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The solution of y ″ + y = 1 − t 2 π 2 if 0 ≤ t ≤ π and 0 if t → ∞ with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 . The solution of y ″ + y = 1 − t 2 π 2 if 0 ≤ t ≤ π and 0 if t → ∞ with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 is y = { ( − 1 − 2 π 2 ) cos t − 1 π 2 t 2 + 1 + 2 π 2 if 0 ≤ t ≤ π ( − 1 − 4 π 2 ) cos t + 2 π sin t if 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 , then the general solution is, y h ( x ) = ( c 1 + c 2 x ) e m 1 . Case 3: If the roots are complex conjugate a ± i b , then the general solution is, y h ( x ) = e a x ( A cos ( b x ) + B sin ( b x ) ) . Calculation: The given differential equation is y ″ + y = 1 − t 2 π 2 . 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 . Here the term r ( x ) is 1 − t 2 π 2 and hence the choice for y p becomes K 0 t 2 + K 1 t + K 2 . That is, y p = K 0 t 2 + K 1 t + K 2 . Differentiate y p = K 0 t 2 + K 1 t + K 2 with respect to t as shown below. y p ′ = d d t ( K 0 t 2 + K 1 t + K 2 ) = 2 K 0 t + K 1 Again differentiate y p ′ with respect to x and obtain y p ′ ′ as follows. y p ′ ′ = d d x ( 2 K 0 t + K 1 ) = 2 K 0 Since the particular solution satisfies the given ordinary differential equation, substitute y p , y p ′ , and y p ′ ′ in y ″ + y = 1 − t 2 π 2 as shown below. y ″ + y = 1 − t 2 π 2 2 K 0 + K 0 t 2 + K 1 t + K 2 = 1 − t 2 π 2 K 0 t 2 + K 1 t + 2 K 0 + K 2 = 1 − t 2 π 2 Equate the coefficients of like terms on both sides as follows. 2 K 0 + K 2 = 1 … … ( 1 ) K 0 = − 1 π 2 … … ( 2 ) K 1 = 0 … … ( 3 ) From equation (1), obtain the value of K 2 as follows. 2 K 0 + K 2 = 1 − 2 π 2 + K 2 = 1 K 2 = 1 + 2 π 2 Then the particular solution of the given ordinary differential equation becomes, y p = K 0 t 2 + K 1 t + K 2 = − 1 π 2 t 2 + 1 + 2 π 2 . Now the general solution of the given differential equation is, y ( t ) = y h ( t ) + y p ( t ) = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 Compute the values of the constants A and B using initial conditions as follows. y ( t ) = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 y ( 0 ) = A cos ( 0 ) + B sin ( 0 ) − 1 π 2 ( 0 ) + 1 + 2 π 2 0 = A + 1 + 2 π 2 A = − 1 − 2 π 2 Differentiate y ( t ) and obtain the first derivative as shown below. y ( t ) = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 y ′ ( t ) = − A sin t + B cos t − 2 t π 2 Now obtain the value of c 2 as follows. y ′ ( 0 ) = − A sin ( 0 ) + B cos ( 0 ) − 2 ( 0 ) π 2 0 = − 0 + B B = 0 Hence the solution of the initial value problem y ″ + y = 1 − t 2 π 2 is, y = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 = ( − 1 − 2 π 2 ) cos t − 1 π 2 t 2 + 1 + 2 π 2 = ( 1 + 2 π 2 ) ( 1 − cos t ) − 1 π 2 t 2 Then, y ′ ( t ) = ( 1 + 2 π 2 ) ( sin t ) − 2 t π 2 . Now obtain the solution of y ″ + y = 0 as follows. The general solution of y ″ + y = 0 is y 2 = C cos t + D sin t . By continuity conditions, the initial conditions are y ( π ) = y 2 ( π ) and y ′ ( π ) = y 2 ′ ( π ) . Evaluate the constant C using y ( π ) = y 2 ( π ) as follows. y ( π ) = y 2 ( π ) ( 1 + 2 π 2 ) ( 1 − cos π ) − 1 π 2 π 2 = C cos π + D sin π ( 1 + 2 π 2 ) ( 2 ) − 1 = − C − C = 1 + 4 π 2 C = − 1 − 4 π 2 Compute the constant C using y ′ ( π ) = y 2 ′ ( π ) as follows. y ′ ( π ) = y 2 ′ ( π ) ( 1 + 2 π 2 ) ( sin π ) − 2 π π 2 = − C sin π + D cos π − 2 π = − D D = 2 π Hence, the solution of y ″ + y = 0 is y 2 = ( − 1 − 4 π 2 ) cos t + 2 π sin t . Therefore, the solution of y ″ + y = 1 − t 2 π 2 if 0 ≤ t ≤ π and 0 if t → ∞ with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 is y = { ( − 1 − 2 π 2 ) cos t − 1 π 2 t 2 + 1 + 2 π 2 if 0 ≤ t ≤ π ( − 1 − 4 π 2 ) cos t + 2 π sin t if t → ∞ _ .

The solution of y ″ + y = 1 − t 2 π 2 if 0 ≤ t ≤ π and 0 if t → ∞ with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 . The solution of y ″ + y = 1 − t 2 π 2 if 0 ≤ t ≤ π and 0 if t → ∞ with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 is y = { ( − 1 − 2 π 2 ) cos t − 1 π 2 t 2 + 1 + 2 π 2 if 0 ≤ t ≤ π ( − 1 − 4 π 2 ) cos t + 2 π sin t if 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 , then the general solution is, y h ( x ) = ( c 1 + c 2 x ) e m 1 . Case 3: If the roots are complex conjugate a ± i b , then the general solution is, y h ( x ) = e a x ( A cos ( b x ) + B sin ( b x ) ) . Calculation: The given differential equation is y ″ + y = 1 − t 2 π 2 . 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 . Here the term r ( x ) is 1 − t 2 π 2 and hence the choice for y p becomes K 0 t 2 + K 1 t + K 2 . That is, y p = K 0 t 2 + K 1 t + K 2 . Differentiate y p = K 0 t 2 + K 1 t + K 2 with respect to t as shown below. y p ′ = d d t ( K 0 t 2 + K 1 t + K 2 ) = 2 K 0 t + K 1 Again differentiate y p ′ with respect to x and obtain y p ′ ′ as follows. y p ′ ′ = d d x ( 2 K 0 t + K 1 ) = 2 K 0 Since the particular solution satisfies the given ordinary differential equation, substitute y p , y p ′ , and y p ′ ′ in y ″ + y = 1 − t 2 π 2 as shown below. y ″ + y = 1 − t 2 π 2 2 K 0 + K 0 t 2 + K 1 t + K 2 = 1 − t 2 π 2 K 0 t 2 + K 1 t + 2 K 0 + K 2 = 1 − t 2 π 2 Equate the coefficients of like terms on both sides as follows. 2 K 0 + K 2 = 1 … … ( 1 ) K 0 = − 1 π 2 … … ( 2 ) K 1 = 0 … … ( 3 ) From equation (1), obtain the value of K 2 as follows. 2 K 0 + K 2 = 1 − 2 π 2 + K 2 = 1 K 2 = 1 + 2 π 2 Then the particular solution of the given ordinary differential equation becomes, y p = K 0 t 2 + K 1 t + K 2 = − 1 π 2 t 2 + 1 + 2 π 2 . Now the general solution of the given differential equation is, y ( t ) = y h ( t ) + y p ( t ) = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 Compute the values of the constants A and B using initial conditions as follows. y ( t ) = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 y ( 0 ) = A cos ( 0 ) + B sin ( 0 ) − 1 π 2 ( 0 ) + 1 + 2 π 2 0 = A + 1 + 2 π 2 A = − 1 − 2 π 2 Differentiate y ( t ) and obtain the first derivative as shown below. y ( t ) = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 y ′ ( t ) = − A sin t + B cos t − 2 t π 2 Now obtain the value of c 2 as follows. y ′ ( 0 ) = − A sin ( 0 ) + B cos ( 0 ) − 2 ( 0 ) π 2 0 = − 0 + B B = 0 Hence the solution of the initial value problem y ″ + y = 1 − t 2 π 2 is, y = A cos t + B sin t − 1 π 2 t 2 + 1 + 2 π 2 = ( − 1 − 2 π 2 ) cos t − 1 π 2 t 2 + 1 + 2 π 2 = ( 1 + 2 π 2 ) ( 1 − cos t ) − 1 π 2 t 2 Then, y ′ ( t ) = ( 1 + 2 π 2 ) ( sin t ) − 2 t π 2 . Now obtain the solution of y ″ + y = 0 as follows. The general solution of y ″ + y = 0 is y 2 = C cos t + D sin t . By continuity conditions, the initial conditions are y ( π ) = y 2 ( π ) and y ′ ( π ) = y 2 ′ ( π ) . Evaluate the constant C using y ( π ) = y 2 ( π ) as follows. y ( π ) = y 2 ( π ) ( 1 + 2 π 2 ) ( 1 − cos π ) − 1 π 2 π 2 = C cos π + D sin π ( 1 + 2 π 2 ) ( 2 ) − 1 = − C − C = 1 + 4 π 2 C = − 1 − 4 π 2 Compute the constant C using y ′ ( π ) = y 2 ′ ( π ) as follows. y ′ ( π ) = y 2 ′ ( π ) ( 1 + 2 π 2 ) ( sin π ) − 2 π π 2 = − C sin π + D cos π − 2 π = − D D = 2 π Hence, the solution of y ″ + y = 0 is y 2 = ( − 1 − 4 π 2 ) cos t + 2 π sin t . Therefore, the solution of y ″ + y = 1 − t 2 π 2 if 0 ≤ t ≤ π and 0 if t → ∞ with the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 0 is y = { ( − 1 − 2 π 2 ) cos t − 1 π 2 t 2 + 1 + 2 π 2 if 0 ≤ t ≤ π ( − 1 − 4 π 2 ) cos t + 2 π sin t if t → ∞ _ .

Advanced Engineering Mathematics

Advanced Engineering Mathematics

10th Edition

ISBN: 9780470458365

Author: Erwin Kreyszig

Publisher: Wiley, John & Sons, Incorporated

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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. 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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