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: 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 ″ + y = cos ω t , ω 2 ≠ 1 . First obtain the solution of the homogeneous part as follows. The auxiliary equation of y ″ + y = 0 is m 2 + 1 = 0 . Obtain the roots of the equation m 2 + 1 = 0 as shown below. m 2 + 1 = 0 m 2 = − 1 m = ± i Here the roots are complex conjugate. Therefore, the solution y h is y h = A cos t + B sin t . Compare the equation y ″ + y = cos ω t , ω 2 ≠ 1 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 = 0 , k = 1 , F 0 = 1 and ω = ω

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Chapter 2.8, Problem 13P

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The transient motion of the mass-spring system modeled by the ordinary differential equation (D2+I)y=cosωt, ω2≠1.

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The transient motion of the mass-spring system modeled by the ordinary differential equation ( D 2 + I ) y = cos ω t , ω 2 ≠ 1 .

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