Explanation Given: The differential equation is, x 2 y ″ + 2 x y ′ − 2 y = 6 x − 2 + 3 x . Approach: If a n y ( n ) + a n − 1 y ( n − 1 ) + ⋅ ⋅ ⋅ + a 1 y ( 1 ) + a 0 = p ( t ) , then y = y c +y p . Here, y c denotes the complementary solution, y p denotes the particular solution. The complementary solution is the solution of the differential equation a n y ( n ) + a n − 1 y ( n − 1 ) + ⋅ ⋅ ⋅ + a 1 y ( 1 ) + a 0 = 0 . The particular solution is the possible linear combinations of the function p ( t ) . The auxiliary equation of a differential equation a n y ( n ) + a n − 1 y ( n − 1 ) + ⋅ ⋅ ⋅ + a 1 y ( 1 ) + a 0 is a n r n + a n − 1 r n − 1 + ⋅ ⋅ ⋅ + a 1 r 1 + a 0 ... ( 1 ) . Here, y ( n ) denotes the n th derivative of y . If x = a is a zero of the polynomial a n x n + a n − 1 x n − 1 + ⋅ ⋅ ⋅ + a 1 x 1 + a 0 , then ( x − a ) is a factor of the polynomial ... ( 2 ) . If the auxiliary equation has complex roots r 1 = a + b i , r 2 = a − b i then the solution of the differential equation is y = A e a t cos b t + B e a t sin b t ... ( 3 ) Calculation: The differential equation is x 2 y ″ + 2 x y ′ − 2 y = 6 x − 2 + 3 x . Let x = e t . Differentiate the above equation with respect to t . d x d t = e t Now, d y d x = d y d t d x d t Substitute e t for d x d t in the above equation. d y d x = d y d t e t = d y d t e − t Differentiate it with respect to x on both sides. d 2 y d x 2 = d d x ( d y d t e − t ) = d d t ( d y d t e − t ) d t d x = e − t d d t ( d y d t e − t ) = e − t ( d y d t ( − e − t ) + e − t × d 2 y d t 2 ) d 2 y d x 2 = e − t ( d y d t ( − e − t ) + e − t × d 2 y d t 2 ) = e − 2 t ( d 2 y d t 2 − d y d t ) Substitute e − 2 t ( d 2 y d t 2 − d y d t ) for d 2 y d x 2 and d y d t e − t for d y d x in equation x 2 y ″ + 2 x y ′ − 2 y = 6 x − 2 + 3 x e 2 t ( e − 2 t ( − d y d t + d 2 y d t 2 ) ) + 2 e t d y d t e − t − 2 y = 6 e − 2 t + 3 e t ( d 2 y d t 2 − d y d t ) + 2 d y d t − 2 y = 6 e − 2 t + 3 e t d 2 y d t 2 + d y d t − 2 y = 6 e − 2 t + 3 e t The equation becomes y ″ + y ′ − 2 y = 6 e − 2 t + 3 e t . The auxiliary equation of the differential equation is y 2 + y − 2 = 0 . Let p ( r ) = r 2 + r − 2 Put p ( r ) = 0 r 2 + r − 2 = 0 r 2 + 2 r − r − 2 = 0 r ( r + 2 ) − 1 ( r + 2 ) = 0 ( r − 1 ) ( r + 2 ) = 0 The roots of p ( r ) are 1 , − 2 . By equation ( 2 ) , y h ( t ) = A e − 2 t + B e t

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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A general solution to the differential equation

Solution Summary: The author explains the solution of the differential equation, which is x2y′′+2x

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A general solution to the differential equation