Reduction of Order
Theorem: Let f(x) be a non-trivial solution of the n-th order H.L.D.E.
a 0 (x)y( n ) + a1 (x)y( n!1) + ..... + a n!1 (x)y" + a n (x)y = 0 (1)
Conclusion:
The transformation y = f(x) v(x) reduces the equation (1) to a (n-1)st-order H.L.D.E. in
dv
the dependent variable w =
.
dx
Let’s consider the case n = 2.
We have the 2nd order equation
a0(x)y’’ + a1(x) y’ + a2(x)y = 0
Assume f(x) is a solution of the equation. Take the transformation y = f(x) v(x) where v
is a function of x that will be determine at the end of the process.
Then
dy
dv
= f(x) + f !(x)v
dx
dx
2
d y
d2v
dv
=
f(x)
+ 2 f !(x) + f !!(x)v(x)
2
2
dx
dx
dx
Substituting in the equation
"
%
d2v
dv
dv
"
%
a 0 (x) $ f(x) 2 + 2 f !(x) + f !!(x)v(x) ' + a1 (x) $ f(x) + f !(x)v(x) ' + a 2 (x)f(x)v(x) = 0
dx
dx
dx
#
&
#
&
or
d2v
dv
a 0 (x)f(x) 2 + [ 2a 0 (x)f !(x) + a1 (x)f(x)] + [ a 0 (x)f !!(x) + a1 (x)f !(x) + a 2 (x)f(x)] v(x) = 0
dx
dx
since f(x) is a solution of the given equation, then
a 0 (x)f !!(x) + a1 (x)f !(x) + a 2 (x)f(x) = 0
and the equation reduces to:
d2v
dv
a 0 (x)f(x) 2 + [ 2a 0 (x)f !(x) + a1 (x)f(x)] = 0
dx
dx
dw
Letting w =
, we have
dx
dw
a 0 (x)f(x)
+ [ 2a 0 (x)f !(x) + a1 (x)f(x)] w = 0
dx
it is a first order homogeneous linear differential equation in the dependent variable w.
The equation is also separable. Assuming f(x) ≠ 0 and a0(x) ≠ 0, we can rewrite the
equation as:
# f "(x) a1 (x) &
dw
= ! %2
+
( dx (2)
w
f(x)
a
(x)
$
'
0
integrating,
ln w = ! ln f(x) ! "
2
!!!!!w =
a1 (x)
dx + ln c
a 2 (x)
#
a (x) &
c exp % ! " 1 dx (
$ a 2 (x) '
( f(x))2
This is a general solution of equation (2). If we choose c = 1, and replace w by dv/dx,
#
a (x) &
exp % ! " 1 dx (
$ a 2 (x) '
dv
!!! =
dx
( f(x))2
integrating,
v(x) =
Then, g(x) = f(x) v(x) = f(x)
"
"
#
a (x) &
exp % ! " 1 dx (
$ a 0 (x) '
( f(x))2
#
a (x) &
exp % ! " 1 dx (
$ a 0 (x) '
( f(x))
2
dx
dx is a solution of the 2nd order original
differential equation. Moreover f(x) and g(x) are linearly independent:
f(x) g(x) f(x)
f(x)v(x)
2
W ( f,g ) =
=
= ( f(x)) v!(x) =
f !(x) g!(x) f !(x) f !(x)v(x) + f(x)v!(x)
( f(x))
$
a (x) '
exp & " # 1 dx )
% a 0 (x) (
2
( f(x))
2
$
a (x) '
= exp & " # 1 dx ) * 0
% a 0 (x) (
Therefore, the combination c1f(x) + c2g(x) is a general solution of the equation.
Example:
1) Given that f(x) = x is a solution of the equation
d2y
dy
x2 + 1
! 2x + 2y = 0
2
dx
dx
find another linearly independent solution by reduction of order.
Solution:
Let y = xv(x), then
dy
dv
d2y
d2v
dv
= x + v!!and!! 2 = x 2 + 2
dx
dx
dx
dx
dx
substituting in the equation,
(
)
! d2v
dv $
d2v
dv
dv
! dv
$
(x 2 + 1) # x 2 + 2 & ' 2x # x + v& + 2xv = (x 2 + 1)x 2 + 2 + 2x 2 ' 2x 2
+ ( 2xv ' 2xv) =
" dx
%
dx %
dx
dx
dx
" dx
(
d2v
dv
+2
=0
2
dx
dx
Letting w = dv/dx, we obtain the equation
dw
x x2 + 1
+ 2w = 0
dx
or
(x 2 + 1)x
(
)
dw
2
2x %
" 2
=!
dx
=
!
+
$# x x 2 + 1 '& dx
w
x x2 + 1
(
)
integrating
ln w = !2 ln x + ln(x 2 + 1) + ln c
w=
(
)
c x2 + 1
2
x
Choosing c =1 and recalling w = dv/dx
dv x 2 + 1
1
=
=
1
+
dx
x2
x2
int egrating
1
v=x!
x
So the second solution is g(x) = f(x)v(x) = x(x - 1/x) = x2 – 1.
The general solution of the equation is: y(x) = c1x + c2(x2 – 1).
2) Given that f(x) = e2x is a solution of the equation
d2y
dy
2x
+
1
(
) 2 ! 4 ( x + 1) + 4y = 0
dx
dx
find another linearly independent solution by reduction of order.
Solution:
Let y = e2xv(x), then
dy
dv
d2y
d2v
dv
= e 2x
+ 2e 2x v!!and!! 2 = e 2x 2 + 4e 2x
+ 4e 2x v
dx
dx
dx
dx
dx
substituting in the equation,
2
$
dv
! dv
$
2x ! d v
(2x + 1)e # 2 + 4
+ 4v& ' 4 ( x + 1) e 2x #
+ 2v& + 4e 2x v =
"
%
dx
dx
" dx
%
d2v
2x dv
+
8x
+
4
'
4x
'
4
e
+ ( 4x + 2 ' 4x ' 4 + 2 ) 4e 2x v = 0
(
)
2
dx
dx
Divide by e2x,
(2x + 1)e 2x
)
d2v
dv
+ 4x
=0
2
dx
dx
Letting w = dv/dx, we obtain the equation
dw
( 2x + 1) + 4xw = 0
dx
or
(2x + 1)
dw
4x
1 %
" 2x %
" 2x + 1 ! 1 %
"
=!
dx = !2 $
dx = !2 $
dx = !2 $1 !
dx
'
'
w
2x + 1
# 2x + 1 &
# 2x + 1 &
# 2x + 1 '&
integrating
ln w = !2x + ln(2x + 1) + ln c
w = e !2x (2x + 1)c
Choosing c =1 and recalling w = dv/dx
dv
= (2x + 1)e !2x
dx
int egrating
!e !2x (1 + x ) e !2x
e !2x ( 2x + 2 )
v = " 2xe + e
dx =
!
=!
= !e !2x (x + 1)
2
2
2
2x
!2x
So the second solution is g(x) = f(x)v(x) = e (!e )(x + 1) = !x ! 1
The general solution of the equation is: y(x) = c1e-2x + c2(-x – 1).
(
!2x
!2x
)