082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

19.7 Series solution of differential equations 585
These expressions for y and d2 y
are substituted into the given differential equation
dx2 d 2 y
dx +y=0.Thus 2
(2a 2
+ (2)(3)a 3
x + (3)(4)a 4
x 2 + (4)(5)a 5
x 3 + ...) +
(a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...) =0.
We now use the technique of equating coefficients. This was first introduced in
Section 1.7 Partial fractions. Equating the constant terms:
2a 2
+a 0
=0 sothat a 2
=− 1 2 a 0 .
Equating coefficients ofx:
(2)(3)a 3
+a 1
=0 sothat a 3
= − 1
(2)(3) a 1 .
Equating coefficients ofx 2 :
(3)(4)a 4
+a 2
=0 sothat a 4
= − 1
(3)(4) a 2 = 1
(2)(3)(4) a 0 .
Equating coefficients ofx 3 :
(4)(5)a 5
+a 3
=0 sothat a 5
= − 1
(4)(5) a 3 = 1
(2)(3)(4)(5) a 1 .
Note that by using factorial notation we can express these coefficients concisely as:
a 2
=− 1 2! a 0 , a 3 =−1 3! a 1 , a 4 = 1 4! a 0 , a 5 = 1 5! a 1 .
Byequatinghigherordertermsinthesamewayfurthercoefficientsa m
canbedetermined
∞∑
and the power seriesy = a m
x m can then be written out explicitly. Observe that all
m=0
coefficientsa m
are multiples ofa 0
whenmis even, and multiples ofa 1
whenmis odd.
We can therefore collect terms together as follows:
∞∑
y = a m
x m
m=0
=a 0
+a 1
x+a 2
x 2 +a 3
x 3 +a 4
x 4 +a 5
x 5 +...
=a 0
+a 1
x− 1 2! a 0 x2 − 1 3! a 1 x3 + 1 4! a 0 x4 + 1 5! a 1 x5 +...
=a 0
(1− 1 2! x2 + 1 4! x4 −...)+a 1
(x− 1 3! x3 + 1 5! x5 − ...) (19.24)
ObservethattheseriesinthebracketsinEquation19.24arethepowerseriesexpansions
of cosx and sinx as discussed inSection 18.6. Hence wecan writethe solution as
y=a 0
cosx+a 1
sinx.
We see that this is simply the general solution of the differential equation as given
in the question but where the coefficients a 0
and a 1
are the arbitrary constants. Had
we not recognised the series expansions for the sine and cosine functions we would
still have obtained the general solution as expressed in the form of the power series in
Equation 19.24.