Math Problem Statement

Find the first four nonzero terms in the power series expansion for the solution of the differential equation (x^2 + 18)y'' + y = 0 about x = 0.

Solution

To solve the given problem, we are tasked with finding the first four nonzero terms in a power series expansion for the solution of the differential equation

(x2+18)y+y=0(x^2 + 18)y'' + y = 0

about x=0x = 0.


1. Power Series Assumption

Assume y(x)y(x) is given as a power series: y(x)=n=0anxn=a0+a1x+a2x2+a3x3+y(x) = \sum_{n=0}^\infty a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots

The derivatives are: y(x)=n=1nanxn1=a1+2a2x+3a3x2+y'(x) = \sum_{n=1}^\infty n a_n x^{n-1} = a_1 + 2a_2 x + 3a_3 x^2 + \cdots y(x)=n=2n(n1)anxn2=2a2+6a3x+12a4x2+y''(x) = \sum_{n=2}^\infty n(n-1)a_n x^{n-2} = 2a_2 + 6a_3 x + 12a_4 x^2 + \cdots


2. Substitution into the Differential Equation

Substitute y(x)y(x), y(x)y'(x), and y(x)y''(x) into the equation (x2+18)y+y=0(x^2 + 18)y'' + y = 0: (x2+18)(n=2n(n1)anxn2)+n=0anxn=0(x^2 + 18)\left(\sum_{n=2}^\infty n(n-1)a_n x^{n-2}\right) + \sum_{n=0}^\infty a_n x^n = 0

Expand (x2+18)y(x^2 + 18)y'': (x2+18)y=x2(n=2n(n1)anxn2)+18(n=2n(n1)anxn2)(x^2 + 18)y'' = x^2 \left(\sum_{n=2}^\infty n(n-1)a_n x^{n-2}\right) + 18\left(\sum_{n=2}^\infty n(n-1)a_n x^{n-2}\right) =n=2n(n1)anxn+n=218n(n1)anxn2= \sum_{n=2}^\infty n(n-1)a_n x^n + \sum_{n=2}^\infty 18n(n-1)a_n x^{n-2}

Rewriting both terms with the yy series, we combine like powers of xx.


3. Simplifying and Recurrence Relation

Matching coefficients of xnx^n, we find a recurrence relation: n(n1)an+18(n+2)(n+1)an+2+an=0n(n-1)a_n + 18(n+2)(n+1)a_{n+2} + a_n = 0

For specific terms:

  1. n=0n = 0: a0+18a2=0    a2=a018a_0 + 18a_2 = 0 \implies a_2 = -\frac{a_0}{18}

  2. n=1n = 1: a1+1832a3=0    a3=a1108a_1 + 18 \cdot 3 \cdot 2 a_3 = 0 \implies a_3 = -\frac{a_1}{108}

  3. n=2n = 2: Solve for a4a_4 in terms of a0a_0, using the pattern established above.


4. First Four Nonzero Terms

Using the recurrence relations, the first four terms are: y(x)=a0(1x218)+a1(xx3108)y(x) = a_0 \left(1 - \frac{x^2}{18}\right) + a_1 \left(x - \frac{x^3}{108}\right)

Thus, the solution is: y(x)=a0a018x2+a1xa1108x3y(x) = a_0 - \frac{a_0}{18}x^2 + a_1 x - \frac{a_1}{108}x^3

Would you like me to provide further details on deriving the recurrence relation or verifying the solution?


Relative Questions:

  1. How does the recurrence relation help in solving power series expansions?
  2. What are the conditions for a power series solution to converge?
  3. How would the solution change if additional terms (higher order) were needed?
  4. Why is it necessary to solve for terms in terms of a0a_0 and a1a_1?
  5. How does the presence of the x2+18x^2 + 18 factor affect the recurrence relation?

Tip:

When solving differential equations using power series, carefully expand each term and match coefficients systematically to avoid algebraic

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Math Problem Analysis

Mathematical Concepts

Power Series
Differential Equations
Recurrence Relations
Series Expansions

Formulas

y(x) = sum_{n=0}^{infinity} a_n x^n
y'(x) = sum_{n=1}^{infinity} n a_n x^{n-1}
y''(x) = sum_{n=2}^{infinity} n(n-1) a_n x^{n-2}
(x^2 + 18)y'' + y = 0

Theorems

Method of solving differential equations using power series
Recurrence relations for power series coefficients

Suitable Grade Level

Grades 11-12