Explanation Given: The given equation is 3 x ( 1 − x ) y ″ + ( 1 − 27 x ) y ′ − 45 y = 0 . Approach: The standard hypergeometric equation is given as, x ( 1 − x ) y ″ + [ γ − ( α + β + 1 ) ] y ′ − α β y = 0 ... ( 1 ) Here, α , β , and γ are fixed parameters. The first solution to the hypergeometric equation is called as Gaussian hypergeometric function and it is given as, y 1 ( x ) = 1 + ∑ n = 1 ∞ ( α ) n ( β ) n n ! ( γ ) n x n It is also represented as, F ( α , β ; γ ; x ) = 1 + ∑ n = 1 ∞ ( α ) n ( β ) n n ! ( γ ) n x n ... ( 2 ) The other solution to the hypergeometric equation is given as, y 2 ( x ) = x 1 − γ + ∑ n = 1 ∞ ( α + 1 − γ ) n ( β + 1 − γ ) n n ! ( 2 − γ ) n x n + 1 − γ . This solution is also represented as, y 2 ( x ) = x 1 − γ F ( α + 1 − γ , β + 1 − γ ; 2 − γ ; x ) ... ( 3 ) Calculation: Consider the given equation, 3 x ( 1 − x ) y ″ + ( 1 − 27 x ) y ′ − 45 y = 0 . Rewrite the above equation as, x ( 1 − x ) y ″ + ( 1 3 − 9 x ) y ′ − 15 y = 0 Compare the above equation with the standard equation ( 1 ) , this gives, γ = 1 3

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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