Explanation Result used: The general solution of ordinary differential equation: “Suppose that, the linear ordinary differential equation of the form y ″ + a y ′ + b y = 0 . The auxiliary equation is m 2 + a m + b = 0 . Then, then the general solution of the equation is of the form: Case 1: If the roots are real distinct m 1 and m 2 , then the general solution is, y h ( x ) = c 1 e m 1 + c 2 e m 2 . Case 2: If the roots are real double roots m 1 and m 1 , then the general solution is, y h ( x ) = ( c 1 + c 2 x ) e m 1 . Case 2: If the roots are complex a ± i b , then the general solution is, y h ( x ) = e a x ( A cos ( b x ) + B sin ( b x ) ) . Formula used: 1. The particular solution of nonhomogeneous ordinary differential equation of the form m y ″ + c y ′ + k y = F 0 sin ω t is given by, y p ( t ) = a cos ω t + b sin ω t where a = − F 0 ω c m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 , b = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 and ω 0 = k m 2. The particular solution of nonhomogeneous ordinary differential equation of the form m y ″ + c y ′ + k y = F 0 cos ω t is given by, y p ( t ) = a cos ω t + b sin ω t where a = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 , b = F 0 ω c m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 and ω 0 = k m Calculation: The given differential equation is y ″ + 2 y ′ + 5 y = 4 cos t + 8 sin t . First obtain the solution of the homogeneous part as follows. The auxiliary equation of y ″ + 2 y ′ + 5 y = 0 is m 2 + 2 m + 5 = 0 . Obtain the roots of the equation m 2 + 2 m + 5 = 0 as shown below. m = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 5 ) 2 ( 1 ) = − 2 ± 4 − 20 2 = − 2 ± − 16 2 = − 2 ± 4 i 2 = − 1 ± 2 i Here the roots are complex conjugate. Therefore, the solution y h is y h = e − t ( A cos 2 t + B sin 2 t ) . By sum rule, the particular solution is y p ( t ) = y p 1 ( t ) + y p 2 ( t ) . Compare the equation y ″ + 2 y ′ + 5 y = 4 cos t with that of m y ″ + c y ′ + k y = F 0 cos ω t and obtain the values of m , c , k , F 0 and ω as m = 1 , c = 2 , k = 5 , F 0 = 4 and ω = 1 . First compute the value of ω 0 as shown below. ω 0 = k m = 5 1 = 5 Now obtain the values of a and b as follows. a = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = 4 ( 1 ) ( ( 5 ) 2 − ( 1 ) 2 ) ( 1 ) 2 ( ( 5 ) 2 − ( 1 ) 2 ) 2 + ( 1 ) 2 ( 2 ) 2 = 4 ( 5 − 1 ) ( 5 − 1 ) 2 + ( 1 ) ( 4 ) = 4 ( 4 ) ( 4 ) 2 + 4 On further simplification, a = 16 16 + 4 = 16 20 = 0

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ISBN: 9781119809210

Author: Kreyszig

Publisher: WILEY

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Chapter 2.8, Problem 12P

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The transient motion of the mass-spring system modeled by the ordinary differential equation (D2+2D+5I)y=4cost+8sint.

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The transient motion of the mass-spring system modeled by the ordinary differential equation ( D 2 + 2 D + 5 I ) y = 4 cos t + 8 sin t .

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The transient motion of the mass-spring system modeled by the ordinary differential equation (D2+2D+5I)y=4cost+8sint.

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