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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

The equation of motion for y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 , with k = 20 , 25 and 30 and sketch the graphs. The equation of motion for k = 20 , 25 and 30 is y ( t ) = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 e ( − 5 − 5 ) t _ , y ( t ) = e − 5 t + 5 t e − 5 t _ and y ( t ) = e − 5 t cos ( 5 t ) + 5 e − 5 t sin ( 5 t ) _ , respectively. Formula used: The governing equation of a mass-spring system is: F e x t = [ i n e r t i a ] y ″ + [ d a m p i n g ] y ′ + [ s t i f f n e s s ] y = m y ″ + b y ′ + k y Calculation: The given equation is y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 and the initial conditions are y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Let k = 20 . The equation becomes y ″ ( t ) + 10 y ′ ( t ) + 20 y ( t ) = 0 . The associated auxiliary equation is r 2 + 10 r + 20 = 0 . Factorize r 2 + 10 r + 20 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 10 ± 10 2 − 4 ( 1 ) ( 20 ) 2 ( 1 ) = − 10 ± 100 − 80 2 ( 1 ) = − 10 ± 20 2 = − 10 ± 2 5 2 = − 5 ± 5 That is, the roots are r 1 = − 5 + 5 and r 2 = − 5 − 5 . Note that, if the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = C 1 e r 1 t + C 2 e r 2 t is a general solution. Thus, the required solution will be y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t . Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t , y ( 0 ) = C 1 e ( − 5 + 5 ) × 0 + C 2 e ( − 5 − 5 ) × 0 1 = C 1 ( 1 ) + C 2 ( 1 ) C 1 + C 2 = 1 C 1 = 1 − C 2 Differentiate y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t with respect to t as, y ′ ( t ) = ( − 5 + 5 ) C 1 e ( − 5 + 5 ) t + ( − 5 − 5 ) C 2 e ( − 5 − 5 ) t . Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 5 + 5 ) C 1 e ( − 5 + 5 ) t + ( − 5 − 5 ) C 2 e ( − 5 − 5 ) t , y ′ ( 0 ) = ( − 5 + 5 ) C 1 e ( − 5 + 5 ) × 0 + ( − 5 − 5 ) C 2 e ( − 5 − 5 ) × 0 0 = ( − 5 + 5 ) C 1 ( 1 ) + ( − 5 − 5 ) C 2 ( 1 ) ( − 5 + 5 ) C 1 + ( − 5 − 5 ) C 2 = 0 Since C 1 = 1 − C 2 , the above equation becomes, ( − 5 + 5 ) ( 1 − C 2 ) + ( − 5 − 5 ) C 2 = 0 − 5 + 5 + 5 C 2 − 5 C 2 − 5 C 2 − 5 C 2 = 0 − 5 + 5 − 2 5 C 2 = 0 2 5 C 2 = − 5 + 5 Simplify further as, C 2 = − 5 + 5 2 5 = 5 ( − 5 + 1 ) 2 5 = 1 − 5 2 Thus, C 1 = 1 − C 2 becomes, C 1 = 1 − 1 − 5 2 = 1 + − 1 + 5 2 = 1 + 5 2 That is, C 1 = 1 + 5 2 and C 2 = 1 − 5 2 . Thus, the solution y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t becomes, y ( t ) = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 e ( − 5 − 5 ) t _ . Use online graphing calculator and sketch the graph of the solution y ( t ) = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 e ( − 5 − 5 ) t as shown below in Figure 1. From Figure 1, it is observed that the maximum value is 1. Let k = 25 . The equation y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 becomes y ″ ( t ) + 10 y ′ ( t ) + 25 y ( t ) = 0 . The associated auxiliary equation is r 2 + 10 r + 25 = 0 . Simplify r 2 + 10 r + 25 = 0 as, r 2 + 5 r + 5 r + 64 = 0 ( r + 5 ) 2 = 0 r = − 5 , − 5 That is, the repeated roots are r = − 5 , − 5 . Note that, the solution form for repeated roots is y ( t ) = C 1 e r t + C 2 t e r t . Thus, the required solution will be y ( t ) = C 1 e − 5 t + C 2 t e − 5 t . Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = C 1 e − 5 t + C 2 t e − 5 t , y ( 0 ) = C 1 e − 5 × 0 + C 2 ( 0 ) e − 5 × 0 1 = C 1 + C 2 ( 0 ) C 1 = 1 Differentiate y ( t ) = C 1 e − 5 t + C 2 t e − 5 t with respect to t as, y ′ ( t ) = − 5 C 1 e − 5 t + C 2 [ − 5 t e − 5 t + e − 5 t ] = − 5 C 1 e − 5 t − 5 C 2 t e − 5 t + C 2 e − 5 t = ( − 5 C 1 + C 2 ) e − 5 t − 5 C 2 t e − 5 t Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 5 C 1 + C 2 ) e − 5 t − 5 C 2 t e − 5 t , y ′ ( 0 ) = ( − 5 C 1 + C 2 ) e − 5 × 0 − 5 C 2 ( 0 ) e − 5 × 0 0 = ( − 5 C 1 + C 2 ) − 5 C 2 ( 0 ) − 5 C 1 + C 2 = 0 C 2 = 5 C 1 Since C 1 = 1 , the above equation becomes, C 2 = 5 ( 1 ) = 5 That is, C 1 = 1 and C 2 = 5 . Thus, the solution y ( t ) = C 1 e − 5 t + C 2 t e − 5 t becomes, y ( t ) = e − 5 t + 5 t e − 5 t _ . Use online graphing calculator and sketch the graph of the solution y ( t ) = e − 5 t + 5 t e − 5 t as shown below in Figure 2. From Figure 2, it is observed that the maximum value is 1. Let k = 30 . The equation y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 becomes y ″ ( t ) + 10 y ′ ( t ) + 30 y ( t ) = 0 . The associated auxiliary equation is r 2 + 10 r + 30 = 0 . Factorize r 2 + 10 r + 30 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 10 ± 10 2 − 4 ( 1 ) ( 30 ) 2 ( 1 ) = − 10 ± 100 − 120 2 = − 10 ± − 20 2 = − 10 ± 2 5 i 2 ( ∵ i = − 1 ) = − 5 ± 5 i That is, the complex roots are r 1 = − 5 + 5 i and r 2 = − 5 − 5 i . Here, α = − 5 and β = 5 . Note that, the solution form for complex root α ± i β is y ( t ) = C 1 e α t cos β t + C 2 e α t sin β t . Thus, the required solution will be y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) . Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) , y ( 0 ) = C 1 e − 5 × 0 cos ( 5 × 0 ) + C 2 e − 5 × 0 sin ( 5 × 0 ) 1 = C 1 cos 0 + C 2 sin 0 1 = C 1 ( 1 ) + C 2 ( 0 ) C 1 = 1 Differentiate y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) with respect to t as, y ′ ( t ) = C 1 [ − 5 e − 5 t sin ( 5 t ) − 5 e − 5 t cos ( 5 t ) ] + C 2 [ 5 e − 5 t cos ( 5 t ) − 5 e − 5 t sin ( 5 t ) ] = ( − 5 C 1 − 5 C 2 ) e − 5 t sin ( 5 t ) + ( − 5 C 1 + 5 C 2 ) e − 5 t cos ( 5 t ) Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 5 C 1 − 5 C 2 ) e − 5 t sin ( 5 t ) + ( − 5 C 1 + 5 C 2 ) e − 5 t cos ( 5 t ) , y ′ ( 0 ) = ( − 5 C 1 − 5 C 2 ) e − 5 × 0 sin ( 5 × 0 ) + ( − 5 C 1 + 5 C 2 ) e − 5 × 0 cos ( 5 × 0 ) 0 = ( − 5 C 1 − 5 C 2 ) sin 0 + ( − 5 C 1 + 5 C 2 ) cos 0 0 = ( − 5 C 1 − 5 C 2 ) ( 0 ) + ( − 5 C 1 + 5 C 2 ) ( 1 ) − 5 C 1 + 5 C 2 = 0 Since C 1 = 1 , the above equation becomes, − 5 ( 1 ) + 5 C 2 5 C 2 = 5 C 2 = 5 5 C 2 = 5 That is, C 1 = 1 and C 2 = 5 . Thus, the solution y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) becomes, y ( t ) = e − 5 t cos ( 5 t ) + 5 e − 5 t sin ( 5 t ) _ . Use online graphing calculator and sketch the graph of the solution y ( t ) = e − 5 t cos ( 5 t ) + 5 e − 5 t sin ( 5 t ) as shown below in Figure 3. From Figure 3, it is observed that the maximum value is 1.

The equation of motion for y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 , with k = 20 , 25 and 30 and sketch the graphs. The equation of motion for k = 20 , 25 and 30 is y ( t ) = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 e ( − 5 − 5 ) t _ , y ( t ) = e − 5 t + 5 t e − 5 t _ and y ( t ) = e − 5 t cos ( 5 t ) + 5 e − 5 t sin ( 5 t ) _ , respectively. Formula used: The governing equation of a mass-spring system is: F e x t = [ i n e r t i a ] y ″ + [ d a m p i n g ] y ′ + [ s t i f f n e s s ] y = m y ″ + b y ′ + k y Calculation: The given equation is y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 and the initial conditions are y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Let k = 20 . The equation becomes y ″ ( t ) + 10 y ′ ( t ) + 20 y ( t ) = 0 . The associated auxiliary equation is r 2 + 10 r + 20 = 0 . Factorize r 2 + 10 r + 20 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 10 ± 10 2 − 4 ( 1 ) ( 20 ) 2 ( 1 ) = − 10 ± 100 − 80 2 ( 1 ) = − 10 ± 20 2 = − 10 ± 2 5 2 = − 5 ± 5 That is, the roots are r 1 = − 5 + 5 and r 2 = − 5 − 5 . Note that, if the auxiliary equation has distinct real roots r 1 and r 2 , then y ( t ) = C 1 e r 1 t + C 2 e r 2 t is a general solution. Thus, the required solution will be y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t . Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t , y ( 0 ) = C 1 e ( − 5 + 5 ) × 0 + C 2 e ( − 5 − 5 ) × 0 1 = C 1 ( 1 ) + C 2 ( 1 ) C 1 + C 2 = 1 C 1 = 1 − C 2 Differentiate y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t with respect to t as, y ′ ( t ) = ( − 5 + 5 ) C 1 e ( − 5 + 5 ) t + ( − 5 − 5 ) C 2 e ( − 5 − 5 ) t . Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 5 + 5 ) C 1 e ( − 5 + 5 ) t + ( − 5 − 5 ) C 2 e ( − 5 − 5 ) t , y ′ ( 0 ) = ( − 5 + 5 ) C 1 e ( − 5 + 5 ) × 0 + ( − 5 − 5 ) C 2 e ( − 5 − 5 ) × 0 0 = ( − 5 + 5 ) C 1 ( 1 ) + ( − 5 − 5 ) C 2 ( 1 ) ( − 5 + 5 ) C 1 + ( − 5 − 5 ) C 2 = 0 Since C 1 = 1 − C 2 , the above equation becomes, ( − 5 + 5 ) ( 1 − C 2 ) + ( − 5 − 5 ) C 2 = 0 − 5 + 5 + 5 C 2 − 5 C 2 − 5 C 2 − 5 C 2 = 0 − 5 + 5 − 2 5 C 2 = 0 2 5 C 2 = − 5 + 5 Simplify further as, C 2 = − 5 + 5 2 5 = 5 ( − 5 + 1 ) 2 5 = 1 − 5 2 Thus, C 1 = 1 − C 2 becomes, C 1 = 1 − 1 − 5 2 = 1 + − 1 + 5 2 = 1 + 5 2 That is, C 1 = 1 + 5 2 and C 2 = 1 − 5 2 . Thus, the solution y ( t ) = C 1 e ( − 5 + 5 ) t + C 2 e ( − 5 − 5 ) t becomes, y ( t ) = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 e ( − 5 − 5 ) t _ . Use online graphing calculator and sketch the graph of the solution y ( t ) = 1 + 5 2 e ( − 5 + 5 ) t + 1 − 5 2 e ( − 5 − 5 ) t as shown below in Figure 1. From Figure 1, it is observed that the maximum value is 1. Let k = 25 . The equation y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 becomes y ″ ( t ) + 10 y ′ ( t ) + 25 y ( t ) = 0 . The associated auxiliary equation is r 2 + 10 r + 25 = 0 . Simplify r 2 + 10 r + 25 = 0 as, r 2 + 5 r + 5 r + 64 = 0 ( r + 5 ) 2 = 0 r = − 5 , − 5 That is, the repeated roots are r = − 5 , − 5 . Note that, the solution form for repeated roots is y ( t ) = C 1 e r t + C 2 t e r t . Thus, the required solution will be y ( t ) = C 1 e − 5 t + C 2 t e − 5 t . Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = C 1 e − 5 t + C 2 t e − 5 t , y ( 0 ) = C 1 e − 5 × 0 + C 2 ( 0 ) e − 5 × 0 1 = C 1 + C 2 ( 0 ) C 1 = 1 Differentiate y ( t ) = C 1 e − 5 t + C 2 t e − 5 t with respect to t as, y ′ ( t ) = − 5 C 1 e − 5 t + C 2 [ − 5 t e − 5 t + e − 5 t ] = − 5 C 1 e − 5 t − 5 C 2 t e − 5 t + C 2 e − 5 t = ( − 5 C 1 + C 2 ) e − 5 t − 5 C 2 t e − 5 t Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 5 C 1 + C 2 ) e − 5 t − 5 C 2 t e − 5 t , y ′ ( 0 ) = ( − 5 C 1 + C 2 ) e − 5 × 0 − 5 C 2 ( 0 ) e − 5 × 0 0 = ( − 5 C 1 + C 2 ) − 5 C 2 ( 0 ) − 5 C 1 + C 2 = 0 C 2 = 5 C 1 Since C 1 = 1 , the above equation becomes, C 2 = 5 ( 1 ) = 5 That is, C 1 = 1 and C 2 = 5 . Thus, the solution y ( t ) = C 1 e − 5 t + C 2 t e − 5 t becomes, y ( t ) = e − 5 t + 5 t e − 5 t _ . Use online graphing calculator and sketch the graph of the solution y ( t ) = e − 5 t + 5 t e − 5 t as shown below in Figure 2. From Figure 2, it is observed that the maximum value is 1. Let k = 30 . The equation y ″ ( t ) + 10 y ′ ( t ) + k y ( t ) = 0 becomes y ″ ( t ) + 10 y ′ ( t ) + 30 y ( t ) = 0 . The associated auxiliary equation is r 2 + 10 r + 30 = 0 . Factorize r 2 + 10 r + 30 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as, r = − 10 ± 10 2 − 4 ( 1 ) ( 30 ) 2 ( 1 ) = − 10 ± 100 − 120 2 = − 10 ± − 20 2 = − 10 ± 2 5 i 2 ( ∵ i = − 1 ) = − 5 ± 5 i That is, the complex roots are r 1 = − 5 + 5 i and r 2 = − 5 − 5 i . Here, α = − 5 and β = 5 . Note that, the solution form for complex root α ± i β is y ( t ) = C 1 e α t cos β t + C 2 e α t sin β t . Thus, the required solution will be y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) . Apply the initial conditions to obtain the value of the constants as follows: Substitute t = 0 and y ( 0 ) = 1 in y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) , y ( 0 ) = C 1 e − 5 × 0 cos ( 5 × 0 ) + C 2 e − 5 × 0 sin ( 5 × 0 ) 1 = C 1 cos 0 + C 2 sin 0 1 = C 1 ( 1 ) + C 2 ( 0 ) C 1 = 1 Differentiate y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) with respect to t as, y ′ ( t ) = C 1 [ − 5 e − 5 t sin ( 5 t ) − 5 e − 5 t cos ( 5 t ) ] + C 2 [ 5 e − 5 t cos ( 5 t ) − 5 e − 5 t sin ( 5 t ) ] = ( − 5 C 1 − 5 C 2 ) e − 5 t sin ( 5 t ) + ( − 5 C 1 + 5 C 2 ) e − 5 t cos ( 5 t ) Substitute t = 0 and y ′ ( 0 ) = 0 in y ′ ( t ) = ( − 5 C 1 − 5 C 2 ) e − 5 t sin ( 5 t ) + ( − 5 C 1 + 5 C 2 ) e − 5 t cos ( 5 t ) , y ′ ( 0 ) = ( − 5 C 1 − 5 C 2 ) e − 5 × 0 sin ( 5 × 0 ) + ( − 5 C 1 + 5 C 2 ) e − 5 × 0 cos ( 5 × 0 ) 0 = ( − 5 C 1 − 5 C 2 ) sin 0 + ( − 5 C 1 + 5 C 2 ) cos 0 0 = ( − 5 C 1 − 5 C 2 ) ( 0 ) + ( − 5 C 1 + 5 C 2 ) ( 1 ) − 5 C 1 + 5 C 2 = 0 Since C 1 = 1 , the above equation becomes, − 5 ( 1 ) + 5 C 2 5 C 2 = 5 C 2 = 5 5 C 2 = 5 That is, C 1 = 1 and C 2 = 5 . Thus, the solution y ( t ) = C 1 e − 5 t cos ( 5 t ) + C 2 e − 5 t sin ( 5 t ) becomes, y ( t ) = e − 5 t cos ( 5 t ) + 5 e − 5 t sin ( 5 t ) _ . Use online graphing calculator and sketch the graph of the solution y ( t ) = e − 5 t cos ( 5 t ) + 5 e − 5 t sin ( 5 t ) as shown below in Figure 3. From Figure 3, it is observed that the maximum value is 1.

Solution Summary: The author calculates the equation of motion for y′′(t)+10 y

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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Chapter 4 Solutions

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

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