Explanation Given: The values t = 2 π n , n = 0 , 1 , ... , 20 for equation ( 1 ) , taking A = F = 1 , ϕ = 0 , ω = 1 , and b = − 0.1 and equation is x ″ ( t ) + b x ′ ( t ) + ω 2 x ( t ) = F cos t . Approach: The linear second-order constant-coefficient differential equation a y ″ + b y ′ + c y = f ( t ) here ( a ≠ 0 ) with the case that f ( t ) = 0 , that is the equation a y ″ + b y ′ + c y = 0 is called the homogeneous equation. To find the solution of second order homogeneous linear differential equation, some steps are given below. Put y = e r t in the equation a y ″ + b y ′ + c y = 0 . Get a equation of the form a r 2 + b r + c = 0 which is called the auxiliary equation or characteristic equation of the equation a y ″ + b y ′ + c y = 0 . The characteristic equation is quadratic equation. Let r 1 and r 2 be two roots of the characteristic equation, let y 1 ( t ) and y 2 ( t ) be the pair of solutions. If the roots of the characteristic are real and distinct then general solution of the differential equation is given as, y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) . If the characteristic equation has repeated roots then general solution of the differential equation is given as, y ( t ) = c 1 y 1 ( t ) + c 2 t y 2 ( t ) . If the characteristic equation has complex roots say α ± i β then general solution of the differential equation is given as, y ( t ) = c 1 y 1 ( t ) + c 2 y 2 ( t ) y ( t ) = c 1 e α t ( cos β t + i sin β t ) + c 2 e α t ( cos β t − i sin β t ) = e α t ( c 1 cos ( β t ) + c 2 sin ( β t ) ) The general solution of the second order non-homogeneous linear differential equation is the sum of the complementary solution of the corresponding homogeneous equation and particular solution. Calculation: Consider the equation x ″ ( t ) + b x ′ ( t ) + ω 2 x ( t ) = F cos t … ( 1 ) The homogeneous equation is given as, x ″ ( t ) + b x ′ ( t ) + ω 2 x ( t ) = 0 … ( 2 ) Substitute e r t for x in equation ( 2 ) . r 2 e r t + b r e r t + ω 2 e r t = 0 e r t ( r 2 + b r + ω 2 ) = 0 The auxiliary equation is given as, r 2 + b r + ω 2 = 0 Suppose that b 2 < 4 ω 2 The roots of the equation r 2 + b r + ω 2 = 0 are calculated as, r = − b ± i b 2 − 4 ω 2 2 Thus, the roots of the auxiliary equation are r = − b ± i b 2 − 4 ω 2 2 . The solution of the equation ( 2 ) is given as, x h ( t ) = c 1 e ( − b + i b 2 − 4 ω 2 2 ) t + c 2 e ( − b − i b 2 − 4 ω 2 2 ) t = e − b t 2 [ c 1 e i ( b 2 − 4 ω 2 2 ) t + c 2 e − i ( b 2 − 4 ω 2 2 ) t ] = e − b t 2 [ c 1 ( cos ( t b 2 − 4 ω 2 2 ) + i sin ( t b 2 − 4 ω 2 2 ) ) + c 2 ( cos ( t b 2 − 4 ω 2 2 ) − i sin ( t b 2 − 4 ω 2 2 ) ) ] = e − b t 2 [ ( c 1 + c 2 ) cos ( t 4 ω 2 − b 2 2 ) + i ( c 1 − c 2 ) sin ( t 4 ω 2 − b 2 2 ) ] … ( 3 ) Suppose c 1 + c 2 = A sin ϕ , i ( c 1 − c 2 ) = A cos ϕ The equation ( 3 ) becomes, x h ( t ) = e − b t 2 [ A sin ϕ cos ( 4 ω 2 − b 2 2 t ) + A cos ϕ sin ( 4 ω 2 − b 2 2 t ) ] = A e − b t 2 sin ( 4 ω 2 − b 2 2 t + ϕ ) … ( 4 ) For the particular solution, x p = A cos t Substitute x p = A cos t in equation ( 1 ) . ( A cos t ) ′ ′ + b ( A cos t ) ′ + ω 2 ( A cos t ) = F cos t − A cos t − b A sin t + ω 2 A cos t = F cos t A ( ω 2 − 1 ) + b 2 [ ( ω 2 − 1 ) ( ω 2 − 1 ) 2 + b 2 cos t − b ( ω 2 − 1 ) 2 + b 2 sin t ] = F cos t Suppose ( ω 2 − 1 ) ( ω 2 − 1 ) 2 + b 2 = sin θ , b ( ω 2 − 1 ) 2 + b 2 = cos θ Therefore, the general solution is given as, x ( t ) = A e − b t 2 sin ( b 2 − 4 ω 2 2 t + ϕ ) + F ( ω 2 − 1 ) 2 + b 2 sin ( t + θ ) tan θ = sin θ cos θ = ( ( ω 2 − 1 ) ( ω 2 − 1 ) 2 + b 2 ) ( b ( ω 2 − 1 ) 2 + b 2 ) = ( 1 9 − 1 ) 0.22 θ = − 1

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 5.8, Problem 5E

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The points of Poincare map, graph them with the given values for the equation, and describe the attractor for the given system.

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Chapter 5.8, Problem 4E

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Chapter 5 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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The points of Poincare map, graph them with the given values for the equation, and describe the attractor for the given system.

Solution Summary: The author describes Poincare's points, graph them, and describe the attractor for the given system. The solution of the second-order homogeneous linear differential equation is given.

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To find: The points of Poincare map, graph them with the given values for the equation, and describe the attractor for the given system.

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Chapter 5 Solutions

To find:

The points of Poincare map, graph them with the given values for the equation, and describe the attractor for the given system.