Consider the ODE - y" − k²y = f(x) 0 ≤ x ≤1 k > 0 with boundary conditions (a) y(0) = 0 y(1) = 0 Compute the Green's function G(x, x') for this ODE. That is, solve the boundary value problem (b) d²G − k²G(x, x') = d(x − x') dx² with boundary conditions G(x = 0, x') = 0 G(x = 1, x') = 0 and where 6(x) is the delta function. Use the Green's function developed in part (a) to solve the inhomo- geneous boundary value problem y" - k²y = 1 0 ≤ x ≤ 1 k > 0 with boundary conditions y(0) = 0 y(1) = 0 You may leave your answer in terms of integrals.
Consider the ODE - y" − k²y = f(x) 0 ≤ x ≤1 k > 0 with boundary conditions (a) y(0) = 0 y(1) = 0 Compute the Green's function G(x, x') for this ODE. That is, solve the boundary value problem (b) d²G − k²G(x, x') = d(x − x') dx² with boundary conditions G(x = 0, x') = 0 G(x = 1, x') = 0 and where 6(x) is the delta function. Use the Green's function developed in part (a) to solve the inhomo- geneous boundary value problem y" - k²y = 1 0 ≤ x ≤ 1 k > 0 with boundary conditions y(0) = 0 y(1) = 0 You may leave your answer in terms of integrals.
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter14: Discrete Dynamical Systems
Section14.3: Determining Stability
Problem 13E: Repeat the instruction of Exercise 11 for the function. f(x)=x3+x For part d, use i. a1=0.1 ii...
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Help me with a and b please. please handwrite detailed solution; u can type if u can make the formula format clear. . do not use chatgpt or other ai tools. those AI tools give wrong answer; please DO NOTt waste my chances to ask question here. thank you
if you don't have time or don't know how to do it, PLEASE DO NOT paste a random answer; last time the expert gave an answer for a statistics quesiton, which is horrible and bleh; i am asking the same question as attached here, but that expert answered w a random statistics questino; please, if you don't know how to do it, just don't answer.
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