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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a) Suppose that we are carrying out the 1-phase simplex algorithm on a linear program in standard inequality form (with 3 variables and 4 constraints) and suppose that we have reached a point where we have obtained the following tableau. Apply one more pivot operation, indicating the highlighted row and column and the row operations you carry out. What can you conclude from your updated tableau? x1 x2 x3 81 82 83 84 81 -2 0 1 1 0 0 0 3 82 3 0 -2 0 1 2 0 6 12 1 1 -3 0 0 1 0 2 84 -3 0 2 0 0 -1 1 4 -2 -2 0 11 0 0-4 0 -8

b) Solve the following linear program using the 2-phase simplex algorithm. You should give the initial tableau, and each further tableau produced during the execution of the algorithm. If the program has an optimal solution, give this solution and state its objective value. If it does not have an optimal solution, say why. maximize ₁ - 2x2+x34x4 subject to 2x1+x22x3x41, 5x1 + x2-x3-×4 ≤ −1, 2x1+x2-x3-34 2, 1, 2, 3, 40.

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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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