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Q21E

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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

# 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 x=0 of a general solution to the given differential equation. y'-xy=sinx

The first four nonzero terms in a power series expansion to the given differential equation y'-xy=sinx are y(x)=a0 (1+x2/2+x4/8+x6/48+...)+(x2/2+x4/12+11x6/720+...).

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,

y'+xy=sinx

Use the formula,

y(x)=Σn=0 anxn

Taking derivative of the above equation,

y'(x)=Σn=0 n anxn-1

The Maclaurin series for sinx is

sinx=x-x3/3!+x5/5! - ...

Substitute the values in the above formula you get,

Σn=0 n anxn-1 -x Σn=0 anxn = x-x3/3! + x5/5! - ...

Σn=0 n anxn-1 - Σn=0 anxn+1 = x-x3/3! + x5/5! - ...

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

Σk=0 (k+1) ak+1 xk - Σk=0 ak-1 xk= x-x3/3! + x5/5! - ...

a1k=1 (k+1) ak+1 xk - Σk=1 ak-1 xk= x-x3/3! + x5/5! - ...

a1k=1 [(k+1) ak+1 - ak-1 ] xk= x-x3/3! + x5/5! - ...

Hence, the expression is a1k=1 [(k+1) ak+1 - ak-1 ] xk= x-x3/3! + x5/5! - ... .

## Find the first four nonzero terms:

Expand the expression given in the previous step.

a1+(2a2-a0)x+(3a3-a1)x2+(4a4-a2)x3+(5a5-a3)x4+(6a6-a4) x5+.....=x-x3/6+x5/120-x7/5040+.....

By equating the coefficients you get,

a1=0

2a2-a0=1

a2 = (1+a0)/2

3a3-a1=0

3a3-0=0

a3=0

4a4-a2=-1/6

a4=1/12+a0/8

5a5-a3=0

a5=0

6a6-a4=1/120

a6=11/720+a0/48

Substitute the coefficient.

y(x)=a0+(1+a0)/2 x2+ (1/12+a0/8) x4=(11/720+a0/48) x6+...

y(x)=a0+(1+x2/2+x4/8+x6/48 +...) + (x2/2+x4/12+11x6/720 +...)

Hence the first four nonzero terms are,

y(x)=a0+(1+x2/2+x4/8+x6/48 +...) + (x2/2+x4/12+11x6/720 +...)

## Most popular questions for Math Textbooks

Question 18: In Problems, find a power series expansion for $f\left(X\right)$ , given the expansion for f(x).

role="math" localid="1664283848012" $f\left(x\right)=\mathrm{sin}x=\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}}{\left(2k+1\right)!}{x}^{2k+1}$