Download Solution of linear differential equations by power series. The case ρ1

Solution of linear di erential equations by
power series. The case 1
= Z.
2 2
Solve the linear di erential equation
2x2 y 00 + 3x
2x2 y 0
(x + 1) y = 0
(1)
by power series in the vicinity of x0 = 0.
1
Solution.
We have
2
= 3 2x2x ; p1 (x) = xP1 (x) = 3 22x ;
P1 (x) = 3x2x2x
2
P2 (x) = x+1
;
p 2 (x) = x2 P2 (x) = x+1
; jxj < +1:
2x2
2
In accordance with the Theorem of Existence 2 there exists a solution
y (x) of equation (1) in the form:
y=x
1
X
n=0
an xn ; a0 6= 0;
where the series (2) converges for any x.
2
2.1
Formulae for derivatives.
The
rst derivative.
y = x
1
X
an x n ;
n=0
1
(2)
y0 =
1
X
0
an xn+
=
n=0
= x
xy = x
1
X
an (n + ) xn+
1
=
n=0
1
X
n=0
1
X
an (n + ) xn
1
an xn+1
(3)
n=0
1
X
y0 = x
(n + ) an xn
1
(4)
(n + ) an xn
(5)
(n + ) an xn+1
(6)
n=0
xy 0 = x
x2 y 0 = x
1
X
n=0
1
X
n=0
2.2
The second derivative.
y 00 = x
1
X
1) an xn
(n + ) (n +
2
(7)
1) an xn
(8)
n=0
x2 y 00 = x
1
X
(n + ) (n +
n=0
3
Computations.
Substitute (8), (5), (4) and (2) to (1). We obtain:
2x2 y 00 + (3 2x) xy 0 (x + 1) y =
P
1) an xn +
=x 1
n=0 2 (n + ) (n +
P1
P
n+1
+x n=0 3 (n + ) an xn x 1
n=0 2 (n + ) an x
P1
P
n
x
a xn+1 x 1
n=0 an x = 0;
P1 n=0 n
(2 (n + ) (n +
1) + 3 (n + ) 1) an xn
n=0
P1
n+1
= 0;
n=0 (2 (n + ) + 1) an x
P1
n=0
(n + + 1) (2 (n + )
1) an xn
;
P1
n=0
(n + + 1) (2n + 2
1) an xn
2
P1
n=0
P1
n=1
(2 (n + ) + 1) an xn+1 = 0;
(2 (n + )
1) an 1 xn = 0;
P1
( + 1) (2
1) a0 +
0;
( + 1) (2
1) a0 +
1
X
n=1
(n + + 1) (2n + 2
(n + + 1) (2n + 2
1) an xn
1) an
P1
n=1
(2 (n + )
n=1
(2 (n + )
1) an
1
!
(9)
If the series (9) equals to zero for all x; jxj < 1; then all the coe cients
of (9) are zeros, i.e.
(n + + 1) (2 (n + )
( + 1) (2
1) a0 = 0; a0 6= 0; (10)
(2 (n + ) 1) an 1 = 0;
(11)
n = 1; 2; :::
1) an
It follows from (10), that
( + 1) (2
1) = 0
(12)
;i.e.
= 21 ; 2 = 1:
The equation (12) is said to be the characteristic equation of source differential equation (1).
Formula (11) yields the recursion of the rst order
1
an =
1
1
(n + + 1)
Since 1 = 12 is greater then
= 12 .
Substitute = 12 to (13):
an =
2
an 1 ; n = 1; 2; :::
=
1; we
rstly investigate the case
2
an 1 ; n = 1; 2; :::
(2n + 3)
We have:
2
n = 1 : a1 = 25 a0 = 523 3a0 = 5!!
3a0 ; 5!! = 1 3 5
2
2
2
22
3a0 ; 7!! = 1 3 5 7
n = 2 : a2 = (2 2+3) a1 = 7 5!! 3a0 = 7!!
n = 3 : a3 =
2
a
(2 3+3) 2
=
2
9
22
3a0
7!!
=
3
23
3a0 ; 9!!
9!!
(13)
=1 3 5 7 9
1) an 1 xn =
xn = 0:
:::
2n
an = (2n+3)!!
3a0 ; n = 1; 2; ::::;where (2n+3)!! = 1 3 5 7 ::: (2n+1) (2n+3)
Hence
y1 (x) = 3a0
p
!
1
X
2n
x
xn ; 8a0 2 R.
(2n
+
3)!!
n=0
(14)
The second solution y2 (x) linear independent of y1 (x) ; is obtained by
the formula:
R
Z
e P1 (x)dx
dx;
(15)
y2 (x) = y1 (x)
y12 (x)
but it is too di cult to calculate integral (15).
4
How to nd y2 (x) by power series.
Theorem 1 If
1
2
2
= Z (the set of entire numbers), then
y1 (x) = x
y2 (x) = x
1
2
1
X
n=0
1
X
an x n ;
(16)
bn x n :
(17)
n=0
Substitute
2
=
1 to (13):
bn =
We have:
bn = n1 bn
Hence
1
=
1
n
1
b
n 1 n 2
=
1
bn 1 ; n = 1; 2; :::
n
1
n
1
n 1
1
b
n 2 n 3
= ::: =
!
1
b;
n! 0
1
1 X
1 n
ex
y2 (x) = b0
x = b0 ; 8b0 2 R.
x n=0 n!
x
The general solution of the equation (1) is:
y (x) = C1 y1 (x) + C2 y2 (x) ;
4
n = 0; 1; 2; :::
(18)
where y1 (x) is obtained by (14) and y2 (x) is obtained by (18). Hence the
general solution of equation (1) is:
y (x) = C1
p
!
1
X
2n
ex
n
x
x + C2 :
x
n=0 (2n + 3)!!
5
(19)