Explanation Given: The differential equation is 2 y ″ − y = t sin t . 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 roots r 1 , r 2 then the solution of the differential equation is y = A e r 1 t + B e r 2 t ... ( 3 ) If the auxiliary equation has same root r 1 = r 2 = r , then the solution of the differential equation is y = A e r t + t B e r t Calculation: The differential equation is 2 y ″ − y = t sin t . Consider 2 y ″ − y = 0 The auxiliary equation of the differential equation is 2 r 2 − 1 = 0 Let p ( r ) = 2 r 2 − 1 Put p ( r ) = 0 2 r 2 − 1 = 0 r 2 = 1 2 r = ± 1 2 The roots of p ( r ) are 1 2 , − 1 2 . y c ( t ) = C 1 e 1 2 t + C 2 e − 1 2 t p ( t ) = t sin t . The particular solution is the linear combination of t sin t and its possible derivatives. y p ( t ) = A t sin t + B t cos t + C sin t + D cos t Differentiate y p ( t ) 2 times with respect to t . y ′ p ( t ) = A ( t cos t + sin t ) + B ( − t sin t + cos t ) + C cos t − D sin t = A t cos t − B t sin t + ( A − D ) sin t + ( B + C ) cos t y ″ p ( t ) = A ( − t sin t + cos t ) − B ( t cos t + sin t ) + ( A − D ) cos t − ( B + C ) sin t = − A t sin t − B t cos t + ( 2 A − D ) cos t − ( 2 B + C ) sin t Since, y p is a solution of 2 y ″ − y = t sin t By ( 2 ) , 2 y ″ p − y p = t sin 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

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A general solution to the differential equation

Solution Summary: The author explains how to find a general solution to the differential equation: y(t)=C_1e

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To find:

A general solution to the differential equation