(a) Explanation Given: The Mathematical model is, x v d v d x + v 2 = 32 x . Here, velocity of chain is v and the length of chain at an instant is x . Initial length of the chain is 3 ft and the initial velocity is 0 (rest). Procedure used: Compare the given differential equation with the standard form of exact differential equation M d x + N d v = 0 . To check whether the given differential equation is exact or not, check the condition ∂ M ∂ v = ∂ N ∂ x is satisfied or not. If the given differential equation is not exact, then make it exact by using an appropriate integrating factor. Calculate ∂ M ∂ v − ∂ N ∂ x N , if this term is a function of x only or a constant then integrating factor is e ∫ f ( x ) d x . Standard solution of the exact differential equation: ∫ M d x + ∫ ( t e r m s o f N f r e e f r o m x ) d v = C Calculation: Simplify the given mathematical model as, x v d v d x + v 2 = 32 x x v d v d x = 32 x − v 2 x v d v = ( 32 x − v 2 ) d x ( v 2 − 32 x ) d x + x v d v = 0 (1) Compare equation (1) with the standard form of exact differential equation. Thus, M = v 2 − 32 and N = x v . Calculate ∂ M ∂ v as follows. ∂ M ∂ v = ∂ ∂ v ( v 2 − 32 x ) = 2 v Calculate ∂ N ∂ x as follows. ∂ N ∂ x = ∂ ∂ x ( x v ) = v Since ∂ M ∂ y ≠ ∂ N ∂ x , the given differential equation is not exact. Calculate ∂ M ∂ y − ∂ N ∂ x N , by substituting 2 v for ∂ M ∂ v , v for ∂ N ∂ x and x v for N . ∂ M ∂ v − ∂ N ∂ x N = 2 v − v x v = v x v = 1 x Since the value of ∂ M ∂ v − ∂ N ∂ x N is the function of x only, integrating factor for the given differential equation is, I F = e ∫ 1 x d x = e ln x = x Multiply equation (1) by the integrating factor (b) To determine The velocity by which the chain leaves the platform.

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.4, Problem 45E

(a)

To determine

An explicit solution for the given mathematical model.

(b)

To determine

The velocity by which the chain leaves the platform.

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Chapter 2.4, Problem 44E

Chapter 2.5, Problem 1E

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4. Consider Chebychev's equation (1 - x²)y" - xy + λy = 0 with boundary conditions y(-1) = 0 and y(1) = 0, where X is a constant. (a) Show that Chebychev's equation can be expressed in Sturm-Liouville form d · (py') + qy + Ary = 0, dx y(1) = 0, y(-1) = 0, where p(x) = (1 = x²) 1/2, q(x) = 0 and r(x) = (1 − x²)-1/2 (b) Show that the eigenfunctions of the Sturm-Liouville equation are extremals of the functional A[y], where A[y] = I[y] J[y]' and I[y] and [y] are defined by - I [y] = √, (my² — qy²) dx and J[y] = [[", ry² dx. Explain briefly how to use this to obtain estimates of the smallest eigenvalue >1. 1 (c) Let k > be a parameter. Explain why the functions y(x) = (1-x²) are suitable 4 trial functions for estimating the smallest eigenvalue. Show that the value of A[y] for these trial functions is 4k2 A[y] = = 4k - 1' and use this to estimate the smallest eigenvalue \1. Hint: L₁ x²(1 − ²)³¹ dr = 1 (1 - x²)³ dx (ẞ > 0). 2ẞ

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An explicit solution for the given mathematical model.

Solution Summary: The author explains how to simplify the given mathematical model by comparing it with the standard form of exact differential equation Mdx+Ndv.

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An explicit solution for the given mathematical model.

The velocity by which the chain leaves the platform.

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Chapter 2 Solutions