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Expert-verified 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.(1+x2)y"-xy'+y=e-x

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

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"-xy'+y=e-x

The above equation is written in the standard form as,

y"-x/1+x2 y'+1/1+x2 y=e-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,

e-x=1-x+x2/2!-x3/3!+x4/4!+...

Substitute the values in the above formula you get,

(1+x2n=2 n(n-1)anxn-2 +x Σn=1 nanxn-1 + Σn=0 anxn =1-x+x2/2!-x3/3!+x4/4!+.....

Σn=2 n(n-1)anxn-2n=2 n(n-1)anxn

n=1 nanxn-1 + Σn=0 anxn =1-x+x2/2!-x3/3!+x4/4!+...

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 k=n+2 in the other term.

Σk=0 (k+2) (k+1)akxk+ Σk=2 k(k-1)akxk - Σk=1 kakxk

=1-x+x2/2!-x3/3!+x4/4!+...

(2a2+a0+6a3+a1x-a1x)+Σk=2 ((k+2) (k+1)ak+2 +k(k-1)ak-kak+ak) xk

=1-x+x2/2!-x3/3!+x4/4!+...

(2a2+a0+6a3)+Σk=2 ((k+2) (k+1)ak+2 +(k2-2k+1)ak) xk

=1-x+x2/2!-x3/3!+x4/4!+...

(2a2+a0+6a3)+Σk=2 ((k+2) (k+1)ak+2 +(k-1)2ak) xk

=1-x+x2/2!-x3/3!+x4/4!+...

Hence the expression is (2a2+a0+6a3)+Σk=2 ((k+2) (k+1)ak+2 +(k-1)2ak) xk =1-x+x2/2!-x3/3!+x4/4!+....

## Find the first four non-zero terms:

Expand the expression given in the previous step.

(2a2+a0+6a3)+(12a4+a1) x2+(20a5+4a3) x2+...=1-x+x2/2!-x3/3!+x4/4!+...

By equating the coefficients you get,

2a2+a0=1

a2=(1-a0)/2

6a3 = -1

a3= -1/6

12a4+a1=1/2

a4=1/24-a1/12

Substitute the coefficient.

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

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

Hence, the first four nonzero terms is y(x)=a0 (1-x2/2+...)+a1 (x-x4/12+...) + (x2/2-x3/6+x4/24+...). ### Want to see more solutions like these? 