Explanation Theorem used: Let r 1 and r 2 be the roots of the indicial equation F ( r ) = r ( r − 1 ) + p 0 r + q 0 with r 1 ≥ r 2 if r 1 and r 2 are real. Then in either the interval − ρ < x < 0 or the interval 0 < x < ρ , there exists a solution of the form y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] where a n ( r 1 ) are the given by recurrence relation with a 0 = 1 and r = r 1 . If r 1 − r 2 is not zero or a positive integer, then in either the interval − ρ < x < 0 or the interval 0 < x < ρ , there exists a second solution of the form y 2 ( x ) = | x | r 2 [ 1 + ∑ n = 1 ∞ a n ( r 2 ) x n ] a n ( r 2 ) are also determined by the recurrence relation with a 0 = 1 and r = r 2 . The power series converges at least for | x | < ρ . If r 1 = r 2 , then the second solution is y 2 ( x ) = y 1 ( x ) ln | x | + | x | r 1 [ 1 + ∑ n = 1 ∞ b n ( r 1 ) x n ] . If r 1 − r 2 = N , positive integer, then y 2 ( x ) = a y 1 ( x ) ln | x | + | x | r 2 [ 1 + ∑ n = 1 ∞ c n ( r 2 ) x n ] . Calculation: The given differential equation is given as x 2 y " + x y ' + 2 x y = 0 . Compare the equation x 2 y " + x y ' + 2 x y = 0 with P ( x ) y " + Q ( x ) y ' + R ( x ) y = 0 here P ( x ) = x 2 , Q ( x ) = x and R ( x ) = 2 x . The singular points occur when P ( x 0 ) = 0 thus the singular point of the given equation is x 0 = 0 . If x 0 = 0 is a regular singular point, check if the following limits are finite. p 0 = lim x → 0 x Q ( x ) P ( x ) and q 0 = lim x → 0 x 2 R ( x ) P ( x ) For x 0 = 0 , therefore the value of p 0 = lim x → 0 x x x 2 = 1 and q 0 = lim x → 0 x 2 2 x x 2 = 0 . Since both are finite, x 0 = 0 is a regular singular point of the given differential equation. Because p 0 = 1 and q 0 = 0 , the corresponding indicial equation is as follows: F ( r ) = r ( r − 1 ) + p 0 r + q 0 Substitute the value of p 0 = 1 and q 0 = 0 in F ( r ) = r ( r − 1 ) + p 0 r + q 0 = 0 . F ( r ) = r ( r − 1 ) + r = r 2 − r + r = r 2 Let F ( r ) = 0 in the above equation it becomes as follows: r 2 = 0 r 1 = r 2 = 0 Therefore, the roots of the equation r 2 = 0 is r 1 = 0 , r 2 = 0 Hence, x 0 is a regular singular point. From the above Theorem, the first solution is given as y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] . Substitute the value of r 1 = 0 in y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] . y 1 ( x ) = | x | r 1 [ 1 + ∑ n = 1 ∞ a n ( r 1 ) x n ] = 1 + ∑ n = 1 ∞ a n ( 0 ) x n Differentiate the equation y 1 ( x ) = 1 + ∑ n = 1 ∞ a n ( 0 ) x n . y 1 ' ( x ) = ∑ n = 0 ∞ ( n + 1 ) a n + 1 ( 0 ) x n The coefficients a n ( 0 ) , apart from a 0 ( 0 ) = 1 are determined by the recurrence relation F ( r + n ) a n + ∑ k = 1 n − 1 a k [ ( r + k ) p n − k + q n − k ] = 0 in case of r = r 1 . n 2 a n ( 0 ) + ∑ k = 0 n − 1 a k ( 0 ) [ k p n − k + q n − k ] = 0 , where ∑ n = 0 ∞ p n x n = x Q ( x ) P ( x ) = 1 and ∑ n = 0 ∞ q n x n = x 2 R ( x ) P ( x ) = 2 x

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 5.7, Problem 3P

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To show: That the given differential equation has a regular singular point at x=0 is and determine two solutions for x>0 by using the given data.

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Chapter 5.7, Problem 2P

Chapter 5.7, Problem 4P

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Chapter 5 Solutions

Elementary Differential Equations

Knowledge Booster

That the given differential equation has a regular singular point at x = 0 is and determine two solutions for x &gt; 0 by using the given data.

To show: That the given differential equation has a regular singular point at x=0 is and determine two solutions for x>0 by using the given data.

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Chapter 5 Solutions

That the given differential equation has a regular singular point at x = 0 is and determine two solutions for x > 0 by using the given data.