Explanation Given: The linear-third equation is, 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x 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 ) Do the following steps. 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 termwise 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 ) here a 0 is arbitrary and the a n ’s are defined in terms of a 0 and r 1 . Calculation: Compare the equation 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x y = 0 with equation ( 1 ) . This gives, a 3 ( x ) = 6 x 3 a 2 ( x ) = ( 13 x 2 − x 3 ) a 1 ( x ) = x a 0 ( x ) = − x And, p ( x ) = ( 13 x 2 − x 3 ) 6 x 3 q ( x ) = x 6 x 3 r ( x ) = − x 6 x 3 Calculate p 0 , q 0 , and r 0 as, p 0 = lim x → 0 x ⋅ ( 13 x 2 − x 3 ) 6 x 3 = 13 6 q 0 = lim δ x → 0 x 2 ⋅ x 6 x 3 = lim δ x → 0 x 3 6 x 3 = 1 6 r 0 = lim x → 0 x 3 ⋅ − x 6 x 3 = 0 The indicial equation is given as, r ( r − 1 ) ( r − 2 ) + p 0 r ( r − 1 ) + q 0 r + r 0 = 0 r ( r − 1 ) ( r − 2 ) + 13 6 r ( r − 1 ) + r 6 + 0 = 0 6 r ( r − 1 ) ( r − 2 ) + 13 r ( r − 1 ) + r = 0 r ( 6 r 2 − 5 r ) = 0 Further simplify, r 2 ( 6 r − 5 ) = 0 The roots of the indicial equation are 0 , 0 , and − 1 2 , and the larger root is 5 6 . Substitute ∑ n = 0 ∞ a n ( x − x 0 ) n + r for y , ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 for y ′ , ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 for y ″ , and ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) ( n + r − 2 ) a n x n + r − 3 for y ‴ in equation 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x y = 0

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Chapter 8.6, Problem 38E

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At least the first four nonzero terms in the series expansion about x=0 for a solution to the given linear third-order equation 6x3y‴+(13x2−x3)y″+xy′−xy=0.

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Chapter 8.6, Problem 37E

Chapter 8.6, Problem 39E

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At least the first four nonzero terms in the series expansion about x = 0 for a solution to the given linear third-order equation 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x y = 0 .

Solution Summary: The author explains how to find the first four nonzero terms in the series expansion about x=0 for a solution to the given linear third-order equation.

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To find:

At least the first four nonzero terms in the series expansion about x=0 for a solution to the given linear third-order equation 6x3y‴+(13x2−x3)y″+xy′−xy=0.

Want to see the full answer?

Chapter 8 Solutions

At least the first four nonzero terms in the series expansion about x = 0 for a solution to the given linear third-order equation 6 x 3 y ‴ + ( 13 x 2 − x 3 ) y ″ + x y ′ − x y = 0 .

Solution Summary: The author explains how to find the first four nonzero terms in the series expansion about x=0 for a solution to the given linear third-order equation.

Videos

To find: At least the first four nonzero terms in the series expansion about x=0 for a solution to the given linear third-order equation 6x3y‴+(13x2−x3)y″+xy′−xy=0.

Want to see the full answer?

Chapter 8 Solutions

To find:

At least the first four nonzero terms in the series expansion about x=0 for a solution to the given linear third-order equation 6x3y‴+(13x2−x3)y″+xy′−xy=0.