(a) Explanation Property used: (1) If y ″ + λ y = 0 with y ( 0 ) = 0 , y ( L ) = 0 , then for λ > 0 , then write λ = α 2 , where α is a positive number. The solution of the differential equation is y = c 1 cos α x + c 2 sin α x . (2) If y ″ + λ y = 0 with y ( 0 ) = 0 , y ( L ) = 0 , then for λ = 0 , then write y ″ = 0 . The solution of the differential equation is y = c 1 x + c 2 . Formula used: The quadratic formula for the quadratic equation is given by, x = − b ± b 2 − 4 a c 2 a Calculation: Consider the given differential equation d d x [ T ( x ) d y d x ] + ρ ω 2 y = 0 . Equation above can be written in simplified format as follows. d d x [ T ( x ) d y d x ] + ρ ω 2 y = 0 T ( x ) d 2 y d x 2 + T ′ ( x ) d y d x + ρ ω 2 y = 0 T ( x ) y ″ + T ′ ( x ) y ′ + ρ ω 2 y = 0 It is given that T ( x ) = x 2 . The above equation becomes an Cauchy-Euler equation. x 2 y ″ + 2 x y ′ + ρ ω 2 y = 0 Assume y = x m . Differentiate the above assumption twice with respect to x . y ′ = m x m − 1 y ″ = m ( m − 1 ) x m − 2 Now, substitute these values in the given differential equation. x 2 ( m ( m − 1 ) x m − 2 ) + 2 x ( m x m − 1 ) + ρ ω 2 x m = 0 ( m ( m − 1 ) + 2 m + ρ ω 2 ) x m = 0 The auxiliary equation of the differential equation is m 2 + m + ρ ω 2 = 0 . Here, a = 1 , b = 1 , c = ρ ω 2 . Quadratic formula is used here to solve the auxiliary equation. m = − 1 ± 1 − 4 ⋅ 1 ⋅ ρ ω 2 2 = − 1 ± 4 ρ ω 2 − 1 i 2 = − 1 2 − 1 2 4 ρ ω 2 − 1 i or − 1 2 + 1 2 4 ρ ω 2 − 1 i The roots of the auxiliary equation are m = − 1 2 − 1 2 4 ρ ω 2 − 1 i and m = − 1 2 + 1 2 4 ρ ω 2 − 1 i when ρ ω 2 > 0.25 . The general solution of the Cauchy-Euler equation is, y ( x ) = c 1 x α cos ( β ln ( x ) ) + c 2 x α sin ( β ln ( x ) ) (b) To determine To sketch: The graph of the deflection curves on the interval [ 1 , e ] for n = 1 , 2 , 3 with c 2 = 1 .

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Chapter 5.2, Problem 28E

(a)

To determine

To prove: That the critical speeds of angular rotation are ωn=12(4n2π2+1)/ρ and the corresponding deflections are yn(x)=c2x−1/2sin(nπlnx), n=1,2,3,….

(b)

To determine

To sketch: The graph of the deflection curves on the interval [1,e] for n=1,2,3 with c2=1.

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Chapter 5 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 5.1 - 5.1.1 Spring/Mass systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A force...Ch. 5.1 - Prob. 7E Ch. 5.1 - Spring/Mass Systems: Free Undamped Motion A mass...Ch. 5.1 - Prob. 9E Ch. 5.1 - 5.1.1Spring/Mass Systems: Free Undamped Motion A...

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To prove: That the critical speeds of angular rotation are ω n = 1 2 ( 4 n 2 π 2 + 1 ) / ρ and the corresponding deflections are y n ( x ) = c 2 x − 1 / 2 sin ( n π ln x ) , n = 1 , 2 , 3 , … .

Solution Summary: The author demonstrates the critical speeds of angular rotation and the corresponding deflections.

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To prove: That the critical speeds of angular rotation are ωn=12(4n2π2+1)/ρ and the corresponding deflections are yn(x)=c2x−1/2sin(nπlnx), n=1,2,3,….

To sketch: The graph of the deflection curves on the interval [1,e] for n=1,2,3 with c2=1.

Chapter 5 Solutions