(a) Explanation Given: Initial value problem is, d y d x = y 2 − 4 y ( 0 ) = 2 , y ( 0 ) = − 2 , y ( 1 4 ) = 1 . Procedure used: Use variable separable method to solve given initial value problem. Calculation: Separate the variables of given initial value problem. d y d x = y 2 − 4 d y y 2 − 4 = d x d y y 2 − 2 2 = d x d y ( y − 2 ) ( y + 2 ) = d x Further, simplify the above equation, [ ( 1 4 ) y − 2 − ( 1 4 ) y + 2 ] d y = d x [ 1 4 ( y − 2 ) − 1 4 ( y + 2 ) ] d y = d x 1 4 [ 1 ( y − 2 ) − 1 ( y + 2 ) ] d y = d x (1) Integrate equation (1), 1 4 ∫ [ 1 ( y − 2 ) − 1 ( y + 2 ) ] d y = ∫ d x 1 4 [ ln | y − 2 | − ln | y + 2 | ] = x + C 1 1 4 ln | y − 2 y + 2 | = x + C 1 ln | y − 2 y + 2 | = 4 x + 4 C 1 (2) Here, constant of integration is C 1 . Substitute C 2 for 4 C 1 in equation (2). ln | y − 2 y + 2 | = 4 x + C 2 | y − 2 y + 2 | = e 4 x + C 2 y − 2 y + 2 = ± e 4 x + C 2 y − 2 y + 2 = ± e C 2 e 4 x (3) Substitute K for ± e C 2 in equation (3) (b) To determine The explicit solution of given initial value problem.

6th Edition

ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 2.2, Problem 35E

(a)

To determine

The explicit solution of given initial value problem.

(b)

To determine

The explicit solution of given initial value problem.

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Chapter 2.2, Problem 34E

Chapter 2.2, Problem 36E

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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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