Explanation The differential given is given below. x 2 R ″ + 2 x R ′ + [ α 2 x 2 − n ( n + 1 ) ] R = 0 ...... ( 1 ) Function R ( x ) is given as follows: R ( x ) = ( α x ) − 1 2 Z ( x ) ...... ( 2 ) Differentiate equation (2). R ′ ( x ) = α − 1 2 ( − 1 2 x − 3 2 Z ( x ) + x − 1 2 Z ′ ( x ) ) ...... ( 3 ) Again differentiate the above equation. R ″ ( x ) = α − 1 2 ( 1 2 x − 3 2 Z ′ ( x ) + 3 4 x − 5 2 Z ( x ) − 1 2 x − 3 2 Z ′ + x − 1 2 Z ″ ) ...... ( 4 ) Now substitute the value of R ′ and R ″ in equation (1). x 2 R ″ + 2 x R ′ + [ α 2 x 2 − n ( n + 1 ) ] R = 0 [ x 2 α − 1 2 ( 1 2 x − 3 2 Z ′ ( x ) + 3 4 x − 5 2 Z ( x ) − 1 2 x − 3 2 Z ′ + x − 1 2 Z ″ ) + 2 x α − 1 2 ( − 1 2 x − 3 2 Z ( x ) + x − 1 2 Z ′ ( x ) ) + [ α 2 x 2 − n ( n + 1 ) ] ( α x ) − 1 2 Z ( x ) ] = 0 [ x 2 ( 1 2 x − 3 2 Z ′ ( x ) + 3 4 x − 5 2 Z ( x ) − 1 2 x − 3 2 Z ′ + x − 1 2 Z ″ ) + 2 x ( − 1 2 x − 3 2 Z ( x ) + x − 1 2 Z ′ ( x ) ) + [ α 2 x 2 − n ( n + 1 ) ] x − 1 2 Z ( x ) ] = 0 ( Divide by α − 1 2 ) Further simplify the above equation

Student Solutions Manual For Zill's A First Course In Differential Equations With Modeling Applications, 11th

11th Edition

ISBN: 9781305965737

Author: Dennis G. Zill

Publisher: Brooks Cole

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Chapter 6.4, Problem 56E

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To show: The general solution of the differential equation x2R″+2xR′+[α2x2−n(n+1)]R=0 on the interval (0,∞) is R(x)=c1jn(αx)+c2yn(αx) using the substitution R(x)=(αx)−12Z(x).

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Chapter 6.4, Problem 55E

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The general solution of the differential equation x 2 R ″ + 2 x R ′ + [ α 2 x 2 − n ( n + 1 ) ] R = 0 on the interval ( 0 , ∞ ) is R ( x ) = c 1 j n ( α x ) + c 2 y n ( α x ) using the substitution R ( x ) = ( α x ) − 1 2 Z ( x ) .

Solution Summary: The author shows the general solution of the differential equation x2R′′+2x

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To show: The general solution of the differential equation x2R″+2xR′+[α2x2−n(n+1)]R=0 on the interval (0,∞) is R(x)=c1jn(αx)+c2yn(αx) using the substitution R(x)=(αx)−12Z(x).

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Chapter 6 Solutions