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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 abouts x=0 of a general solution to the given differential equation.y"2xy'+3y=x2

The first four non-zero terms in a power series expansion to the given differential equation y"2xy'+3y=x2 are y(x)=a0 (1-3x2/2-x4/8+...)+a1 (x-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,

y"2xy'+3y=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

Substitute the values in the above formula you get,

Σn=2 n(n-1)anxn-2 -2x Σn=1 nanxn-1 + 3Σn=0 anxn =x2

In order to make the exponent common for all terms you 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)ak+2 xk-2Σk=1 k ak xk+3 Σk=0 ak xk =x2

2a2+3a0k=1 [(k+2)(k+1)ak+2+(3-2k)ak] xk =x2

Hence the expression is 2a2+3a0k=1 [(k+2)(k+1)ak+2+(3-2k)ak] xk =x2.

## Find the first four nonzero terms:

Expand the expression given in the previous step.

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

By equating the coefficients you get,

2a2+3a0=0

a2= -3a0/2

6a3+a1=0

a3=-a1/6

12a4-a2=1

a4=1/12-a0/8

Substitute the coefficient.

y(x)=a0+a1x+((-3a0)/2)x2+((-a1)/6)x3+(1/12-a0/8) x4+...

=a0 (1-3x2/2-x4/8+...)+a1 (x-x3/6+...) + (x4/12+...)

Hence, the first four non-zero terms is y(x)=a0 (1-3x2/2-x4/8+...)+a1 (x-x3/6+...) + (x4/12+...).

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