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The solution of the initial value problem for the RLC -circuit when R = 18 Ω , L = 1 H , C = 12.5 ⋅ 10 − 3 F , and E ( t ) = 820 cos 10 t V and sketch the solution. The solution of the initial value problem is I = e − 3 t ( − 100 cos 4 t + 75 sin 4 t ) + 100 cos t _ . Formula used: The current in an RLC -circuit is the solution of nonhomogeneous second-order ordinary differential equation, L I ″ + R I ′ + 1 C I = E ′ ( t ) ⋯ ⋯ ( 1 ) Calculation: It is given that, R = 18 Ω , L = 1 H , C = 12.5 ⋅ 10 − 3 F , and E ( t ) = 820 cos 10 t V . First compute the derivative of E ( t ) = 820 cos 10 t V with respect to t as follows. E ′ ( t ) = d d t ( 820 cos 10 t ) = 820 ( − 10 sin 10 t ) Now substitute R = 18 Ω , L = 1 H , C = 0.0125 F , and E ′ ( t ) = 820 ( − 10 sin 10 t ) in equation (1) and obtain a second order nonhomogeneous ordinary differential equation as shown below. L I ″ + R I ′ + 1 C I = E ′ ( t ) I ″ + 18 I ′ + 1 0.0125 I = 820 ( − 10 sin 10 t ) I ″ + 18 I ′ + 80 I = 820 ( − 10 sin 10 t ) Hence the current in the given RLC circuit is the solution of second order ordinary differential equation I ″ + 18 I ′ + 80 I = 820 ( − 10 sin 10 t ) . The auxiliary equation of I ″ + 18 I ′ + 80 I = 0 is m 2 + 18 m + 80 = 0 . Obtain the roots of m 2 + 18 m + 80 = 0 as shown below. m 2 + 18 m + 80 = 0 m 2 + 10 m + 8 m + 80 = 0 m ( m + 10 ) + 8 ( m + 10 ) ( m + 10 ) ( m + 8 ) = 0 m = − 10 , m = − 8 Here the roots are real and distinct. Hence the solution I h ( t ) is I h ( t ) = c 1 e − 10 t + c 2 e − 8 t . Now compute the particular solution of I ″ + 18 I ′ + 80 I = 820 ( − 10 sin 10 t ) as shown below. Here the term r ( x ) is 820 ( − 10 sin 10 t ) and hence the choice for I p becomes ( K cos ω t + M sin ω t ) . That is, I p = ( K cos 10 t + M sin 10 t ) . Differentiate I p = ( K cos 10 t + M sin 10 t ) with respect to t as shown below. I p ′ = d d t ( K cos 10 t + M sin 10 t ) = − 10 K sin 10 t + 10 M cos 10 t Again differentiate I ′ p with respect to x and obtain I ″ p as follows. I p ′ ′ = d d x ( − 10 K sin 10 t + 10 M cos 10 t ) = − 100 K cos 10 t − 100 M sin 10 t Since the particular solution satisfies the given ordinary differential equation, substitute I p , I p ′ , and I p ′ ′ in I p ′ ′ + 18 I p ′ + 80 I p = 820 ( − 10 sin 10 t ) as shown below. { − 100 K cos 10 t − 100 M sin 10 t + 18 ( − 10 K sin 10 t + 10 M cos 10 t ) + 80 ( K cos 10 t + M sin 10 t ) } = 820 ( − 10 sin 10 t ) { − 100 K cos 10 t − 100 M sin 10 t − 180 K sin 10 t + 180 M cos 10 t + 80 K cos 10 t + 80 M sin 10 t } = − 8200 sin 10 t { cos 10 t ( − 20 K + 180 M ) + sin 10 t ( − 20 M − 180 K ) } = − 8200 sin 10 t Equate the coefficients of like terms on both sides as follows. − 20 K + 180 M = 0 … … ( 1 ) − 20 M − 180 K = − 8200 … … ( 2 ) From equation (1), solve for K as follows. − 20 K + 180 M = 0 − 20 K = − 180 M K = 9 M From equation (2) obtain the value of K as shown below. − 20 M − 180 K = − 8200 M + 9 K = 410 M + 9 ( 9 M ) = 410 82 M = 410 M = 5 Now obtain the value of K as follows. K = 9 M = 9 ( 5 ) = 45 Then the particular solution of the given ordinary differential equation becomes, I p = ( K cos 10 t + M sin 10 t ) = 45 cos 10 t + 5 sin 10 t . Therefore, the solution of the initial value problem is I = c 1 e − 10 t + c 2 e − 8 t + 45 cos 10 t + 5 sin 10 t . Now obtain the values of the constants c 1 and c 2 by using the initial conditions I ( 0 ) = 0 and I ′ ( 0 ) = 0 as follows. I ( 0 ) = c 1 e 0 + c 2 e 0 + 45 cos 0 + 5 sin 0 c 1 + c 2 + 45 = 0 c 2 = − c 1 − 45 Differentiate I = c 1 e − 10 t + c 2 e − 8 t + 45 cos 10 t + 5 sin 10 t and obtain the first derivative as shown below. I ′ ( t ) = − 10 c 1 e − 10 t − 8 c 2 e − 8 t − 45 ( 10 ) sin 10 t + 5 ( 10 ) cos 10 t = − 10 c 1 e − 10 t − 8 c 2 e − 8 t − 450 sin 10 t + 50 cos 10 t Obtain the value of c 1 as follows. I ′ ( 0 ) = − 10 c 1 e 0 − 8 c 2 e 0 − 450 sin 0 + 50 cos 0 − 10 c 1 − 8 c 2 + 50 = 0 − 10 c 1 − 8 ( − c 1 − 45 ) + 50 = 0 − 2 c 1 = − 410 c 1 = 205 Now obtain the value of c 2 as shown below. Therefore, the solution of the initial value problem is I = e − 3 t ( − 100 cos 4 t + 75 sin 4 t ) + 100 cos t _ . Sketch the solution I = e − 3 t ( − 100 cos 4 t + 75 sin 4 t ) + 100 cos t as shown in below Figure 1. From Figure 1, it is observed that the domain of the solution is ( 0 , ∞ ) .

The solution of the initial value problem for the RLC -circuit when R = 18 Ω , L = 1 H , C = 12.5 ⋅ 10 − 3 F , and E ( t ) = 820 cos 10 t V and sketch the solution. The solution of the initial value problem is I = e − 3 t ( − 100 cos 4 t + 75 sin 4 t ) + 100 cos t _ . Formula used: The current in an RLC -circuit is the solution of nonhomogeneous second-order ordinary differential equation, L I ″ + R I ′ + 1 C I = E ′ ( t ) ⋯ ⋯ ( 1 ) Calculation: It is given that, R = 18 Ω , L = 1 H , C = 12.5 ⋅ 10 − 3 F , and E ( t ) = 820 cos 10 t V . First compute the derivative of E ( t ) = 820 cos 10 t V with respect to t as follows. E ′ ( t ) = d d t ( 820 cos 10 t ) = 820 ( − 10 sin 10 t ) Now substitute R = 18 Ω , L = 1 H , C = 0.0125 F , and E ′ ( t ) = 820 ( − 10 sin 10 t ) in equation (1) and obtain a second order nonhomogeneous ordinary differential equation as shown below. L I ″ + R I ′ + 1 C I = E ′ ( t ) I ″ + 18 I ′ + 1 0.0125 I = 820 ( − 10 sin 10 t ) I ″ + 18 I ′ + 80 I = 820 ( − 10 sin 10 t ) Hence the current in the given RLC circuit is the solution of second order ordinary differential equation I ″ + 18 I ′ + 80 I = 820 ( − 10 sin 10 t ) . The auxiliary equation of I ″ + 18 I ′ + 80 I = 0 is m 2 + 18 m + 80 = 0 . Obtain the roots of m 2 + 18 m + 80 = 0 as shown below. m 2 + 18 m + 80 = 0 m 2 + 10 m + 8 m + 80 = 0 m ( m + 10 ) + 8 ( m + 10 ) ( m + 10 ) ( m + 8 ) = 0 m = − 10 , m = − 8 Here the roots are real and distinct. Hence the solution I h ( t ) is I h ( t ) = c 1 e − 10 t + c 2 e − 8 t . Now compute the particular solution of I ″ + 18 I ′ + 80 I = 820 ( − 10 sin 10 t ) as shown below. Here the term r ( x ) is 820 ( − 10 sin 10 t ) and hence the choice for I p becomes ( K cos ω t + M sin ω t ) . That is, I p = ( K cos 10 t + M sin 10 t ) . Differentiate I p = ( K cos 10 t + M sin 10 t ) with respect to t as shown below. I p ′ = d d t ( K cos 10 t + M sin 10 t ) = − 10 K sin 10 t + 10 M cos 10 t Again differentiate I ′ p with respect to x and obtain I ″ p as follows. I p ′ ′ = d d x ( − 10 K sin 10 t + 10 M cos 10 t ) = − 100 K cos 10 t − 100 M sin 10 t Since the particular solution satisfies the given ordinary differential equation, substitute I p , I p ′ , and I p ′ ′ in I p ′ ′ + 18 I p ′ + 80 I p = 820 ( − 10 sin 10 t ) as shown below. { − 100 K cos 10 t − 100 M sin 10 t + 18 ( − 10 K sin 10 t + 10 M cos 10 t ) + 80 ( K cos 10 t + M sin 10 t ) } = 820 ( − 10 sin 10 t ) { − 100 K cos 10 t − 100 M sin 10 t − 180 K sin 10 t + 180 M cos 10 t + 80 K cos 10 t + 80 M sin 10 t } = − 8200 sin 10 t { cos 10 t ( − 20 K + 180 M ) + sin 10 t ( − 20 M − 180 K ) } = − 8200 sin 10 t Equate the coefficients of like terms on both sides as follows. − 20 K + 180 M = 0 … … ( 1 ) − 20 M − 180 K = − 8200 … … ( 2 ) From equation (1), solve for K as follows. − 20 K + 180 M = 0 − 20 K = − 180 M K = 9 M From equation (2) obtain the value of K as shown below. − 20 M − 180 K = − 8200 M + 9 K = 410 M + 9 ( 9 M ) = 410 82 M = 410 M = 5 Now obtain the value of K as follows. K = 9 M = 9 ( 5 ) = 45 Then the particular solution of the given ordinary differential equation becomes, I p = ( K cos 10 t + M sin 10 t ) = 45 cos 10 t + 5 sin 10 t . Therefore, the solution of the initial value problem is I = c 1 e − 10 t + c 2 e − 8 t + 45 cos 10 t + 5 sin 10 t . Now obtain the values of the constants c 1 and c 2 by using the initial conditions I ( 0 ) = 0 and I ′ ( 0 ) = 0 as follows. I ( 0 ) = c 1 e 0 + c 2 e 0 + 45 cos 0 + 5 sin 0 c 1 + c 2 + 45 = 0 c 2 = − c 1 − 45 Differentiate I = c 1 e − 10 t + c 2 e − 8 t + 45 cos 10 t + 5 sin 10 t and obtain the first derivative as shown below. I ′ ( t ) = − 10 c 1 e − 10 t − 8 c 2 e − 8 t − 45 ( 10 ) sin 10 t + 5 ( 10 ) cos 10 t = − 10 c 1 e − 10 t − 8 c 2 e − 8 t − 450 sin 10 t + 50 cos 10 t Obtain the value of c 1 as follows. I ′ ( 0 ) = − 10 c 1 e 0 − 8 c 2 e 0 − 450 sin 0 + 50 cos 0 − 10 c 1 − 8 c 2 + 50 = 0 − 10 c 1 − 8 ( − c 1 − 45 ) + 50 = 0 − 2 c 1 = − 410 c 1 = 205 Now obtain the value of c 2 as shown below. Therefore, the solution of the initial value problem is I = e − 3 t ( − 100 cos 4 t + 75 sin 4 t ) + 100 cos t _ . Sketch the solution I = e − 3 t ( − 100 cos 4 t + 75 sin 4 t ) + 100 cos t as shown in below Figure 1. From Figure 1, it is observed that the domain of the solution is ( 0 , ∞ ) .

ADVANCED ENGINEERING MATH W/ACCESS

ADVANCED ENGINEERING MATH W/ACCESS

10th Edition

ISBN: 9781119096023

Author: Kreyszig

Publisher: WILEY

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ADVANCED ENGINEERING MATH W/ACCESS

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. Show that in the...Ch. 2.4 - DAMPED MOTION 13. Initial value problem. Find the...Ch. 2.4 - DAMPED MOTION 14. Shock absorber. What is the...Ch. 2.4 - DAMPED MOTION 15. Frequency. Find an approximation...Ch. 2.4 - DAMPED MOTION 16. Maxima. Show that the maxima of...Ch. 2.4 - DAMPED MOTION 17. Underdamping. Determine the...Ch. 2.4 - DAMPED MOTION 18. Logarithmic decrement. Show that...Ch. 2.4 - DAMPED MOTION 19. Damping constant. Consider an...Ch. 2.5 - Prob. 1P Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Prob. 9P Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - Find a real general solution. Show the details of...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.5 - INITIAL VALUE PROBLEM Solve and graph the...Ch. 2.6 - Derive (6*) from (6). Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - BASIS OF SOLUTIONS. WRONSKIAN Find the Wronskian....Ch. 2.6 - Prob. 9P Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. IVP (a) Find a...Ch. 2.6 - ODE FOR GIVEN BASIS. WRONSKIAN. 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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