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Q27E

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Found in: Page 450

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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# In Problems 21-28, use the procedure illustrated in Problem 20 to find at least the first four nonzero terms in a power series expansion about’s x=0 of a general solution to the given differential equation.(1-x2) y"-y'+y=tan x

The first four nonzero terms in a power series expansion to the given differential equation (1-x2) y"-y'+y=tan x are y(x)=a0 (1-x2/2-x3/6-x4/12-...)+a1 (x+x2/2+x4/24+...)+(x3/6+x4/12+...)..

See the step by step solution

## Define power series expansion:

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a recurrence relation for the coefficients.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable.

It is generally given by the formula,

y(x)=Σn=0 anxn

## Find the expression:

Given,

(1-x2) y"-y'+y=tan x

The standard form is

y"-1/(1-x2)y'+1/(1-x2)y=tan x/(1+x2)

Use the formula,

y(x)=Σn=0 anxn

Taking derivative of the above equation,

y'(x)=Σn=1 nanxn-1

y"(x)=Σn=2 n(n-1)anxn-2

The series expansion is

tan x=x+x3/3+2x5/15+...

Substitute the values in the above formula you get,

(1+x2n=2 n(n-1)anxn-2n=1 nanxn-1 + Σn=0 anxn =x+x3/3+2x5/15+...

Σn=2 n(n-1)anxn-2n=2 n(n-1)anxnn=1 nanxn-1n=0 anxn =x+x3/3+2x5/15+...

In order to make the exponent common for all terms we will make the substitution n-2=k therefore k=n+2 in first term and n-1=k therefore n=k+1 in the other term.

Σk=0 (k+2)(k+1) ak+2 xkk=2 k(k-1) ak xk - Σk=0 (k+1) ak+1 xk + Σk=0 ak xk =x+x3/3+2x5/15+...

Hence the expression is Σk=0 (k+2)(k+1) ak+2 xkk=2 k(k-1) ak xk - Σk=0 (k+1) ak+1 xk + Σk=0 ak xk =x+x3/3+2x5/15+... .

## Find the first four nonzero terms:

Expand the expression given in the previous step.

(2a2+a0-a1)+(6a3+a1-2a2) x+ (12a4-2a2-3a3+a2) x2+...=x+x3/3+2x5/15+...

By equating the coefficients you get,

2a2+a0-a1=0

a2=(a1-a0)/2

6a3+a1-2a2=1

a3=(1-a0)/6

12a4-2a2-3a3+a2=0

a4=a1/24-a0/12+1/24

Substitute the coefficient.

y(x)=a0+a1x+((a1-a0)/2)x2+((1-a0)/6)x3+(a1/24-a0/12+1/24) x4+...

=a0 (1-x2/2-x3/6-x4/12-...)+a1 (x+x2/2+x4/24+...) + (x3/6+x4/12+...)

Hence, the first four nonzero terms are y(x)=a0 (1-x2/2-x3/6-x4/12-...)+a1 (x+x2/2+x4/24+...)+(x3/6+x4/12+...).

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