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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’s x=0 of a general solution to the given differential equation. y"-xy'+2y=cosx

The first four non-zero terms in a power series expansion to the given differential equation y"-xy'+2y=cosx are y(x)=a0+a1x+(1/2-a0) x2-a1/6 x3 -1/24 x4-a1/120x5+....

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'+2y=cosx

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,

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

Substitute the values in the above formula you get,

Σn=2 n(n-1)anxn-2 -x Σn=1 nanxn-1 + 2 Σn=0 anxn =1-x2/2!+x4/4! -...

(2a2+a0)+Σk=1 [(k+2)(k+1)ak+2+(2-k)ak] xk =1-x2/2!+x4/4! -...

Hence, the expression is (2a2+a0)+Σk=1 [(k+2)(k+1)ak+2+(2-k)ak] xk =1-x2/2!+x4/4! -... .

## Find the first four nonzero terms:

Expand the expression given in the previous step.

(2a2+2a0)+(6a3+a1) x+(12a4)x2+(20a5-a3)+.....=1-x2/2!+x4/4!

By equating the coefficients you get,

2a2+2a0=1

a2=1/2-a0

6a3+a1=0

a3=-a1/6

12a4= -1/2

a4= -1/24

20a5-a3=0

a5=-a1/120

Substitute the coefficient.

y(x)=a0+a1x+(1/2-a0) x2-a1/6 x3 -1/24 x4-a1/120x5+...

Hence, the first four nonzero terms is y(x)=a0+a1x+(1/2-a0) x2-a1/6 x3 -1/24 x4-a1/120x5+...

.