Suppose using the convergent power series ∑ n=0
[infinity]
a n
x n
to solve the differential equation (1+9x 2
)y ′′
−18y=0, has led to obtaining the relation ∑ n=0
[infinity]
[(n+2)(n+1)a n+2
+9(n−2)(n+1)a n
]x n
=0 You do not need to prove this relation. Use this to find the solution to the differential equation. Find the radius of convergence and introduce the interval in which the solution is valid. Write your solution using sigma notation, and simplify as much as possible. Find a power series solution y(x)=∑ n=0
[infinity]
a n
x n
for the differential equation (x 2
+2)y ′′
+4xy ′
+2y=0. Solution We assume that the solution of the differential equation has a Maclaurin series y(x)=∑ n=0
[infinity]
a n
x n
with positive radius of convergence R. If we substitute the power series into the differential equation, 0
= n=0
∑
[infinity]
n(n−1)a n
x n
+ n=0
∑
[infinity]
2n(n−1)a n
x n−2
+ n=0
∑
[infinity]
4na n
x n
+ n=0
∑
[infinity]
2a n
x n
= n=0
∑
[infinity]
n(n−1)a n
x n
+ n=−2
∑
[infinity]
2(n+2)(n+1)a n+2
x n
+ n=0
∑
[infinity]
4na n
x n
+ n=0
∑
[infinity]
2a n
x n
= n=0
∑
[infinity]
[2(n+2)(n+1)a n+2
+(n 2
+3n+2)a n
]x n
Equating coefficients to zero gives a n+2
=− 2(n+2)(n+1)
(n 2
+3n+2)a n
=− 2
a n
,n≥0. Iteration of this result leads to a 2
=− 2
a 0
,a 4
= 2 2
a 0
,a 6
=− 2 3
a 0
,a 8
= 2 4
a 0
,…;
a 3
=− 2
a 1
,a 5
= 2 2
a 1
,a 7
=− 2 3
a 1
,a 9
= 2 4
a 1
,….
The solution is therefore y(x)=
=
a 0
(1− 2
x 2
+ 2 2
x 4
− 2 3
x 6
+ 2 4
x 8
−⋯)
+a 1
(x− 2
x 3
+ 2 2
x 5
− 2 3
x 7
+ 2 4
x 9
−⋯)
a 0
n=0
∑
[infinity]
2 n
(−1) n
x 2n
+a 1
n=0
∑
[infinity]
2 n
(−1) n
x 2n+1
.
Both series y 1
(x)=∑ n=0
[infinity]
[(−1) n
/2 n
]x 2n
and y 2
(x)=∑ n=0
[infinity]
[(−1) n
/2 n
]x 2n+1
converge for − 2
, and hence we have a solution of the differential equation on this interval. Alternatively, these are geometric series with sums y 1
(x)= x 2
+2
2
and y 2
(x)= x 2
+2
2x
, and it is straightforward to show that these functions satisfy the differential equation for all x. Thus, a general solution of the differential equation is y(x)=a 0
y 1
(x)+a 1
y 2
(x)= x 2
+2
2(a 0
+a 1
x)
Power series do not always yield general solutions of differential equations. Sometimes they yield solutions, but not general solutions, and sometimes they yield no solution whatsoever. We illustrate in the following two examples.
