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 y ″ + ω 0 y = F 0 m sin ω 0 t is given by, y p ( t ) = − F 0 2 m ω 0 t cos ω 0 t . Calculation: The given differential equation is y ″ + 4 y = sin t + 1 3 sin 3 t + 1 5 sin 5 t . First obtain the solution of the homogeneous part as follows. The auxiliary equation of y ″ + 4 y = 0 is m 2 + 4 = 0 . Obtain the roots of the equation m 2 + 4 = 0 as shown below. m 2 + 4 = 0 m 2 = − 4 m = ± 2 i Here the roots are complex conjugate. Therefore, the solution y h is y h = A cos 2 t + B sin 2 t . By sum rule, the particular solution y p = y p 1 + y p 2 + y p 3 . Compare the equation y ″ + 4 y = sin t with that of m y ″ + c y ′ + k y = F 0 sin ω t and obtain the values of m , c , k , F 0 and ω as m = 1 , c = 0 , k = 4 , F 0 = 1 and ω = 1 . First compute the value of ω 0 as shown below. ω 0 = k m = 4 1 = 4 = 2 Now obtain the values of a and b as follows. a = − F 0 ω c m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = − ( 1 ) ( 2 ) ( 0 ) ( 1 ) 2 ( ( 2 ) 2 − ( 1 ) 2 ) 2 + ( 1 ) 2 ( 0 ) 2 = 0 Compute the value of b as shown below. b = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = ( 1 ) ( ( 2 ) 2 − ( 1 ) 2 ) ( 1 ) 2 ( ( 2 ) 2 − ( 1 ) 2 ) 2 + ( 1 ) 2 ( 0 ) 2 = ( 1 ) ( 4 − 1 ) ( 1 ) 2 ( 4 − 1 ) 2 + ( 1 ) ( 0 ) = 3 ( 3 ) 2 + 0 = 1 3 Therefore, the particular solution y p 1 is y p 1 ( t ) = a cos ω t + b sin ω t = 1 3 sin t . Now compare y ″ + 4 y = 1 3 sin 3 t with that of m y ″ + c y ′ + k y = F 0 sin ω t and obtain the values of m , c , k , F 0 and ω as m = 1 , c = 0 , k = 4 , F 0 = 1 3 and ω = 3 . Obtain the values of a and b as follows. a = − F 0 ω c m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = − ( 1 3 ) ( 3 ) ( 0 ) ( 1 ) 2 ( ( 2 ) 2 − ( 3 ) 2 ) 2 + ( 3 ) 2 ( 0 ) 2 = 0 Compute the value of b as shown below. b = F 0 m ( ω 0 2 − ω 2 ) m 2 ( ω 0 2 − ω 2 ) 2 + ω 2 c 2 = ( 1 3 ) ( ( 2 ) 2 − ( 3 ) 2 ) ( 1 ) 2 ( ( 2 ) 2 − ( 3 ) 2 ) 2 + ( 3 ) 2 ( 0 ) 2 = ( 1 ) ( 4 − 9 ) ( 1 ) 2 ( 4 − 9 ) 2 + ( 9 ) ( 0 ) = ( − 5 3 ) ( − 5 ) 2 + 0 = − 1 15 Therefore, the particular solution y p 2 is y p 2 ( t ) = a cos ω t + b sin ω t = − 1 15 sin 3 t

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ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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Chapter 2.8, Problem 17P

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The motion of the mass-spring system modeled by the ordinary differential equation (D2+4I)y=sint+13sin3t+15sin5t and the initial conditions y(0)=0, y′(0)=335 and sketch the solution.

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Chapter 2.8, Problem 16P

Chapter 2.8, Problem 18P

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The motion of the mass-spring system modeled by the ordinary differential equation ( D 2 + 4 I ) y = sin t + 1 3 sin 3 t + 1 5 sin 5 t and the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 3 35 and sketch the solution.

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The motion of the mass-spring system modeled by the ordinary differential equation (D2+4I)y=sint+13sin3t+15sin5t and the initial conditions y(0)=0, y′(0)=335 and sketch the solution.

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