Explanation Given: The nonhomogeneous boundary value problem is given as: A 2 ( x ) y ″ ( x ) + A 1 y ′ ( x ) + A 0 y ( x ) = − f ( x ) , a < x < b ... ( 1 ) a 1 y ( a ) + a 2 y ′ ( a ) = 0 b 1 y ( b ) + b 2 y ′ ( b ) = 0 Approach: To obtain the required solution to L [ y ] = − f ; B 1 [ y ] = 0 , B 2 ( y ) = 0 , where L [ y ] = ( p y ′ ) ′ + q y and f is continuous on [ a , b ] , steps are given as: Find two independent solutions z 1 and z 2 to L [ y ] = 0 that satisfy B 1 [ z 1 ] = 0 B 2 [ z 2 ] = 0 . Compute C = p ( x ) W [ z 1 , z 2 ] ( x ) , where W [ z 1 , z 2 ] = z 1 z ′ 2 − z ′ 1 z 2 . Construct the Green’s function: K ( x , s ) = { − z 1 ( s ) z 2 ( x ) p ( x ) W [ z 1 , z 2 ] , a ≤ s ≤ x − z 1 ( x ) z 2 ( s ) p ( x ) W [ z 1 , z 2 ] , x ≤ s ≤ b Calculation: Assume that a particular solution to the equation ( 1 ) has the form, y ( x ) = c 1 ( x ) z 1 ( x ) + c 2 ( x ) z 2 ( x ) Differentiate the above equation with respect to x , this gives, y ′ ( x ) = c ′ 1 ( x ) z 1 ( x ) + c 1 ( x ) z ′ 1 ( x ) + c ′ 2 ( x ) z 2 ( x ) + c 2 ( x ) z ′ 2 ( x ) = [ c ′ 1 ( x ) z 1 ( x ) + c ′ 2 ( x ) z 2 ( x ) ] + [ c 1 ( x ) z ′ 1 ( x ) + c 2 ( x ) z ′ 2 ( x ) ] ... ( 2 ) Suppose c ′ 1 ( x ) z 1 ( x ) + c ′ 2 ( x ) z 2 ( x ) = 0 . Thus, the equation ( 2 ) becomes, y ′ ( x ) = c 1 ( x ) z ′ 1 ( x ) + c 2 ( x ) z ′ 2 ( x ) ... ( 3 ) Differentiate the above equation with respect to x , this gives, y ″ ( x ) = c ′ 1 ( x ) z ′ 1 ( x ) + c 1 ( x ) z ″ 1 ( x ) + c ′ 2 ( x ) z ′ 2 ( x ) + c 2 ( x ) z ″ 2 ( x ) Substitute { c 1 ( x ) z 1 ( x ) + c 2 ( x ) z 2 ( x ) } for y ( x ) , { c 1 ( x ) z ′ 1 ( x ) + c 2 ( x ) z ′ 2 ( x ) } for y ′ ( x ) and { c ′ 1 ( x ) z ′ 1 ( x ) + c 1 ( x ) z ″ 1 ( x ) + c ′ 2 ( x ) z ′ 2 ( x ) + c 2 ( x ) z ″ 2 ( x ) } for y ″ ( x ) in the equation ( 1 ) . − f ( x ) = A 2 ( x ) [ c ′ 1 ( x ) z ′ 1 ( x ) + c 1 ( x ) z ″ 1 ( x ) + c ′ 2 ( x ) z ′ 2 ( x ) + c 2 ( x ) z ″ 2 ( x ) ] + A 1 ( x ) [ c 1 ( x ) z ′ 1 ( x ) + c 2 ( x ) z ′ 2 ( x ) ] + A 0 ( x ) [ c 1 ( x ) z 1 ( x ) + c 2 ( x ) z 2 ( x ) ] Rearrange the above equation as,

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Chapter 11.6, Problem 22E

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y(x)=∫abK(x,s)ds.

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Fundamentals of Differential Equations and Boundary Value Problems

Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - Prob. 3E Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - Prob. 7E Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...Ch. 11.2 - Prob. 9E Ch. 11.2 - In Problems 1-12, determine the solutions, if any,...

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That the solution to the given problem is given by the formula y ( x ) = ∫ a b K ( x , s ) d s .

Solution Summary: The author explains that the solution to the nonhomogeneous boundary value problem is given by the formula y(x)=displaystyle int

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That the solution to the given problem is given by the formula

y(x)=∫abK(x,s)ds.

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That the solution to the given problem is given by the formula y ( x ) = ∫ a b K ( x , s ) d s .

Solution Summary: The author explains that the solution to the nonhomogeneous boundary value problem is given by the formula y(x)=displaystyle int

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To show: That the solution to the given problem is given by the formula y(x)=∫abK(x,s)ds.

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Chapter 11 Solutions

To show:

That the solution to the given problem is given by the formula

y(x)=∫abK(x,s)ds.