370 Chapter 13Second-order differential equations. Some special functions
For this to be equal to zero for all values of x(within the radius of convergence of the
power series) the coefficient of each power of xmust be zero:
a
1
1 + 1 a
0
1 = 1 0, 2a
2
1 + 1 a
1
1 = 1 0, 3a
3
1 + 1 a
2
1 = 1 0,=
so that
and, in general,a
m
1 = 1 (−1)
m
a
0
2 m!. Therefore,
We recognize the infinite series as the power-series expansion of the functione
−x
:
Thereforey 1 = 1 a
0
e
−x
wherea
0
is an arbitrary constant (see Example 11.4(ii) withk 1 = 1 − 1 ).
0 Exercises 1, 2
EXAMPLE 13.2Use the power-series method to solve the equation
1
By equation (13.2), we have
We can write
=++
=
+
∑
m
m
m
mmax
0
2
21
∞
()()
′′=− =×+×+×
=
−
∑
ymmax a ax
m
m
m
2
2
23
1213243
∞
() ()() ())ax
4
2
+
yaxy max y mm
m
m
m
m
m
m
m
=,′=,′′=−
==
−
=
∑∑ ∑
01
1
2
1
∞∞ ∞
()aax
m
m− 2
dy
dx
y
2
2
+= 0
ex
xx x
m
x
m
m
−
=
=− +
!
−
!
+=
−
!
∑
1
23
23
0
∞
()
yax
ax
m
a
x
m
m
m
m
m
mm
m
m
==
−
!
=
−
!
== =
∑∑ ∑
00
0
0
0
1
∞∞ ∞
()
()
aaa
aa
a
aa
102
10
3
20
22 33
=− , =− =+
!
,=−=−
!
,
1
Leibniz discovered the differential equation for the sine function in 1693 by a geometric argument. He then
solved the equation by a method equivalent to that given in this example to obtain the power-series representation
of the function. The ability to represent functions as power series was a vital element in the development of the
calculus by both Leibniz and Newton.