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### 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’s x=0 of a general solution to the given differential equation. y"-(sin x)y=cos x

The first four nonzero terms in a power series expansion to the given differential equation y"-(sin x)y=cos x are y(x)=a0+a1x+1/2 x2+a0/6 x3 + (a1/12-1/24) x4+... .

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"-(sin x)y=cos x

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,

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

cos x=1-x2/2!+x4/4!-...

Substitute the values in the above formula you get,

Σn=2 n(n-1)anxn-2 -(x-x3/3!+x5/5!-...) Σn=0 anxn

=1-x2/2!+x4/4!-...

(2a2+6a3x+12a4x2+20a5x3+...) - (x-x3/3!+x5/5!-...) (a0+a1x+a2x2+a3x3+...) 1-x2/2!+x4/4!-...

Hence, the expression is (2a2+6a3x+12a4x2+20a5x3+...) - (x-x3/3!+x5/5!-...) (a0+a1x+a2x2+a3x3+...) 1-x2/2!+x4/4!-... .

## Find the first four nonzero terms:

Expand the expression given in the previous step.

(2a2+6a3x+12a4x2+20a5x3+...) - (a0x2+a1x+a2x3+a3x4+...) + (a0x3/3!+a1x4/3!+a2x5/3!+a3x6/3!+...) - (a0x5/5!+a1x6/5!+a2x7/5!+a3x8/5!+...) + ... =1-x2/2!+x4/4!+...

2a2+(6a3-a0) x+(12a4-a1) x2+...= 1-x2/2!+x4/4!+...

By equating the coefficients you get,

2a2=1

a2=1/2

6a3-a0=0

a3=a0/6

12a4-a1= -1/2

a4=a1/12-1/24

Substitute the coefficient.

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

Hence, the first four nonzero terms are y(x)=a0+a1x+1/2 x2+a0/6 x3 + (a1/12-1/24) x4+... .