Explanation Approach: The green’s function G ( x , s ) of the given form is G ( x , s ) = { − z 1 ( s ) z 2 ( x ) c , 0 ≤ s ≤ x − z 1 ( x ) z 2 ( s ) c , x ≤ s ≤ 2 } ... ( 1 ) Calculation: Consider given value problem ( e 2 x y ' ) ' + e 2 x y = − f ( x ) ; y ( 0 ) = 0 , y ( 1 ) = 0 So, ( e 2 x y ' ) ' + e 2 x y = 0 e 2 x y ' ' + 2 e 2 x y ' + e 2 x y = 0 e 2 x ( y ' ' + 2 y ' + y ) = 0 y ' ' + 2 y ' + y = 0 So auxiliary equation is, r 2 + 2 r + 1 = 0 The roots of auxiliary equation are r = − 1 , 1 The general solution of homogeneous problem y ' ' + 2 y ' + y = 0 is given by y ( x ) = A e − x + B e − x Let z 1 ( x ) = A e − x + B e − x Determine z 1 ( x ) such that z 1 ( x ) = A e − x + B e − x satisfies y ( 0 ) = 0 y ( x ) = A e − x + B e − x y ( 0 ) = A e − 0 + B e − 0 0 = A + B A = − B A = x B = 0 Thus z 1 ( x ) = x e − x Determine z 2 ( x ) such that y ( x ) = A e − x + B e − x satisfies y ( 1 ) = 0 y ( x ) = A e − x + B e − x y ( 1 ) = A ( e ) − 1 + B ( e ) − 1 0 = A ( e ) − 1 + B ( e ) − 1 B = − 1 A = x Thus we get z 2 ( x ) = x e − x − e − x Now using formula w [ z 1 , z 2 ] = z 1 z ' 2 − z 2 z ' 1 We get

Pearson eText Fundamentals of Differential Equations with Boundary Value Problems -- Instant Access (Pearson+)

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Author: R. Nagle, Edward Saff

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Chapter 11.6, Problem 9E

To determine

Green’s function G(x,s) for given boundary value

(e2xy')'+e2xy=−f(x);y(0)=0,y(1)=0.

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Chapter 11.6, Problem 8E

Chapter 11.6, Problem 10E

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Pearson eText Fundamentals of Differential Equations with Boundary Value Problems -- Instant Access (Pearson+)

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Green’s function G ( x , s ) for given boundary value ( e 2 x y ' ) ' + e 2 x y = − f ( x ) ; y ( 0 ) = 0 , y ( 1 ) = 0 .

Solution Summary: The author explains that Green's function G(x,s) for given boundary value is l-z_1(s),0

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Green’s function G(x,s) for given boundary value

(e2xy')'+e2xy=−f(x);y(0)=0,y(1)=0.

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Chapter 11 Solutions

Green’s function G ( x , s ) for given boundary value ( e 2 x y ' ) ' + e 2 x y = − f ( x ) ; y ( 0 ) = 0 , y ( 1 ) = 0 .

Solution Summary: The author explains that Green's function G(x,s) for given boundary value is l-z_1(s),0

Videos

Green’s function G(x,s) for given boundary value (e2xy')'+e2xy=−f(x);y(0)=0,y(1)=0.

Want to see the full answer?

Chapter 11 Solutions

Green’s function G(x,s) for given boundary value

(e2xy')'+e2xy=−f(x);y(0)=0,y(1)=0.