Explanation Given: The differential equation is y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t . The initial value conditions are y ( 0 ) = 2 , y ′ ( 0 ) = − 8 Approach: 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 = a + b i , r 2 = a − b i 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 y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t . The auxiliary equation of the differential equation is r 2 − 2 r + 10 = 0 Let p ( r ) = r 2 − 2 r + 10 Put p ( r ) = 0 r 2 − 2 r + 10 = 0 r = 2 ± 4 − 4 × 1 × 40 2 r = 2 ± 6 i 2 r = 1 ± 3 i The roots are 1 + 3 i , 1 − 3 i . y h ( t ) = C 1 e t cos 3 t + C 2 e t sin 3 t ( 4 ) . p ( t ) = 6 cos 3 t − sin 3 t . The particular solution is the linear combination of 2 cos θ and its possible derivatives. y p ( θ ) = A cos 3 t + B sin 3 t . Differentiate y p ( t ) 2 times with respect to t . y ′ p ( θ ) = − 3 A sin 3 t + 3 B cos 3 t y ″ p ( θ ) = − 9 A cos 3 t − 9 B sin 3 t Since, y p is a solution of y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t By ( 2 ) , y ″ p − 2 y ′ p + 10 y p = 6 cos 3 t − sin 3 t Substitute A cos 3 t + B sin 3 t for y p , − 3 A sin 3 t + 3 B cos 3 t for y ′ p , − 9 A cos 3 t − 9 B sin 3 t for y ″ p in y ″ − 2 y ′ + 10 y = 6 cos 3 t − sin 3 t . − 9 A cos 3 t − 9 B sin 3 t − 2 ( − 3 A sin 3 t + 3 B cos 3 t ) + 10 ( A cos 3 t + B sin 3 t ) = 6 cos 3 t − sin 3 t cos 3 t ( A − 6 B ) + sin 3 t ( 6 A + B ) = 6 cos 3 t − sin 3 t The equations obtained are A − 6 B = 6 ( 4 ) 6 A + B = − 1 ( 5 ) Multiply equation ( 4 ) by 6

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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The solution to the differential equation

Solution Summary: The author explains the solution of the differential equation: y=2etmathrmcos3t-73, and the auxiliary equation is

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The solution to the differential equation