Explanation The equations (1)-(4) are given as, ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 ... ( 1 ) The boundary conditions for this equation is, u ( 0 , t ) = u ( L , t ) = 0 ... ( 2 ) The initial conditions are, u ( x , 0 ) = f ( x ) ... ( 3 ) and ∂ u ∂ t ( x , 0 ) = g ( x ) ... ( 4 ) It is given that g ( x ) = { v 0 x a , 0 < x ≤ a , v 0 ( L − x ) ( L − a ) , a < x < L , and f ( x ) = 0 . The given equations are wave equation and it is known that the solution of a wave equation is given by, u ( x , t ) = ∑ n = 1 ∞ [ a n cos n π α L t + b n sin n π α L t ] sin n π x L ... ( 5 ) where, a n = 2 L ∫ 0 L f ( x ) sin n π x L d x b n = 2 n π α ∫ 0 L g ( x ) sin n π x L d x Since f ( x ) = 0 , a n = 0 . Now solve for b n , Further solve, b n = 2 n π α [ v 0 L 2 a n 2 π 2 sin n π a L − v 0 L n π cos n π a L + v 0 L − a [ ( L ( L − a ) n π cos n π a L ) + L 2 n 2 π 2 sin n π a L ] ] b n = 2 n π

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 10.6, Problem 6E

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The formal solution to the given initial value problem f(x)=0 and g(x)={v0xa,0<x≤a,v0(L−x)(L−a),a<x<L,.

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Chapter 10.6, Problem 5E

Chapter 10.6, Problem 7E

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(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

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Fundamentals of Differential Equations and Boundary Value Problems

Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 6E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...

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The formal solution to the given initial value problem f ( x ) = 0 and g ( x ) = { v 0 x a , 0 < x ≤ a , v 0 ( L − x ) ( L − a ) , a < x < L , .

Solution Summary: The author explains that the formal solution to the given initial value problem is f(x)=0, and

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The formal solution to the given initial value problem f(x)=0 and g(x)={v0xa,0<x≤a,v0(L−x)(L−a),a<x<L,.

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Chapter 10 Solutions

The formal solution to the given initial value problem f ( x ) = 0 and g ( x ) = { v 0 x a , 0 &lt; x ≤ a , v 0 ( L − x ) ( L − a ) , a &lt; x &lt; L , .

Solution Summary: The author explains that the formal solution to the given initial value problem is f(x)=0, and

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The formal solution to the given initial value problem f(x)=0 and g(x)={v0xa,0<x≤a,v0(L−x)(L−a),a<x<L,.

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Chapter 10 Solutions

The formal solution to the given initial value problem f ( x ) = 0 and g ( x ) = { v 0 x a , 0 < x ≤ a , v 0 ( L − x ) ( L − a ) , a < x < L , .