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The steady state motion of the mass spring system modeled by the ordinary differential equation ( 4 D 2 + 12 D + 9 I ) y = 225 − 75 sin 3 t . The steady state motion of the mass spring system is y ( t ) = 25 − 1.333 cos ( 3 t ) + sin ( 3 t ) _ . 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 And 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 Calculation: The given differential equation is ( 4 D 2 + 12 D + 9 I ) y = 225 − 75 sin 3 t . Since d 2 y d x = D 2 y , d y d x = D y and I ( y ) = y , the equation ( 4 D 2 + 12 D + 9 I ) y = 225 cos ( 0 t ) − 75 sin ( 3 t ) becomes, ( 4 D 2 + 12 D + 9 I ) y = 225 cos ( 0 t ) − 75 sin ( 3 t ) 4 D 2 y + 12 D y + 9 I ( y ) = 225 cos ( 0 t ) − 75 sin ( 3 t ) 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) The steady state motion of the system modeled by the ordinary differential equation 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) is the particular solution of 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) . The particular solution of 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) is y p ( t ) = a 1 cos ( ω 1 t ) + b 1 sin ( ω 1 t ) + a 2 cos ( ω 2 t ) + b 2 sin ( ω 2 t ) . Compare the equation 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) with that of m y ″ + c y ′ + k y = F 01 cos ( ω 1 t ) + F 02 sin ( ω 2 t ) and obtain the values m = 4 , c = 12 , k = 9 , F 01 = 225 , F 02 = − 75 , ω 1 = 0 and ω 2 = 3 . First compute the value of ω 0 as shown below. ω 0 = k m = 9 4 = 3 2 = 1.5 Now obtain the value of a 1 as follows. a 1 = F 01 m ( ω 0 2 − ω 1 2 ) m 2 ( ω 0 2 − ω 1 2 ) 2 + ω 1 2 c 2 = 225 ⋅ 4 ⋅ ( 1.5 2 − 0 ) 4 2 ( 1.5 2 − 0 ) 2 + 0 2 ⋅ 12 2 = 225 ⋅ 4 ⋅ ( 2.25 ) 16 ( 2.25 ) 2 + 0 = 2025 81 = 25 Compute the value of a 2 as shown below. a 2 = F 02 ω 2 c m 2 ( ω 0 2 − ω 2 2 ) 2 + ω 2 2 c 2 = − 75 ⋅ 3 ⋅ 12 ( 4 ) 2 ( 1.5 2 − 9 ) 2 + ( 3 ) 2 ( 12 ) 2 = 2700 729 + 1296 = 2700 2025 ≈ 1.333 Obtain the value b 1 as follows. b 1 = F 01 ω 1 c m 2 ( ω 0 2 − ω 1 2 ) 2 + ω 1 2 c 2 = 225 ⋅ 0 ⋅ 12 ( 4 ) 2 ( 2.25 − 0 ) 2 + ( 0 ) 2 ( 12 ) 2 = 0 Compute the value of b 2 as shown below. b 2 = F 02 m ( ω 0 2 − ω 2 2 ) m 2 ( ω 0 2 − ω 2 2 ) 2 + ω 2 2 c 2 = − ( − 75 ) ⋅ 4 ⋅ ( 1.5 2 − 9 ) 4 2 ( 1.5 2 − 3 2 ) 2 + 3 2 ⋅ 12 2 = 300 ⋅ ( 2.25 − 9 ) 16 ( 2.25 − 9 ) 2 + 9 ⋅ 144 = 2025 2025 = 1 Therefore, the particular solution is y p ( t ) = 25 − 1.333 cos ( 3 t ) + sin ( 3 t ) . Thus, the steady state motion of the mass spring system is y ( t ) = 25 − 1.333 cos ( 3 t ) + sin ( 3 t ) _ .

The steady state motion of the mass spring system modeled by the ordinary differential equation ( 4 D 2 + 12 D + 9 I ) y = 225 − 75 sin 3 t . The steady state motion of the mass spring system is y ( t ) = 25 − 1.333 cos ( 3 t ) + sin ( 3 t ) _ . 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 And 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 Calculation: The given differential equation is ( 4 D 2 + 12 D + 9 I ) y = 225 − 75 sin 3 t . Since d 2 y d x = D 2 y , d y d x = D y and I ( y ) = y , the equation ( 4 D 2 + 12 D + 9 I ) y = 225 cos ( 0 t ) − 75 sin ( 3 t ) becomes, ( 4 D 2 + 12 D + 9 I ) y = 225 cos ( 0 t ) − 75 sin ( 3 t ) 4 D 2 y + 12 D y + 9 I ( y ) = 225 cos ( 0 t ) − 75 sin ( 3 t ) 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) The steady state motion of the system modeled by the ordinary differential equation 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) is the particular solution of 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) . The particular solution of 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) is y p ( t ) = a 1 cos ( ω 1 t ) + b 1 sin ( ω 1 t ) + a 2 cos ( ω 2 t ) + b 2 sin ( ω 2 t ) . Compare the equation 4 y ″ + 12 y ′ + 9 y = 225 cos ( 0 t ) − 75 sin ( 3 t ) with that of m y ″ + c y ′ + k y = F 01 cos ( ω 1 t ) + F 02 sin ( ω 2 t ) and obtain the values m = 4 , c = 12 , k = 9 , F 01 = 225 , F 02 = − 75 , ω 1 = 0 and ω 2 = 3 . First compute the value of ω 0 as shown below. ω 0 = k m = 9 4 = 3 2 = 1.5 Now obtain the value of a 1 as follows. a 1 = F 01 m ( ω 0 2 − ω 1 2 ) m 2 ( ω 0 2 − ω 1 2 ) 2 + ω 1 2 c 2 = 225 ⋅ 4 ⋅ ( 1.5 2 − 0 ) 4 2 ( 1.5 2 − 0 ) 2 + 0 2 ⋅ 12 2 = 225 ⋅ 4 ⋅ ( 2.25 ) 16 ( 2.25 ) 2 + 0 = 2025 81 = 25 Compute the value of a 2 as shown below. a 2 = F 02 ω 2 c m 2 ( ω 0 2 − ω 2 2 ) 2 + ω 2 2 c 2 = − 75 ⋅ 3 ⋅ 12 ( 4 ) 2 ( 1.5 2 − 9 ) 2 + ( 3 ) 2 ( 12 ) 2 = 2700 729 + 1296 = 2700 2025 ≈ 1.333 Obtain the value b 1 as follows. b 1 = F 01 ω 1 c m 2 ( ω 0 2 − ω 1 2 ) 2 + ω 1 2 c 2 = 225 ⋅ 0 ⋅ 12 ( 4 ) 2 ( 2.25 − 0 ) 2 + ( 0 ) 2 ( 12 ) 2 = 0 Compute the value of b 2 as shown below. b 2 = F 02 m ( ω 0 2 − ω 2 2 ) m 2 ( ω 0 2 − ω 2 2 ) 2 + ω 2 2 c 2 = − ( − 75 ) ⋅ 4 ⋅ ( 1.5 2 − 9 ) 4 2 ( 1.5 2 − 3 2 ) 2 + 3 2 ⋅ 12 2 = 300 ⋅ ( 2.25 − 9 ) 16 ( 2.25 − 9 ) 2 + 9 ⋅ 144 = 2025 2025 = 1 Therefore, the particular solution is y p ( t ) = 25 − 1.333 cos ( 3 t ) + sin ( 3 t ) . Thus, the steady state motion of the mass spring system is y ( t ) = 25 − 1.333 cos ( 3 t ) + sin ( 3 t ) _ .

ADVANCED ENGINEERING MATHEMATICS

ADVANCED ENGINEERING MATHEMATICS

10th Edition

ISBN: 9781119664697

Author: Kreyszig

Publisher: WILEY

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ADVANCED ENGINEERING MATHEMATICS

Ch. 2.1 - Prob. 1P Ch. 2.1 - Reduction. Show that F(y, y′, y″) = 0 can be...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...Ch. 2.1 - 3–10 REDUCTION OF ORDER Reduce to first order and...

Ch. 2.1 - 11–14 APPLICATIONS OF REDUCIBLE...Ch. 2.1 - 11–14 APPLICATIONS OF REDUCIBLE ODEs 12. Hanging...Ch. 2.1 - APPLICATIONS OF REDUCIBLE ODEs 13. Motion. If, in...Ch. 2.1 - Motion. In a straight-line motion, let the...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.1 - GENERAL SOLUTION. INITIAL VALUE PROBLEM...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - GENERAL SOLUTION Find a general solution. Check...Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - 1–15 GENERAL SOLUTION Find a general solution....Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - Find a general solution. Check your answer by...Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 16. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 17. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 18. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 19. Ch. 2.2 - 16–20 FIND AN ODE for the given basis. 20. Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - INITIAL VALUES PROBLEMS Solve the IVP. Check that...Ch. 2.2 - Solve the IVP. Check that your answer satisfies...Ch. 2.2 - LINEAR INDEPENDENCE is of basic importance, in...Ch. 2.2 - LINEAR INDEPENDENCE is of basic importance, in...Ch. 2.2 - Prob. 33P Ch. 2.2 - Prob. 34P Ch. 2.2 - Prob. 35P Ch. 2.2 - LINEAR INDEPENDENCE is of basic importance, in...Ch. 2.2 - Instability. Solve y″ − y = 0 for the initial...Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Prob. 4P Ch. 2.3 - Apply the given operator to the given functions....Ch. 2.3 - Factor as in the text and solve. (D2 + 4.00D +...Ch. 2.3 - Factor as in the text and solve. (4D2 − I)y = 0 Ch. 2.3 - Factor as in the text and solve. (D2 + 3I)y = 0 Ch. 2.3 - Factor as in the text and solve. (D2 − 4.20D +...Ch. 2.3 - Factor as in the text and solve. (D2 + 4.80D +...Ch. 2.3 - Factor as in the text and solve. (D2 − 4.00D +...Ch. 2.3 - Prob. 12P Ch. 2.3 - Linear operator. Illustrate the linearity of L in...Ch. 2.3 - Double root. If D2 + aD + bI has distinct roots μ...Ch. 2.3 - Definition of linearity. Show that the definition...Ch. 2.4 - Initial value problem. Find the harmonic motion...Ch. 2.4 - Frequency. If a weight of 20 nt (about 4.5 lb)...Ch. 2.4 - Frequency. How does the frequency of the harmonic...Ch. 2.4 - Initial velocity. Could you make a harmonic...Ch. 2.4 - Springs in parallel. What are the frequencies of...Ch. 2.4 - Spring in series. If a body hangs on a spring s1...Ch. 2.4 - Prob. 7P Ch. 2.4 - Prob. 8P Ch. 2.4 - HARMONIC OSCILLATIONS (UNDAMPED MOTION) 9....Ch. 2.4 - Prob. 11P Ch. 2.4 - DAMPED MOTION 12. Overdamping. 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IVP (a) Find a...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: GENERAL SOLUTION Find...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - NONHOMOGENEOUS LINEAR ODEs: IVPs Solve the initial...Ch. 2.7 - CAS PROJECT. Structure of Solutions of Initial...Ch. 2.8 - Prob. 2P Ch. 2.8 - Find the steady-state motion of the mass–spring...Ch. 2.8 - Find the steady-state motion of the mass–spring...Ch. 2.8 - Find the steady-state motion of the mass–spring...Ch. 2.8 - Prob. 6P Ch. 2.8 - Prob. 7P Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - Prob. 12P Ch. 2.8 - Prob. 13P Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - TRANSIENT SOLUTIONS Find the transient motion of...Ch. 2.8 - INITIAL VALUE PROBLEMS Find the motion of the...Ch. 2.8 - Prob. 17P Ch. 2.8 - INITIAL VALUE PROBLEMS Find the motion of the...Ch. 2.8 - Prob. 19P Ch. 2.8 - Prob. 20P Ch. 2.8 - Prob. 21P Ch. 2.8 - Prob. 22P Ch. 2.8 - Prob. 24P Ch. 2.9 - RC-Circuit. Model the RC-circuit in Fig. 64. Find...Ch. 2.9 - RC-Circuit. Solve Prob. 1 when E = E0 sin ωt and...Ch. 2.9 - RL-Circuit. Model the RL-circuit in Fig. 66. Find...Ch. 2.9 - RL-Circuit. Solve Prob. 3 when E = E0 sin ωt and...Ch. 2.9 - LC-Circuit. This is an RLC-circuit with negligibly...Ch. 2.9 - LC-Circuit. Find the current when L = 0.5 H, C =...Ch. 2.9 - Prob. 7P Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - 8–14 Find the steady-state current in the...Ch. 2.9 - Find the steady-state current in the RLC-circuit...Ch. 2.9 - Find the steady-state current in the RLC-circuit...Ch. 2.9 - Prob. 14P Ch. 2.9 - Prob. 15P Ch. 2.9 - Solve the initial value problem for the...Ch. 2.9 - Prob. 17P Ch. 2.9 - Prob. 18P Ch. 2.9 - Complex Solution Method. Solve , by substituting...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Prob. 5P Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2.10 - Prob. 12P Ch. 2.10 - Solve the given nonhomogeneous linear ODE by...Ch. 2 - Prob. 1RQ Ch. 2 - Prob. 2RQ Ch. 2 - By what methods can you get a general solution of...Ch. 2 - Prob. 4RQ Ch. 2 - Prob. 5RQ Ch. 2 - Prob. 6RQ Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Prob. 15RQ Ch. 2 - Prob. 16RQ Ch. 2 - Prob. 17RQ Ch. 2 - Find a general solution. Show the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Solve the problem, showing the details of your...Ch. 2 - Find the steady-state current in the RLC-circuit...Ch. 2 - Find a general solution of the homogeneous linear...Ch. 2 - Find the steady-state current in the RLC-circuit...Ch. 2 - Find the current in the RLC-circuit in Fig. 71...Ch. 2 - Prob. 27RQ Ch. 2 - Prob. 28RQ Ch. 2 - Prob. 29RQ Ch. 2 - Prob. 30RQ

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