Explanation Given: The values t = 2 π n , n = 0 , 1 , ... , 20 for equation ( 1 ) , taking A = F = 1 , ϕ = 0 , and ω = 1 / 3 and equation is 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 ) + ω 2 x ( t ) = F cos t … ( 1 ) The homogeneous equation is given as, x ″ ( t ) + ω 2 x ( t ) = 0 … ( 2 ) Substitute e r t for x in equation ( 2 ) . r 2 e r t + ω 2 e r t = 0 e r t ( r 2 + ω 2 ) = 0 The auxiliary equation is given as, r 2 + ω 2 = 0 Thus, the roots of the auxiliary equation are r = ± i ω . The solution of the equation ( 2 ) is given as, x h ( t ) = C 1 cos ( ω t ) + C 2 sin ( ω t ) … ( 3 ) Suppose C 1 = A sin ϕ , C 2 = A cos ϕ The equation ( 3 ) becomes, x h ( t ) = A sin ϕ cos ω t + A cos ϕ sin ω t = A sin ( ω t + ϕ ) For the particular solution, x p = A cos t Substitute x p = A cos t in equation ( 1 ) . ( A cos t ) ′ ′ + ω 2 ( A cos t ) = F cos t − A cos t + ω 2 A cos t = F cos t ( ω 2 − 1 ) A cos t = F cos t A = F ( ω 2 − 1 ) Therefore, the general solution is given as, x ( t ) = A sin ( ω t + ϕ ) + F ω 2 − 1 … ( 4 ) Here, the amplitude A and the phase angle ϕ are the arbitrary constants. Differentiate the equation ( 4 ) with respect to t . v ( t ) = [ x ( t ) ] ′ = ( A sin ( ω t + ϕ ) + F ω 2 − 1 ) ′ = A ω cos ( ω t + ϕ ) − F ω 2 − 1 sin t Substitute t = 2 π n into x ( t ) and v ( t ) . x n = x ( 2 π n ) = A sin ( 2 π n ω + ϕ ) + F ω 2 − 1 cos ( 2 π n ) = A sin ( 2 π n ω + ϕ ) + F ω 2 − 1

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Author: Nagle

Publisher: PEARSON CO

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Chapter 5.8, Problem 2E

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

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

Solution Summary: The author explains the Poincare map and the solution of the second-order homogeneous linear differential equation.

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