Explanation Given: The second order differential equation ( 1 − x cot x ) y ″ − x y ′ + y = 0 and y 1 ( x ) = x , y 2 ( x ) = sin x . Theorem used: Suppose that y 1 and y 2 are two solutions of equation L [ y ] = y ″ + p ( t ) y ′ + q ( t ) y = 0 and that the initial conditions y ( t 0 ) = y 0 , y ′ ( t 0 ) = y ′ 0 are assigned. Then it is always possible to choose the constants c 1 , c 2 so that y = c 1 y 1 ( t ) + c 2 y 2 ( t ) satisfies the differential equation L [ y ] = y ″ + p ( t ) y ′ + q ( t ) y = 0 and the initial conditions y ( t 0 ) = y 0 , y ′ ( t 0 ) = y ′ 0 if and only if the Wronskian W ( y 1 , y 2 ) = y 1 y ′ 2 − y ′ 1 y 2 is not zero at t 0 . Formula used: The Wronskian determinant of two functions y 1 and y 2 is given by W ( y 1 , y 2 ) = y 1 y ′ 2 − y ′ 1 y 2 . Calculation: The given differential equation is ( 1 − x cot x ) y ″ − x y ′ + y = 0 . Differentiate y 1 ( x ) = x and obtain the value of y ″ 1 as shown below: y 1 ( x ) = x y 1 ′ ( x ) = 1 y 1 ′ ′ ( x ) = 0 Substitute y 1 ′ ′ ( x ) = 0 , y 1 ( x ) = x in ( 1 − x cot x ) y ″ − x y ′ + y = 0 as, ( 1 − x cot x ) y ″ − x y ′ + y = ( 1 − x cot x ) ( 0 ) − x ( 1 ) + x = 0 Thus, y 1 ( x ) = x is the solution of the given differential equation

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3.2, Problem 27P

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To verify: The functions y1 and y2 are solution of the given differential equation and check whether the functions constitute a fundamental set of solution.

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Elementary Differential Equations

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To verify: The functions y 1 and y 2 are solution of the given differential equation and check whether the functions constitute a fundamental set of solution.

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To verify: The functions y1 and y2 are solution of the given differential equation and check whether the functions constitute a fundamental set of solution.

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