Explanation Given: The differential equation is y ″ ′ − 12 y ″ + 27 y ′ + 40 y = 0 . The initial value conditions are y ( 0 ) = − 3 y ′ ( 0 ) = − 6 y ″ ( 0 ) = − 12 . 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 ″ ′ − 12 y ″ + 27 y ′ + 40 y = 0 . The auxiliary equation of the differential equation is r 3 − 12 r 2 + 27 r + 40 = 0 Let p ( r ) = r 3 − 12 r 2 + 27 r + 40 Put p ( r ) = 0 r 3 − 12 r 2 + 27 r + 40 = 0 Calculate p ( − 1 ) Substitute − 1 for r in p ( r ) = r 3 − 12 r 2 + 27 r + 40 . p ( − 1 ) = ( − 1 ) 3 − 12 × ( − 1 ) 2 + 27 × ( − 1 ) + 40 = − 1 − 12 − 27 + 40 = 0 This implies ( r + 1 ) is a factor of p ( r ) . Divide p ( r ) by ( r + 1 ) . ( r + 1 ) r 2 − 13 r + 40 r 3 − 12 r 2 + 27 r + 40 − r 3 − r 2 _ − 13 r 2 + 27 r + 40 + 13 r 2 + 13 r _ 40 r + 40 − 40 r − 40 _ 0 _ Now, D ividend = Divisor × Quotient + Remainder . Substitute r 3 − 12 r 2 + 27 r + 40 for Dividend , ( r + 1 ) for Divisor , r 2 − 13 r + 40 for Quotient and 0 for remainder in the above equation. r 3 − 12 r 2 + 27 r + 40 = ( r + 1 ) ( r 2 − 13 r + 40 ) + 0 = ( r + 1 ) ( r 2 − 8 r − 5 r + 40 ) = ( r + 1 ) ( r ( r − 8 ) − 5 ( r − 8 ) ) = ( r + 1 ) ( r − 5 ) ( r − 8 ) The zeroes of p ( r ) are − 1 , 5 , 8 . y ( t ) = C 1 e − t + C 2 e 5 t + C 3 e 8 t ( 4 ) Now, y ( 0 ) = − 3 Substitute 0 for t , − 3 for y in the equation ( 4 ) C 1 + C 2 + C 3 = − 3

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ISBN: 9780135902738

Author: Nagle

Publisher: PEARSON

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Chapter 4.RP, Problem 33RP

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Chapter 4.RP, Problem 32RP

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The solution to the differential equation

Solution Summary: The author explains the solution of the differential equation: y(t)=-e-t

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To find: The solution to the differential equation

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Chapter 4 Solutions

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The solution to the differential equation