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Math Advanced Math Fundamentals of Differential Equations (9th Edition)

The solution to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 and sketch the solutions. The solutions to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 are y ( t ) = − 1 3 e − 4 t + 4 3 e − t _ , y ( t ) = e − 2 t + 2 t e − 2 t _ and y ( t ) = e − t cos ( 3 t ) + 1 3 e − t sin ( 3 t ) _ , respectively. Procedure used: Consider the differential equation a y ″ + b y ′ + c y = 0 . The associated auxiliary equation is, a r 2 + b r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) 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. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 e r t + c 2 t e r t is a general solution. (c) If the auxiliary equation has complex conjugate roots α ± i β , then y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t is a general solution. Calculation: The given initial value problem is y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Consider b = 5 . Then, the differential equation will be, y ″ + 5 y ′ + 4 y = 0 . The associated auxiliary equation is r 2 + 5 r + 4 = 0 . Factorize the polynomial r 2 + 5 r + 4 = 0 as follows: r 2 + r + 4 r + 4 = 0 r ( r + 1 ) + 4 ( r + 1 ) = 0 ( r + 4 ) ( r + 1 ) = 0 r = − 4 , − 1 Note that, the roots r = − 4 and r = − 1 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e − 4 t + c 2 e − t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − 4 t + c 2 e − t , y ( 0 ) = c 1 e − 4 × 0 + c 2 e 0 1 = c 1 ( 1 ) + c 2 ( 1 ) c 1 + c 2 = 1 Differentiate y ( t ) = c 1 e − 4 t + c 2 e − t with respect to t as, y ′ ( t ) = − 4 c 1 e − 4 t − c 2 e − t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 4 c 1 e − 4 t − c 2 e − t , y ′ ( 0 ) = − 4 c 1 e − 4 × 0 − c 2 e 0 0 = − 4 c 1 − c 2 c 2 = − 4 c 1 Substitute c 2 = − 4 c 1 in c 1 + c 2 = 1 , c 1 + c 2 = 1 c 1 − 4 c 1 = 1 − 3 c 1 = 1 c 1 = − 1 3 Thus, c 2 = − 4 c 1 becomes, c 2 = 4 3 . That is, c 1 = − 1 3 and c 2 = 4 3 . The general solution y ( t ) = c 1 e − 4 t + c 2 e − t becomes y ( t ) = − 1 3 e − 4 t + 4 3 e − t . Consider b = 4 . Then, the differential equation will be, y ″ + 4 y ′ + 4 y = 0 . The associated auxiliary equation is r 2 + 4 r + 4 = 0 . Factorize the polynomial r 2 + 4 r + 4 = 0 as follows: r 2 + 2 r + 2 r + 4 = 0 r ( r + 2 ) + 2 ( r + 2 ) = 0 ( r + 2 ) ( r + 2 ) = 0 r = − 2 , − 2 Note that, the roots r = − 2 and r = − 2 are repeated. Thus, the general solution will be y ( t ) = c 1 e − 2 t + c 2 t e − 2 t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − 2 t + c 2 t e − 2 t , y ( 0 ) = c 1 e − 2 × 0 + c 2 ( 0 ) e − 2 × 0 1 = c 1 ( 1 ) c 1 = 1 Differentiate y ( t ) = c 1 e − 2 t + c 2 t e − 2 t with respect to t to obtain y ′ ( t ) = − 2 c 1 e − 2 t + c 2 ( − 2 t e − 2 t + e − 2 t ) . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 2 c 1 e − 2 t + c 2 ( − 2 t e − 2 t + e − 2 t ) , y ′ ( 0 ) = − 2 c 1 e − 2 × 0 + c 2 ( − 2 ( 0 ) e − 2 × 0 + e − 2 × 0 ) 0 = − 2 c 1 + c 2 c 2 = 2 c 1 Substitute c 1 = 1 in c 2 = 2 c 1 to obtain c 2 = 2 . That is, c 1 = 1 and c 2 = 2 . The general solution y ( t ) = c 1 e − 2 t + c 2 t e − 2 t becomes y ( t ) = e − 2 t + 2 t e − 2 t . Consider b = 2 . Then, the differential equation will be, y ″ + 2 y ′ + 4 y = 0 . The associated auxiliary equation is r 2 + 2 r + 4 = 0 . Factorize the polynomial r 2 + 2 r + 4 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as follows: r = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 4 ) 2 ( 1 ) = − 2 ± 4 − 16 2 = − 2 ± − 12 2 = − 2 ± 2 3 i 2 ( − 1 = i ) = − 1 ± 3 i That is, r = − 1 ± 3 i . Here, α = − 1 and β = 3 . Thus, the general solution will be y ( t ) = c 1 e − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) , y ( 0 ) = c 1 e − ( 0 ) cos ( 3 × 0 ) + c 2 e − ( 0 ) sin ( 3 × 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 − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) with respect to t as, y ′ ( t ) = c 1 [ − 3 e − t sin ( 3 t ) − e − t cos ( 3 t ) ] + c 2 [ 3 e − t cos ( 3 t ) − e − t sin ( 3 t ) ] = − ( 3 c 1 + c 2 ) e − t sin ( 3 t ) + ( 3 c 2 − c 1 ) e − t cos ( 3 t ) Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − ( 3 c 1 + c 2 ) e − t sin ( 3 t ) + ( 3 c 2 − c 1 ) e − t cos ( 3 t ) , y ′ ( 0 ) = − ( 3 c 1 + c 2 ) e 0 sin ( 3 × 0 ) + ( 3 c 2 − c 1 ) e 0 cos ( 3 × 0 ) 0 = − ( 3 c 1 + c 2 ) sin ( 0 ) + ( 3 c 2 − c 1 ) cos ( 0 ) 0 = ( 3 c 2 − c 1 ) c 1 = 3 c 2 c 2 = c 1 3 Substitute c 1 = 1 in c 2 = c 1 3 to obtain c 2 = 1 3 . That is, c 1 = 1 and c 2 = 1 3 . The general solution y ( t ) = c 1 e − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) becomes y ( t ) = e − t cos ( 3 t ) + 1 3 e − t sin ( 3 t ) . Therefore, the solutions to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 are y ( t ) = − 1 3 e − 4 t + 4 3 e − t _ , y ( t ) = e − 2 t + 2 t e − 2 t _ and y ( t ) = e − t cos ( 3 t ) + 1 3 e − t sin ( 3 t ) _ , respectively. Use online graphing calculator and graph the solutions to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 as shown below in Figure 1. From Figure 1, it is observed that for b = 5 , 4 and 2 , the solution tends to 0 as t → ∞ . This can be explained due to the presence of negative powers of the exponential.

The solution to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 and sketch the solutions. The solutions to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 are y ( t ) = − 1 3 e − 4 t + 4 3 e − t _ , y ( t ) = e − 2 t + 2 t e − 2 t _ and y ( t ) = e − t cos ( 3 t ) + 1 3 e − t sin ( 3 t ) _ , respectively. Procedure used: Consider the differential equation a y ″ + b y ′ + c y = 0 . The associated auxiliary equation is, a r 2 + b r + c = 0 . Solving the auxiliary equation gives the roots r 1 and r 2 . (a) 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. (b) If the auxiliary equation has repeated root r , then y ( t ) = c 1 e r t + c 2 t e r t is a general solution. (c) If the auxiliary equation has complex conjugate roots α ± i β , then y ( t ) = c 1 e α t cos β t + c 2 e α t sin β t is a general solution. Calculation: The given initial value problem is y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 . Consider b = 5 . Then, the differential equation will be, y ″ + 5 y ′ + 4 y = 0 . The associated auxiliary equation is r 2 + 5 r + 4 = 0 . Factorize the polynomial r 2 + 5 r + 4 = 0 as follows: r 2 + r + 4 r + 4 = 0 r ( r + 1 ) + 4 ( r + 1 ) = 0 ( r + 4 ) ( r + 1 ) = 0 r = − 4 , − 1 Note that, the roots r = − 4 and r = − 1 are distinct and real. Thus, the general solution will be y ( t ) = c 1 e − 4 t + c 2 e − t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − 4 t + c 2 e − t , y ( 0 ) = c 1 e − 4 × 0 + c 2 e 0 1 = c 1 ( 1 ) + c 2 ( 1 ) c 1 + c 2 = 1 Differentiate y ( t ) = c 1 e − 4 t + c 2 e − t with respect to t as, y ′ ( t ) = − 4 c 1 e − 4 t − c 2 e − t . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 4 c 1 e − 4 t − c 2 e − t , y ′ ( 0 ) = − 4 c 1 e − 4 × 0 − c 2 e 0 0 = − 4 c 1 − c 2 c 2 = − 4 c 1 Substitute c 2 = − 4 c 1 in c 1 + c 2 = 1 , c 1 + c 2 = 1 c 1 − 4 c 1 = 1 − 3 c 1 = 1 c 1 = − 1 3 Thus, c 2 = − 4 c 1 becomes, c 2 = 4 3 . That is, c 1 = − 1 3 and c 2 = 4 3 . The general solution y ( t ) = c 1 e − 4 t + c 2 e − t becomes y ( t ) = − 1 3 e − 4 t + 4 3 e − t . Consider b = 4 . Then, the differential equation will be, y ″ + 4 y ′ + 4 y = 0 . The associated auxiliary equation is r 2 + 4 r + 4 = 0 . Factorize the polynomial r 2 + 4 r + 4 = 0 as follows: r 2 + 2 r + 2 r + 4 = 0 r ( r + 2 ) + 2 ( r + 2 ) = 0 ( r + 2 ) ( r + 2 ) = 0 r = − 2 , − 2 Note that, the roots r = − 2 and r = − 2 are repeated. Thus, the general solution will be y ( t ) = c 1 e − 2 t + c 2 t e − 2 t . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − 2 t + c 2 t e − 2 t , y ( 0 ) = c 1 e − 2 × 0 + c 2 ( 0 ) e − 2 × 0 1 = c 1 ( 1 ) c 1 = 1 Differentiate y ( t ) = c 1 e − 2 t + c 2 t e − 2 t with respect to t to obtain y ′ ( t ) = − 2 c 1 e − 2 t + c 2 ( − 2 t e − 2 t + e − 2 t ) . Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − 2 c 1 e − 2 t + c 2 ( − 2 t e − 2 t + e − 2 t ) , y ′ ( 0 ) = − 2 c 1 e − 2 × 0 + c 2 ( − 2 ( 0 ) e − 2 × 0 + e − 2 × 0 ) 0 = − 2 c 1 + c 2 c 2 = 2 c 1 Substitute c 1 = 1 in c 2 = 2 c 1 to obtain c 2 = 2 . That is, c 1 = 1 and c 2 = 2 . The general solution y ( t ) = c 1 e − 2 t + c 2 t e − 2 t becomes y ( t ) = e − 2 t + 2 t e − 2 t . Consider b = 2 . Then, the differential equation will be, y ″ + 2 y ′ + 4 y = 0 . The associated auxiliary equation is r 2 + 2 r + 4 = 0 . Factorize the polynomial r 2 + 2 r + 4 = 0 using quadratic formula − b ± b 2 − 4 a c 2 a as follows: r = − 2 ± ( 2 ) 2 − 4 ( 1 ) ( 4 ) 2 ( 1 ) = − 2 ± 4 − 16 2 = − 2 ± − 12 2 = − 2 ± 2 3 i 2 ( − 1 = i ) = − 1 ± 3 i That is, r = − 1 ± 3 i . Here, α = − 1 and β = 3 . Thus, the general solution will be y ( t ) = c 1 e − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) . Apply the condition y ( 0 ) = 1 in y ( t ) = c 1 e − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) , y ( 0 ) = c 1 e − ( 0 ) cos ( 3 × 0 ) + c 2 e − ( 0 ) sin ( 3 × 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 − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) with respect to t as, y ′ ( t ) = c 1 [ − 3 e − t sin ( 3 t ) − e − t cos ( 3 t ) ] + c 2 [ 3 e − t cos ( 3 t ) − e − t sin ( 3 t ) ] = − ( 3 c 1 + c 2 ) e − t sin ( 3 t ) + ( 3 c 2 − c 1 ) e − t cos ( 3 t ) Apply the condition y ′ ( 0 ) = 0 in y ′ ( t ) = − ( 3 c 1 + c 2 ) e − t sin ( 3 t ) + ( 3 c 2 − c 1 ) e − t cos ( 3 t ) , y ′ ( 0 ) = − ( 3 c 1 + c 2 ) e 0 sin ( 3 × 0 ) + ( 3 c 2 − c 1 ) e 0 cos ( 3 × 0 ) 0 = − ( 3 c 1 + c 2 ) sin ( 0 ) + ( 3 c 2 − c 1 ) cos ( 0 ) 0 = ( 3 c 2 − c 1 ) c 1 = 3 c 2 c 2 = c 1 3 Substitute c 1 = 1 in c 2 = c 1 3 to obtain c 2 = 1 3 . That is, c 1 = 1 and c 2 = 1 3 . The general solution y ( t ) = c 1 e − t cos ( 3 t ) + c 2 e − t sin ( 3 t ) becomes y ( t ) = e − t cos ( 3 t ) + 1 3 e − t sin ( 3 t ) . Therefore, the solutions to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 are y ( t ) = − 1 3 e − 4 t + 4 3 e − t _ , y ( t ) = e − 2 t + 2 t e − 2 t _ and y ( t ) = e − t cos ( 3 t ) + 1 3 e − t sin ( 3 t ) _ , respectively. Use online graphing calculator and graph the solutions to the initial value problem y ″ + b y ′ + 4 y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = 0 for b = 5 , 4 and 2 as shown below in Figure 1. From Figure 1, it is observed that for b = 5 , 4 and 2 , the solution tends to 0 as t → ∞ . This can be explained due to the presence of negative powers of the exponential.

Solution Summary: The author illustrates the solution to the initial value problem y′′+b

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

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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