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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 450
Fundamentals Of Differential Equations And Boundary Value Problems

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

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Short Answer

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

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+...

.

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