12/2/21, 2:06 PM
Section 8.3: Power Series Solution to Linear Diffe-Huzaifa Karimjee
Instructor: Nelson De La Rosa
Course: MAP 2302: Differential
Equations (FIU Online) Fall 2021 FIU
Student: Huzaifa Karimjee
Date: 12/02/21
Assignment: Section 8.3: Power Series
Solution to Linear Diffe
Find at least the first four nonzero terms in a power series expansion about x = 0 for a general solution to the given differential
equation.
y′ + (x + 10)y = 0
∞
Let y be given by the sum
∑
∞
n
an x and the derivative y′ be given by the sum
n=0
∑
n−1
nan x
. Substitute the sums into the
n=1
differential equation and apply the distributive property.
y′ + (x + 10)y = 0
∞
∞
nan x
∑
n−1
+ ​(x + 10​)
n=1
∞
nan x
n−1
n=1
∑
+x
∑
an x + 10
n−1
+
n=1
= 0
∑
∑
an x
n
= 0
n=0
∞
nan x
n
∞
n
n=0
∞
an x
n=0
∞
∑
∑
∞
an x
n+1
+
n=0
10an x
∑
n
= 0
n=0
k
Rewrite the summands so they have the generic term x .
∞
∑
∞
nan x
n−1
+
n=1
an x
+
n=0
∞
∑
∑
∞
n+1
(k + 1)ak + 1 x +
k=0
∑
n
= 0
n=0
∞
k
10an x
∑
∞
k
ak − 1 x +
k=1
10ak x
∑
k
= 0
k=0
Combine the two summands with a starting index of k = 0.
∞
∑
∞
k
(k + 1)ak + 1 x +
k=0
∑
∞
k
ak − 1 x +
k=1
k
= 0
k=0
∞
∑
10ak x
∑
∞
k
(k + 1)ak + 1 + 10ak x +
k=0
∑
ak − 1 x
k
= 0
k=1
Evaluate the first sum at k = 0 so that the resulting sum has a starting index at k = ​1, then combine the sums.
∞
∑
∞
k
(k + 1)ak + 1 + 10ak x +
k=0
∑
ak − 1 x
k
= 0
k=1
∞
a1 + 10a0 +
∑
(k + 1)ak + 1 + 10ak + ak − 1 x
k
= 0
k=1
Set the constant terms equal to 0 and solve for a1 .
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12/2/21, 2:06 PM
Section 8.3: Power Series Solution to Linear Diffe-Huzaifa Karimjee
a1 + 10a0 = 0
a1 = − 10a0
∞
Next, solve for a2 by substituting k = 1 into the equation
∑
k
(k + 1)ak + 1 + 10ak + ak − 1 x = 0.
k=1
2a2 + 10a1 + a0 = 0
2a2 + 10 − 10a0 + a0 = 0
a2 =
99
2
a0
Similarly, solve for a3 by substituting k = 2 into the equation.
3a3 + 10a2 + a1 = 0
3a3 + 10
99
2
a0 +
= 0
− 10a0
a3 = −
485
3
Use the four coefficients a0 ​, a1 = − 10a0 ​, a2 =
y(x) = a0 − 10a0 x +
99
2
2
a0 x −
485
3
a0
99
2
a0 ​, and a3 = −
485
3
a0 to find the first four terms of the solution.
3
a0 x + ⋯
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