Explanation Given: The given differential equation is, 18 ( x − 4 ) 2 ( x − 6 ) y ″ + 9 x ( x − 4 ) y ′ − 32 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 differential equation, 18 ( x − 4 ) 2 ( x − 6 ) y ″ + 9 x ( x − 4 ) y ′ − 32 y = 0 Change the variable of the differential equation bye, z = 1 x y ′ = d y d z d z d x = d y d z ( − 1 x 2 ) = − z 2 d y d z y ″ = d d x ( − 1 x 2 d y d z ) = − 1 x 2 d 2 y d z 2 ( − 1 x 2 ) − d y d z ( − 2 x 3 ) = z 4 d 2 y d z 2 + 2 z 3 d y d z Substitute the values of y ′ and y ″ in the given differential equation to get, 18 ( x − 4 ) 2 ( x − 6 ) ( z 4 d 2 y d z 2 + 2 z 3 d y d z ) + 9 x ( x − 4 ) ( − z 2 d y d z ) − 32 y = 0 Substitute 1 z for x to get, 18 ( 1 z − 4 ) 2 ( 1 z − 6 ) ( z 4 d 2 y d z 2 + 2 z 3 d y d z ) + 9 ( 1 z ) ( 1 z − 4 ) ( − z 2 d y d z ) − 32 y = 0 18 z ( 1 − 4 z ) 2 ( 1 − 6 z ) d 2 y d z 2 + [ 4 ( 1 − 4 z ) ( 1 − 6 z ) − 1 ] 9 ( 1 − 4 z ) d y d z − 32 y = 0 18 z ( 1 − 4 z ) 2 ( 6 z − 1 ) d 2 y d z 2 + 9 ( 96 z 2 − 40 z + 3 ) ( 1 − 4 z ) d y d z − 32 y = 0 18 z ( 4 z − 1 ) 2 ( 6 z − 1 ) d 2 y d z 2 + 9 ( 96 z 2 − 40 z + 3 ) ( 4 z − 1 ) d y d z + 32 y = 0 d 2 y d z 2 + 96 z 2 − 40 z + 3 2 z ( 4 z − 1 ) ( 6 z − 1 ) d y d z + 32 18 z ( 4 z − 1 ) 2 ( 6 z − 1 ) y = 0 … ( 4 ) So, p ( z ) = 96 z 2 − 40 z + 3 2 z ( 4 z − 1 ) ( 6 z − 1 ) and q ( z ) = 32 18 z ( 4 z − 1 ) 2 ( 6 z − 1 )

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Chapter 8.6, Problem 42E

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Infinity is a singular point of the given differential equation and find the first nonzero terms in the series expansion of the solution about infinity.

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Infinity is a singular point of the given differential equation and find the first nonzero terms in the series expansion of the solution about infinity.

Solution Summary: The author explains how to find the first nonzero terms in the series expansion of the solution about infinity.

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To show: Infinity is a singular point of the given differential equation and find the first nonzero terms in the series expansion of the solution about infinity.

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Infinity is a singular point of the given differential equation and find the first nonzero terms in the series expansion of the solution about infinity.