting it into

standard form,

! y "2

t y =

C

t 2 (10.146)

An integrating factor is µ = e(! 2 / t )dt "

= e ! 2 ln t = t ! 2 , hence

d

dt y / t 2!" #$ = Ct %4 (10.147)

Integrating both sides of the equation over t ,

y

t 2= (C / ! 3)t

! 3+ C 1 (10.148)



Adding the two equations (10.153) and (10.155) gives C 1

= 1 , hence C 2

= 0 , so that the

complete solution of the initial value problem is

y=

t 2

!1

t . (10.156)

Exercises.

Historical Notes

Josef Hoëné de Wronski (1778-1853) was born Josef Hoëné but he adopted the name Wronski around1810. An immigrant to France, he became a citizen in 1800. Most of his work was dismissed by hiscontemporaries as worthless. Among other things, he published a not very well received study on thefoundations of mathematics (1810); a “proof” claiming that every equation could be solved algebraically(1812); the book Introduction to a course on mathematics (1821); and several papers claiming that

philosophy should take precedence over logic in mathematical proofs. He was something of an inventor but none of his devices were very successful. Thomas Muir (1844-1934), a noted algebraicist, coined thename Wronskian in 1882 for the determinants that bear his name; Wronski had in fact used these matricesas coefficients in series expansions of functions.

Niels Henrik Abel (1802-1829) was a Norwegian mathematician best known for his work in algebra,inventing group theory. He proved that there was no general solution of the fifth degree equation in termsof roots; published the first known solution of an integral equation; and developed the theory of ellipticfunctions. He proved tha that the bionomial theorem holds for all real numbers at the age of 16. He spentmuch of his adult life in near poverty because he was unable to obtain a university post (his contemporary

Gauss had never heard of him and dismissed his manuscripts, without reading them, as worthless) and poorhealth, and died of tuberculosis after a long sled journey. He was offered a post as professor at Berlin twodays after he died.

10.VariableCoefficients

Documents

Transcript of 10.VariableCoefficients