To determine The general solution of the differential equation 2 x 2 y ″ − 4 x y ′ + 6 y = 0 . Explanation Result used: (1) “Suppose the equation L [ x r ] = x 2 y ″ + α x y ′ + β y = 0 has a solution of the from y = x r , then it becomes L [ x r ] = x r [ r ( r − 1 ) + α r + β ] , where α and β are real constants”. (2) “If r is a root of the quadratic equation F ( r ) = r ( r − 1 ) + α r + β = 0 , then L [ x r ] is zero and y = x r is a solution of x 2 y ″ + α x y ′ + β y = 0 and the roots of the equation are r 1 , r 2 = − ( α − 1 ) ± ( α − 1 ) 2 − 4 β 2 ”. (3) ‘If the roots r 1 = λ + i μ and r 2 = λ − i μ of the equation are complex, then the general solution is in the form of y = c 1 x λ cos ( μ ln x ) + c 2 x λ sin ( μ ln x ) , x > 0 ”. Calculation: The given differential equation 2 x 2 y ″ − 4 x y ′ + 6 y = 0 . It is observed that x = 0 is the only singular point for the given differential equation, so it is not included in that interval. Assume that the solution is in the form of y = x r . Substitute y = x r in the equation 2 x 2 y ″ − 4 x y ′ + 6 y = 0 as follows. By the result(1) stated above, the given equation becomes x r [ 2 r ( r − 1 ) − 4 r + 6 ] = 0

10th Edition

ISBN: 9780470458327

Author: William E. Boyce, Richard C. DiPrima

Publisher: Wiley, John & Sons, Incorporated

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Chapter 5.4, Problem 8P

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The general solution of the differential equation 2x2y″−4xy′+6y=0.

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Elementary Differential Equations

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