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

(a) A solution of the form A cos Ω t + B sin Ω t to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t . The solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t is y ( t ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 cos Ω t + b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 sin Ω t _ . The generic forced oscillator equation is, m y ″ + b y ′ + k y = cos Ω t . Let y = A cos Ω t + B sin Ω t . The aim is to obtain the values of A and B . Differentiate y = A cos Ω t + B sin Ω t with respect to t as, y ′ = − A Ω sin Ω t + B Ω cos Ω t . Differentiate y ′ = − A Ω sin Ω t + B Ω cos Ω t with respect to t as, y ″ = − A Ω 2 cos Ω t − B Ω 2 sin Ω t . Substitute y ″ = − A Ω 2 cos Ω t − B Ω 2 sin Ω t , y ′ = − A Ω sin Ω t + B Ω cos Ω t and y = A cos Ω t + B sin Ω t in m y ″ + b y ′ + k y = cos Ω t , m y ″ + b y ′ + k y = cos Ω t m ( − A Ω 2 cos Ω t − B Ω 2 sin Ω t ) + b ( − A Ω sin Ω t + B Ω cos Ω t ) + k ( A cos Ω t + B sin Ω t ) = cos Ω t − A m Ω 2 cos Ω t − B m Ω 2 sin Ω t − A b Ω sin Ω t + B b Ω cos Ω t + A k cos Ω t + B k sin Ω t = cos Ω t ( − A m Ω 2 + B b Ω + A k ) cos Ω t + ( − B m Ω 2 − A b Ω + B k ) sin Ω t = cos Ω t Comparing the coefficients on either sides, the equations obtained are: − A m Ω 2 + B b Ω + A k = 1 … ( 1 ) − B m Ω 2 − A b Ω + B k = 0 … ( 2 ) From equation (1), A ( k − m Ω 2 ) + B b Ω = 1 A ( k − m Ω 2 ) = 1 − B b Ω A = 1 − B b Ω k − m Ω 2 Substitute A = 1 − B b Ω k − m Ω 2 in equation (2), − B m Ω 2 − A b Ω + B k = 0 ( k − m Ω 2 ) B − A b Ω = 0 ( k − m Ω 2 ) B − ( 1 − B b Ω k − m Ω 2 ) b Ω = 0 ( k − m Ω 2 ) 2 B − ( 1 − B b Ω ) b Ω = 0 Simplify further as, ( k − m Ω 2 ) 2 B = ( 1 − B b Ω ) b Ω ( k − m Ω 2 ) 2 B = b Ω − B b 2 Ω 2 ( k − m Ω 2 ) 2 B + b 2 Ω 2 B = b Ω [ ( k − m Ω 2 ) 2 + b 2 Ω 2 ] B = b Ω B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 Substitute B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 in A = 1 − B b Ω k − m Ω 2 , A = 1 − ( b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 ) b Ω k − m Ω 2 = 1 − ( b 2 Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 ) k − m Ω 2 = ( k − m Ω 2 ) 2 + b 2 Ω 2 − b 2 Ω 2 ( k − m Ω 2 ) ( ( k − m Ω 2 ) 2 + b 2 Ω 2 ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 Substitute A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 in y = A cos Ω t + B sin Ω t to obtain as, y ( t ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 cos Ω t + b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 sin Ω t . Therefore, the solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t is y ( t ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 cos Ω t + b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 sin Ω t _ . (b) To sketch: The graph of the coefficients A and B as a function of Ω for m = 1 , b = 0.1 and k = 25 . From part (a), the coefficients are A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 . Substitute m = 1 , b = 0.1 and k = 25 in A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 , A = 25 − Ω 2 ( 25 − Ω 2 ) 2 + ( 0.1 ) 2 Ω 2 = 25 − Ω 2 625 − 50 Ω 2 + Ω 4 + 0.01 Ω 2 = 25 − Ω 2 625 − 49.99 Ω 2 + Ω 4 Substitute m = 1 , b = 0.1 and k = 25 in B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 , B = 0.1 Ω ( 25 − Ω 2 ) 2 + ( 0.1 ) 2 Ω 2 = 0.1 Ω 625 − 50 Ω 2 + Ω 4 + 0.01 Ω 2 = 0.1 Ω 625 − 49.99 Ω 2 + Ω 4 Use online graphing calculator and sketch the graph of A = 25 − Ω 2 625 − 49.99 Ω 2 + Ω 4 as shown below in Figure 1. From Figure 1, it is observed that the maximum value for A is approximately 1.1. Use online graphing calculator and sketch the graph of B = 0.1 Ω 625 − 49.99 Ω 2 + Ω 4 as shown below in Figure 2. From Figure 2, it is observed that the maximum value for B is 10. (c) To sketch: The graph of the coefficients A and B as a function of Ω for m = 1 , b = 0 and k = 25 and explain the nature of the coefficients at Ω = 5 . From part (a), the coefficients are A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 . Substitute m = 1 , b = 0 and k = 25 in A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 , A = 25 − Ω 2 ( 25 − Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 25 − Ω 2 ( 25 − Ω 2 ) 2 = 1 25 − Ω 2 Substitute m = 1 , b = 0 and k = 25 in B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 , B = ( 0 ) Ω ( 25 − Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 0 Use online graphing calculator and sketch the graph of A = 1 25 − Ω 2 as shown below in Figure 3. From Figure 3, it is observed that at Ω = 5 , the coefficient A is discontinuous. That is, the coefficient A grows out bound as Ω approaches 5. (d) To show: There are no synchronous solution of the form A cos Ω t + B sin Ω t to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m . From part (a), the coefficients are A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 . Substitute b = 0 and Ω = k m in A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 , A = k − m Ω 2 ( k − m Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 1 k − m Ω 2 = 1 k − m ( k m ) 2 = 1 0 = ∞ Substitute b = 0 and Ω = k m in B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 , B = ( 0 ) Ω ( k − m Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 0 The solution A cos Ω t + B sin Ω t becomes, y ( t ) = ( ∞ ) cos Ω t + ( 0 ) sin Ω t . Thus, there are no synchronous solution of the form A cos Ω t + B sin Ω t to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m . (e) To verify: The function y ( t ) = ( 2 m Ω ) − 1 t sin Ω t is a non-synchronous solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m . Consider the given function y ( t ) = ( 2 m Ω ) − 1 t sin Ω t . Differentiate y ( t ) = ( 2 m Ω ) − 1 t sin Ω t with respect to t as, y ′ ( t ) = ( 2 m Ω ) − 1 [ Ω t cos Ω t + sin Ω t ] . Differentiate y ′ ( t ) = ( 2 m Ω ) − 1 [ Ω t cos Ω t + sin Ω t ] with respect to t as, y ″ ( t ) = ( 2 m Ω ) − 1 [ Ω ( t ( − Ω sin Ω t ) + cos Ω t ) + Ω cos Ω t ] = ( 2 m Ω ) − 1 [ − Ω 2 t sin Ω t + Ω cos Ω t + Ω cos Ω t ] = ( 2 m Ω ) − 1 [ − Ω 2 t sin Ω t + 2 Ω cos Ω t ] When b = 0 , the equation m y ″ + b y ′ + k y = cos Ω t gives m y ″ + k y = cos Ω t . Substitute y ″ ( t ) = ( 2 m Ω ) − 1 ( − Ω 2 t sin Ω t + 2 Ω cos Ω t ) and y ( t ) = ( 2 m Ω ) − 1 t sin Ω t in m y ″ + k y = cos Ω t , m y ″ + k y = cos Ω t m [ ( 2 m Ω ) − 1 ( − Ω 2 t sin Ω t + 2 Ω cos Ω t ) ] + k ( ( 2 m Ω ) − 1 t sin Ω t ) = cos Ω t ( 2 Ω ) − 1 ( − Ω 2 t sin Ω t + 2 Ω cos Ω t ) + k ( 2 m Ω ) − 1 t sin Ω t = cos Ω t − ( 2 ) − 1 Ω t sin Ω t + cos Ω t + k ( 2 m Ω ) − 1 t sin Ω t = cos Ω t Substitute k = Ω 2 m in the above equation. − ( 2 ) − 1 Ω t sin Ω t + cos Ω t + ( Ω 2 m ) ( 2 m Ω ) − 1 t sin Ω t = cos Ω t − ( 2 ) − 1 Ω t sin Ω t + cos Ω t + ( 2 ) − 1 Ω t sin Ω t = cos Ω t cos Ω t = cos Ω t Since the equality holds, it is concluded that y ( t ) = ( 2 m Ω ) − 1 t sin Ω t is a non-synchronous solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m .

(a) A solution of the form A cos Ω t + B sin Ω t to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t . The solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t is y ( t ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 cos Ω t + b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 sin Ω t _ . The generic forced oscillator equation is, m y ″ + b y ′ + k y = cos Ω t . Let y = A cos Ω t + B sin Ω t . The aim is to obtain the values of A and B . Differentiate y = A cos Ω t + B sin Ω t with respect to t as, y ′ = − A Ω sin Ω t + B Ω cos Ω t . Differentiate y ′ = − A Ω sin Ω t + B Ω cos Ω t with respect to t as, y ″ = − A Ω 2 cos Ω t − B Ω 2 sin Ω t . Substitute y ″ = − A Ω 2 cos Ω t − B Ω 2 sin Ω t , y ′ = − A Ω sin Ω t + B Ω cos Ω t and y = A cos Ω t + B sin Ω t in m y ″ + b y ′ + k y = cos Ω t , m y ″ + b y ′ + k y = cos Ω t m ( − A Ω 2 cos Ω t − B Ω 2 sin Ω t ) + b ( − A Ω sin Ω t + B Ω cos Ω t ) + k ( A cos Ω t + B sin Ω t ) = cos Ω t − A m Ω 2 cos Ω t − B m Ω 2 sin Ω t − A b Ω sin Ω t + B b Ω cos Ω t + A k cos Ω t + B k sin Ω t = cos Ω t ( − A m Ω 2 + B b Ω + A k ) cos Ω t + ( − B m Ω 2 − A b Ω + B k ) sin Ω t = cos Ω t Comparing the coefficients on either sides, the equations obtained are: − A m Ω 2 + B b Ω + A k = 1 … ( 1 ) − B m Ω 2 − A b Ω + B k = 0 … ( 2 ) From equation (1), A ( k − m Ω 2 ) + B b Ω = 1 A ( k − m Ω 2 ) = 1 − B b Ω A = 1 − B b Ω k − m Ω 2 Substitute A = 1 − B b Ω k − m Ω 2 in equation (2), − B m Ω 2 − A b Ω + B k = 0 ( k − m Ω 2 ) B − A b Ω = 0 ( k − m Ω 2 ) B − ( 1 − B b Ω k − m Ω 2 ) b Ω = 0 ( k − m Ω 2 ) 2 B − ( 1 − B b Ω ) b Ω = 0 Simplify further as, ( k − m Ω 2 ) 2 B = ( 1 − B b Ω ) b Ω ( k − m Ω 2 ) 2 B = b Ω − B b 2 Ω 2 ( k − m Ω 2 ) 2 B + b 2 Ω 2 B = b Ω [ ( k − m Ω 2 ) 2 + b 2 Ω 2 ] B = b Ω B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 Substitute B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 in A = 1 − B b Ω k − m Ω 2 , A = 1 − ( b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 ) b Ω k − m Ω 2 = 1 − ( b 2 Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 ) k − m Ω 2 = ( k − m Ω 2 ) 2 + b 2 Ω 2 − b 2 Ω 2 ( k − m Ω 2 ) ( ( k − m Ω 2 ) 2 + b 2 Ω 2 ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 Substitute A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 in y = A cos Ω t + B sin Ω t to obtain as, y ( t ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 cos Ω t + b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 sin Ω t . Therefore, the solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t is y ( t ) = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 cos Ω t + b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 sin Ω t _ . (b) To sketch: The graph of the coefficients A and B as a function of Ω for m = 1 , b = 0.1 and k = 25 . From part (a), the coefficients are A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 . Substitute m = 1 , b = 0.1 and k = 25 in A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 , A = 25 − Ω 2 ( 25 − Ω 2 ) 2 + ( 0.1 ) 2 Ω 2 = 25 − Ω 2 625 − 50 Ω 2 + Ω 4 + 0.01 Ω 2 = 25 − Ω 2 625 − 49.99 Ω 2 + Ω 4 Substitute m = 1 , b = 0.1 and k = 25 in B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 , B = 0.1 Ω ( 25 − Ω 2 ) 2 + ( 0.1 ) 2 Ω 2 = 0.1 Ω 625 − 50 Ω 2 + Ω 4 + 0.01 Ω 2 = 0.1 Ω 625 − 49.99 Ω 2 + Ω 4 Use online graphing calculator and sketch the graph of A = 25 − Ω 2 625 − 49.99 Ω 2 + Ω 4 as shown below in Figure 1. From Figure 1, it is observed that the maximum value for A is approximately 1.1. Use online graphing calculator and sketch the graph of B = 0.1 Ω 625 − 49.99 Ω 2 + Ω 4 as shown below in Figure 2. From Figure 2, it is observed that the maximum value for B is 10. (c) To sketch: The graph of the coefficients A and B as a function of Ω for m = 1 , b = 0 and k = 25 and explain the nature of the coefficients at Ω = 5 . From part (a), the coefficients are A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 . Substitute m = 1 , b = 0 and k = 25 in A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 , A = 25 − Ω 2 ( 25 − Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 25 − Ω 2 ( 25 − Ω 2 ) 2 = 1 25 − Ω 2 Substitute m = 1 , b = 0 and k = 25 in B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 , B = ( 0 ) Ω ( 25 − Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 0 Use online graphing calculator and sketch the graph of A = 1 25 − Ω 2 as shown below in Figure 3. From Figure 3, it is observed that at Ω = 5 , the coefficient A is discontinuous. That is, the coefficient A grows out bound as Ω approaches 5. (d) To show: There are no synchronous solution of the form A cos Ω t + B sin Ω t to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m . From part (a), the coefficients are A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 and B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 . Substitute b = 0 and Ω = k m in A = k − m Ω 2 ( k − m Ω 2 ) 2 + b 2 Ω 2 , A = k − m Ω 2 ( k − m Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 1 k − m Ω 2 = 1 k − m ( k m ) 2 = 1 0 = ∞ Substitute b = 0 and Ω = k m in B = b Ω ( k − m Ω 2 ) 2 + b 2 Ω 2 , B = ( 0 ) Ω ( k − m Ω 2 ) 2 + ( 0 ) 2 Ω 2 = 0 The solution A cos Ω t + B sin Ω t becomes, y ( t ) = ( ∞ ) cos Ω t + ( 0 ) sin Ω t . Thus, there are no synchronous solution of the form A cos Ω t + B sin Ω t to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m . (e) To verify: The function y ( t ) = ( 2 m Ω ) − 1 t sin Ω t is a non-synchronous solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m . Consider the given function y ( t ) = ( 2 m Ω ) − 1 t sin Ω t . Differentiate y ( t ) = ( 2 m Ω ) − 1 t sin Ω t with respect to t as, y ′ ( t ) = ( 2 m Ω ) − 1 [ Ω t cos Ω t + sin Ω t ] . Differentiate y ′ ( t ) = ( 2 m Ω ) − 1 [ Ω t cos Ω t + sin Ω t ] with respect to t as, y ″ ( t ) = ( 2 m Ω ) − 1 [ Ω ( t ( − Ω sin Ω t ) + cos Ω t ) + Ω cos Ω t ] = ( 2 m Ω ) − 1 [ − Ω 2 t sin Ω t + Ω cos Ω t + Ω cos Ω t ] = ( 2 m Ω ) − 1 [ − Ω 2 t sin Ω t + 2 Ω cos Ω t ] When b = 0 , the equation m y ″ + b y ′ + k y = cos Ω t gives m y ″ + k y = cos Ω t . Substitute y ″ ( t ) = ( 2 m Ω ) − 1 ( − Ω 2 t sin Ω t + 2 Ω cos Ω t ) and y ( t ) = ( 2 m Ω ) − 1 t sin Ω t in m y ″ + k y = cos Ω t , m y ″ + k y = cos Ω t m [ ( 2 m Ω ) − 1 ( − Ω 2 t sin Ω t + 2 Ω cos Ω t ) ] + k ( ( 2 m Ω ) − 1 t sin Ω t ) = cos Ω t ( 2 Ω ) − 1 ( − Ω 2 t sin Ω t + 2 Ω cos Ω t ) + k ( 2 m Ω ) − 1 t sin Ω t = cos Ω t − ( 2 ) − 1 Ω t sin Ω t + cos Ω t + k ( 2 m Ω ) − 1 t sin Ω t = cos Ω t Substitute k = Ω 2 m in the above equation. − ( 2 ) − 1 Ω t sin Ω t + cos Ω t + ( Ω 2 m ) ( 2 m Ω ) − 1 t sin Ω t = cos Ω t − ( 2 ) − 1 Ω t sin Ω t + cos Ω t + ( 2 ) − 1 Ω t sin Ω t = cos Ω t cos Ω t = cos Ω t Since the equality holds, it is concluded that y ( t ) = ( 2 m Ω ) − 1 t sin Ω t is a non-synchronous solution to the generic forced oscillator equation m y ″ + b y ′ + k y = cos Ω t , for b = 0 and Ω = k m .

Solution Summary: The author explains the solution of the form AmathrmcosOmega t+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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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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