Explanation Given: The decay constants are k 1 = − 0.138629 and k 2 = − 0.004951 and x 0 is 20. Procedure used: Radioactive decay of first element x: d x d t = − k 1 x Radioactive decay of second element y: d y d t + k 2 y = k 1 x Radioactive decay of third element z: d z d t = k 2 y Calculation: Write the expression for Radioactive decay of first element x . d x d t = − k 1 x 1 x d x = − k 1 d t (1) Integrate both sides of equation (1). ∫ 1 x d x = ∫ − k 1 d t ln ( x ) = − k 1 t + c 1 c 1 = ln ( x ) + k 1 t c 1 = x e k 1 t (2) Substitute boundary conditions t = 0 and x ( 0 ) = x o in equation (2). c 1 = x o e k 1 × 0 c 1 = x o e 0 c 1 = x o Substitute x o for c 1 and − 0.138629 for k 1 in equation (2). x o = x e − 0.138629 t x = x o e 0.138629 t (3) Write the expression for Radioactive decay of second element y . d y d t + k 2 y = k 1 x (4) Substitute x o e − k 1 t for x in equation (4). d y d t + k 2 y = k 1 x o e − k 1 t Write the expression to calculate the integrating factor. I = e ∫ P d t (5) Compare equation (4) with general differential equation to obtain value of P as k 2 and Q as k 1 x o e − k 1 t . Substitute k 2 for P in equation (5). I = e ∫ k 2 d t I = e k 2 ∫ d t I = e k 2 t Substitute e k 2 t for I in general solution equation. e k 2 t y = ∫ e k 2 t λ 1 x o e − k 1 t d t e k 2 t y = k 1 x o ∫ e ( k 2 − k 1 ) t d t e k 2 t y = k 1 x o ( k 2 − k 1 ) e ( k 2 − k 1 ) t + c 2 y = k 1 x o ( k 2 − k 1 ) e ( − k 1 ) t + c 2 e − k 2 t (6) Substitute boundary condition y ( 0 ) = 0 in equation (6). 0 = k 1 x o ( k 2 − k 1 ) e ( − k 1 ) 0 + c 2 e − k 2 0 0 = k 1 x o ( k 2 − k 1 ) + c 2 c 2 = − k 1 x o ( k 2 − k 1 ) Substitute − k 1 x o ( k 2 − k 1 ) for c 2 , − 0.138629 for k 1 and − 0.004951 for k 2 in equation (6). y = − 0.138629 x o ( − 0.004951 + 0.138629 ) e ( 0.138629 ) t − − 0.138629 x o ( λ 2 − 0.138629 ) e − 0.004951 y = − 0.138629 x o ( − 0.004951 + 0.138629 ) ( e ( − 0.138629 ) t − e ( − 0.004951 ) t ) y = − 0.138629 x o 0133678 ( e ( − 0.138629 ) t − e ( − 0.004951 ) t ) (7) Write the expression for Radioactive decay of third element z

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

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Differential Equations

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Chapter 2.9, Problem 3E

To determine

Determine the approximate time when the amounts x(t) and y(t) are same, the approximate time when the amounts x(t) and z(t) are same and the approximate time when the amounts y(t) and z(t) are same and why does the time that is determined when the amounts y(t) and z(t) are same make intuitive sense?

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Chapter 2.9, Problem 2E

Chapter 2.9, Problem 4E

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Chapter 2 Solutions