Explanation Given: The given differential equation is, x 2 y ″ + y ′ − 2 y = 0 . Approach: To derive a series solution about the singular point x 0 of the equation a 2 ( x ) y ″ ( x ) + a 1 ( x ) y ′ ( x ) + a 0 ( x ) y ( x ) = 0 , x > x 0 … ( 1 ) The following steps are given below. 1. Set p ( x ) = a 1 ( x ) a 2 ( x ) , q ( x ) = a 0 ( x ) a 2 ( x ) . If both ( x − x 0 ) p ( x ) and ( x − x 0 ) 2 q ( x ) are analytic at x 0 , then x 0 is a regular point and remaining steps apply. 2. Let w ( r , x ) = ( x − x 0 ) r ∑ n = 0 ∞ a n ( x − x 0 ) n = ∑ n = 0 ∞ a n ( x − x 0 ) n + r … ( 2 ) And, using term wise differentiation, substitute w ( r , x ) into equation ( 1 ) to obtain an equation of the form A 0 ( x − x 0 ) r + J + A 1 ( x − x 0 ) r + J + 1 + ... = 0 3. Set the coefficients A 0 , A 1 , A 3 , equal to zero. [Notice that the equation A 0 = 0 is just a constant multiple of the indicial equation r ( r − 1 ) + p 0 r + q 0 = 0 here, p 0 = lim x → 0 x p ( x ) , q 0 = lim x → 0 x 2 q ( x ) , and r 0 = lim x → 0 x 3 r ( x ) .] 4. Use the system of equations A 0 = 0 , A 1 = 0 , ... , A k = 0 to find a recurrence relation involving a k and a 0 , a 1 ,…, a k − 1 . 5. Take r = r 1 , the larger root of the indicial equation, and use the obtained in step 4 to determine a 1 , a 2 ,…, recursively in terms of a 0 and r 1 . 6. A series expansion of a solution to equation ( 1 ) is, w ( r 1 , x ) = ( x − x 0 ) r 1 ∑ n = 0 ∞ a n ( x − x 0 ) n , x > x 0 … ( 3 ) Calculation: Consider the given equation, x 2 y ″ + y ′ − 2 y = 0 . p ( x ) = 1 x 2 and q ( x ) = − 2 x 2 p 0 = lim x → 0 x p ( x ) = lim x → 0 x ( 1 x 2 ) = 0 q 0 = lim x → 0 x 2 q ( x ) = lim x → 0 x 2 ( − 2 x 2 ) = − 2 The indicial equation is, r ( r − 1 ) − 2 = 0 r 2 − 2 r + r − 2 = 0 ( r − 2 ) ( r + 1 ) = 0 The roots of the indicial equation are r 1 = − 1 and r 2 = 2 . Let y 1 ( x ) be a solution of the differential equation. y 1 ( x ) = ∑ n = 0 ∞ a n x n + r y ′ 1 ( x ) = ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 y ″ 1 ( x ) = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 Substitute the values of y 1 ( x ) , y ′ 1 ( x ) , and y ″ 1 ( x ) in the given equation to get,

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 8.7, Problem 23E

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The first three nonzero terms in the series expansion about the irregular singular point x=0 for a general solution to the given differential equation.

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Chapter 8.7, Problem 22E

Chapter 8.7, Problem 24E

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4. Negate the following sentences. Then (where appropriate) indicate whether the orig- inal statement is true, or the negation is true. ⚫ If I take linear algebra, then I will do my homework or go to class. • (x > 2 or x < −2) ⇒ |x| ≥ 2 • P⇒ (QVR) ⇒(¬PV QV R) Vn EN Em E Q (nm = 1) • Ex E N Vy & Z (x. y = 1)

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To state: The first three nonzero terms in the series expansion about the irregular singular point x = 0 for a general solution to the given differential equation.

Solution Summary: The author states the three nonzero terms in the series expansion about the irregular singular point x=0 for a general solution to the given differential equation.

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The first three nonzero terms in the series expansion about the irregular singular point x=0 for a general solution to the given differential equation.

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To state: The first three nonzero terms in the series expansion about the irregular singular point x = 0 for a general solution to the given differential equation.

Solution Summary: The author states the three nonzero terms in the series expansion about the irregular singular point x=0 for a general solution to the given differential equation.

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To state: The first three nonzero terms in the series expansion about the irregular singular point x=0 for a general solution to the given differential equation.

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Chapter 8 Solutions

To state:

The first three nonzero terms in the series expansion about the irregular singular point x=0 for a general solution to the given differential equation.