Explanation The corresponding characteristic equation is written as: y ( 4 ) + 6 y ″ ′ + 1 7 y ″ + 2 2 y ′ + 1 4 y = 0 r ( 4 ) + 6 r ( 3 ) + 17 r 2 + 22 r + 14 = 0 ( r − ( − 1 ± i ) ) ( r − ( − 2 ± 3 i ) ) = 0 From the above equation, it can be observed that the roots of the equation are r = ( − 1 ± i ) , ( − 2 ± 3 i ) . So, the general solution of the differential equation is written as: y ( t ) = c 1 e − t cos t + c 2 e − t sin t + c 3 e − 2 t cos ( 3 t ) + c 4 e − 2 t sin ( 3 t ) y ( t ) = e − t ( c 1 cos t + c 2 sin t ) + e − 2 t ( c 3 cos 3 t + c 4 sin 3 t ) y ( 0 ) = e − 0 ( c 1 cos × 0 + c 2 sin × 0 ) + e − 2 × 0 ( c 3 cos 3 × 0 + c 4 sin 3 × 0 ) 1 = c 1 + c 3 . Differentiate the above solution, with respect to t . y ′ ( t ) = − e − t ( c 1 cos t + c 2 sin t ) + e − t ( − c 1 sin t + c 2 cos t ) − e − 2 t ( c 3 cos 3 t + c 4 sin 3 t ) e − 2 t ( − 3 c 3 sin 3 t + 3 c 4 cos 3 t ) y ′ ( t ) = e − t [ ( c 2 − c 1 ) cos t − ( c 2 + c 1 ) sin t ] + e − 2 t [ ( − 2 c 3 + 3 c 4 ) cos 3 t + ( − 3 c 4 − 2 c 4 ) sin 3 t ] y ′ ( 0 ) = e − 0 [ ( c 2 − c 1 ) cos × 0 − ( c 2 + c 1 ) sin × 0 ] + e − 2 × 0 [ ( − 2 c 3 + 3 c 4 ) cos 3 × 0 + ( − 3 c 4 − 2 c 4 ) sin 3 × 0 ] − 2 = c 2 − c 1 − 2 c 3 + 3 c 4 Now differentiate the y ′ ( t ) = e − t [ ( c 2 − c 1 ) cos t − ( c 2 + c 1 ) sin t ] + e − 2 t [ ( − 2 c 3 + 3 c 4 ) cos 3 t + ( − 3 c 4 − 2 c 4 ) sin 3 t ] , with respect to t

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Author: William E. Boyce, Richard C. DiPrima

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Chapter 4.2, Problem 36P

To determine

The general solution of the differential equation y(4)+6y″′+17y″+22y′+14y=0, y(0)=1,y′(0)=−2,y″(0)=0 and y″′(0)=3.

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Chapter 4.2, Problem 35P

Chapter 4.2, Problem 37P

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Chapter 4 Solutions

Elementary Differential Equations

Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - In each of Problems 1 through 6, determine...Ch. 4.1 - Prob. 6P Ch. 4.1 - In each of Problems 7 through 10, determine...Ch. 4.1 - Prob. 8P Ch. 4.1 - In each of Problems 7 through 10, determine...Ch. 4.1 - Prob. 10P

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The general solution of the differential equation y ( 4 ) + 6 y ″ ′ + 1 7 y ″ + 2 2 y ′ + 1 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = − 2 , y ″ ( 0 ) = 0 and y ″ ′ ( 0 ) = 3 .

The general solution of the differential equation y(4)+6y″′+17y″+22y′+14y=0, y(0)=1,y′(0)=−2,y″(0)=0 and y″′(0)=3.

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Chapter 4 Solutions