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The motion of the mass-spring system modeled by the ordinary differential equation ( D 2 + 2 D + 2 I ) y = e − t 2 sin 1 2 t and the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 1 and sketch the solution. The motion of the mass spring system is y = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) _ . 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 ) ) . Calculation: The given differential equation is y ″ + 2 y ′ + 2 y = e − t 2 sin 1 2 t . First obtain the solution of the homogeneous part as follows. The auxiliary equation of y ″ + 2 y ′ + 2 y = 0 is m 2 + 2 m + 2 = 0 . Obtain the roots of the equation m 2 + 2 m + 2 = 0 as shown below. m = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = − 2 ± 4 − 8 2 = − 2 ± − 4 2 = − 2 ± 2 i 2 = − 1 ± i Here the roots are complex conjugate. Therefore, the solution y h is y h = e − t ( A cos t + B sin t ) . Here the term r ( x ) is e − t 2 sin 1 2 t which is of the form he choice for y p becomes e α t ( K cos ω t + M sin ω t ) . That is, y p = e − t 2 ( K cos t 2 + M sin t 2 ) . Differentiate y p = e − t 2 ( K cos t 2 + M sin t 2 ) with respect to t as shown below. y p ′ = d d t ( e − t 2 ( K cos t 2 + M sin t 2 ) ) = e − t 2 ( − K 2 sin t 2 + M 2 cos t 2 ) + ( − e − t 2 2 ) ( K cos t 2 + M sin t 2 ) = e − t 2 { − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 } Again differentiate y p ′ with respect to x and obtain y p ′ ′ as follows. y p ′ ′ = d d x ( e − t 2 { − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 } ) = { e − t 2 ( − K 4 cos t 2 − M 4 sin t 2 + K 4 sin t 2 − M 4 cos t 2 ) − e − t 2 2 ( − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 ) } Since the particular solution satisfies the given ordinary differential equation, substitute y p , y p ′ , and y p ′ ′ in y ″ + 2 y ′ + 2 y = e − t 2 sin 1 2 t as shown below. { e − t 2 ( − K 4 cos t 2 − M 4 sin t 2 + K 4 sin t 2 − M 4 cos t 2 ) − e − t 2 2 ( − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 ) + 2 e − t 2 { − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 } + 2 e − t 2 ( K cos t 2 + M sin t 2 ) } = e − t 2 sin 1 2 t { e − t 2 cos t 2 ( − K 4 − M 4 − M 4 + K 4 + M − K + 2 K ) + e − t 2 sin t 2 ( − M 4 + K 4 + M 4 + K 4 − M − K + 2 M ) } = e − t 2 sin 1 2 t e − t 2 cos t 2 ( K + M 2 ) + e − t 2 sin t 2 ( M − K 2 ) = e − t 2 sin 1 2 t Equate the coefficients of like terms on both sides as follows. K + M 2 = 0 … … ( 1 ) M − K 2 = 1 … … ( 2 ) From equation (1), solve for M as follows. K + M 2 = 0 K = − M 2 Substitute K = − M 2 in equation (2) and obtain the value of M as shown below. M − K 2 = 1 M + M 2 2 = 1 M + M 4 = 1 M = 4 5 = 0.8 Now compute the value of K as follows. K = − M 2 = − 0.8 2 = − 0.4 Then the particular solution of the given ordinary differential equation becomes, y p = e − t 2 ( K cos t 2 + M sin t 2 ) = e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) . Now the motion of the mass spring system is obtained as follows. y ( t ) = y h ( t ) + y p ( t ) = e − t ( A cos t + B sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) Compute the values of the constants A and B using initial conditions as follows. y ( t ) = e − t ( A cos t + B sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) y ( 0 ) = e 0 ( A cos 0 + B sin 0 ) + e 0 ( − 0.4 cos 0 + 0.8 sin 0 ) 0 = A − 0.4 A = 0.4 Differentiate y ( t ) and obtain the first derivative as shown below. y ′ ( t ) = { e − t ( − A sin t + B cos t ) − e − t ( A cos t + B sin t ) + e − t 2 ( 0.2 sin t 2 + 0.4 cos t 2 ) − e − t 2 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) } Now obtain the value of c 2 as follows. y ′ ( 0 ) = { e 0 ( − A sin 0 + B cos 0 ) − e 0 ( A cos 0 + B sin 0 ) + e 0 ( 0.2 sin 0 + 0.4 cos 0 ) − e 0 2 ( − 0.4 cos 0 + 0.8 sin 0 ) } 1 = B − A + 0.4 + 0.2 1 = B − 0.4 + 0.4 + 0.2 1 = B + 0.2 B = 0.8 Hence the motion of the mass spring system is, y = e − t ( A cos t + B sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) Therefore, the motion of the mass spring system is y = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) _ . Sketch the solution y = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) as shown in below Figure 1. From Figure 1, it is observed that the domain of the solution is ( 0 , ∞ ) . Now sketch y − y p = e − t ( 0.4 cos t + 0.8 sin t ) as shown in below Figure 2. From Figure 2, it is observed that the solution approaches the steady state.

The motion of the mass-spring system modeled by the ordinary differential equation ( D 2 + 2 D + 2 I ) y = e − t 2 sin 1 2 t and the initial conditions y ( 0 ) = 0 , y ′ ( 0 ) = 1 and sketch the solution. The motion of the mass spring system is y = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) _ . 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 ) ) . Calculation: The given differential equation is y ″ + 2 y ′ + 2 y = e − t 2 sin 1 2 t . First obtain the solution of the homogeneous part as follows. The auxiliary equation of y ″ + 2 y ′ + 2 y = 0 is m 2 + 2 m + 2 = 0 . Obtain the roots of the equation m 2 + 2 m + 2 = 0 as shown below. m = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 2 ) 2 ( 1 ) = − 2 ± 4 − 8 2 = − 2 ± − 4 2 = − 2 ± 2 i 2 = − 1 ± i Here the roots are complex conjugate. Therefore, the solution y h is y h = e − t ( A cos t + B sin t ) . Here the term r ( x ) is e − t 2 sin 1 2 t which is of the form he choice for y p becomes e α t ( K cos ω t + M sin ω t ) . That is, y p = e − t 2 ( K cos t 2 + M sin t 2 ) . Differentiate y p = e − t 2 ( K cos t 2 + M sin t 2 ) with respect to t as shown below. y p ′ = d d t ( e − t 2 ( K cos t 2 + M sin t 2 ) ) = e − t 2 ( − K 2 sin t 2 + M 2 cos t 2 ) + ( − e − t 2 2 ) ( K cos t 2 + M sin t 2 ) = e − t 2 { − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 } Again differentiate y p ′ with respect to x and obtain y p ′ ′ as follows. y p ′ ′ = d d x ( e − t 2 { − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 } ) = { e − t 2 ( − K 4 cos t 2 − M 4 sin t 2 + K 4 sin t 2 − M 4 cos t 2 ) − e − t 2 2 ( − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 ) } Since the particular solution satisfies the given ordinary differential equation, substitute y p , y p ′ , and y p ′ ′ in y ″ + 2 y ′ + 2 y = e − t 2 sin 1 2 t as shown below. { e − t 2 ( − K 4 cos t 2 − M 4 sin t 2 + K 4 sin t 2 − M 4 cos t 2 ) − e − t 2 2 ( − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 ) + 2 e − t 2 { − K 2 sin t 2 + M 2 cos t 2 − K 2 cos t 2 − M 2 sin t 2 } + 2 e − t 2 ( K cos t 2 + M sin t 2 ) } = e − t 2 sin 1 2 t { e − t 2 cos t 2 ( − K 4 − M 4 − M 4 + K 4 + M − K + 2 K ) + e − t 2 sin t 2 ( − M 4 + K 4 + M 4 + K 4 − M − K + 2 M ) } = e − t 2 sin 1 2 t e − t 2 cos t 2 ( K + M 2 ) + e − t 2 sin t 2 ( M − K 2 ) = e − t 2 sin 1 2 t Equate the coefficients of like terms on both sides as follows. K + M 2 = 0 … … ( 1 ) M − K 2 = 1 … … ( 2 ) From equation (1), solve for M as follows. K + M 2 = 0 K = − M 2 Substitute K = − M 2 in equation (2) and obtain the value of M as shown below. M − K 2 = 1 M + M 2 2 = 1 M + M 4 = 1 M = 4 5 = 0.8 Now compute the value of K as follows. K = − M 2 = − 0.8 2 = − 0.4 Then the particular solution of the given ordinary differential equation becomes, y p = e − t 2 ( K cos t 2 + M sin t 2 ) = e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) . Now the motion of the mass spring system is obtained as follows. y ( t ) = y h ( t ) + y p ( t ) = e − t ( A cos t + B sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) Compute the values of the constants A and B using initial conditions as follows. y ( t ) = e − t ( A cos t + B sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) y ( 0 ) = e 0 ( A cos 0 + B sin 0 ) + e 0 ( − 0.4 cos 0 + 0.8 sin 0 ) 0 = A − 0.4 A = 0.4 Differentiate y ( t ) and obtain the first derivative as shown below. y ′ ( t ) = { e − t ( − A sin t + B cos t ) − e − t ( A cos t + B sin t ) + e − t 2 ( 0.2 sin t 2 + 0.4 cos t 2 ) − e − t 2 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) } Now obtain the value of c 2 as follows. y ′ ( 0 ) = { e 0 ( − A sin 0 + B cos 0 ) − e 0 ( A cos 0 + B sin 0 ) + e 0 ( 0.2 sin 0 + 0.4 cos 0 ) − e 0 2 ( − 0.4 cos 0 + 0.8 sin 0 ) } 1 = B − A + 0.4 + 0.2 1 = B − 0.4 + 0.4 + 0.2 1 = B + 0.2 B = 0.8 Hence the motion of the mass spring system is, y = e − t ( A cos t + B sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) Therefore, the motion of the mass spring system is y = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) _ . Sketch the solution y = e − t ( 0.4 cos t + 0.8 sin t ) + e − t 2 ( − 0.4 cos t 2 + 0.8 sin t 2 ) as shown in below Figure 1. From Figure 1, it is observed that the domain of the solution is ( 0 , ∞ ) . Now sketch y − y p = e − t ( 0.4 cos t + 0.8 sin t ) as shown in below Figure 2. From Figure 2, it is observed that the solution approaches the steady state.

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. 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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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