Explanation The given differential equation is x 2 y ″ − 2 x y ′ + 2 y = 0 . This differential equation is of the form x 2 y ″ + α x y ′ + β y = 0 with α = − 2 and β = 2 . Assume that the differential equation x 2 y ″ − 2 x y ′ + 2 y = 0 has a solution of the form y = x r . Substitute y = x r in x 2 y ″ − 2 x y ′ + 2 y . x 2 y ″ − 2 x y ′ + 2 y = x 2 ( x r ) ′ ′ − 2 x ( x r ) ′ + 2 ( x r ) = x 2 ( r x r − 1 ) ′ − 2 x ( r x r − 1 ) + 2 x r = x 2 ( r ( r − 1 ) x r − 2 ) − 2 r x r + 2 x r = r ( r − 1 ) x r − 2 r x r + 2 x r Simplify further and proceed as follows. x 2 y ″ − 2 x y ′ + 2 y = r ( r − 1 ) x r − 2 r x r + 2 x r = ( r 2 − r ) x r − 2 r x r + 2 x r = ( r 2 − r − 2 r + 2 ) x r = ( r 2 − 3 r + 2 ) x r Solve for r in the expression r 2 − 3 r + 2 . r 2 − 3 r + 2 = 0 ( r − 1 ) ( r − 2 ) = 0 r = 1 or 2 Since r = 1 and r = 2 , x and x 2 are the solutions of x 2 y ″ − 2 x y ′ + 2 y = 0 . Therefore, a linear combination of x and x 2 is the general solution of x 2 y ″ − 2 x y ′ + 2 y = 0 . Hence, the general solution of x 2 y ″ − 2 x y ′ + 2 y = 0 is given by y = c 1 x + c 2 x 2 , c 1 and c 2 are arbitrary constants. Use the given initial conditions y ( 1 ) = − 1 , y ( 2 ) = 1 and evaluate c 1 and c 2 . Substitute x = 1 and y ( 1 ) = − 1 in the complementary function y ( x ) = c 1 x + c 2 x 2 as follows

Elementary Differential Equations and Boundary Value Problems, Enhanced

11th Edition

ISBN: 9781119381648

Author: Boyce

Publisher: WILEY

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Chapter 10.1, Problem 11P

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To solve: The boundary value problem x2y″−2xy′+2y=0, y(1)=−1, y(2)=1.

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Elementary Differential Equations and Boundary Value Problems, Enhanced

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The boundary value problem x 2 y ″ − 2 x y ′ + 2 y = 0 , y ( 1 ) = − 1 , y ( 2 ) = 1 .

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To solve: The boundary value problem x2y″−2xy′+2y=0, y(1)=−1, y(2)=1.

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