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We have found the constants for the equation relating X to t. Now we can solve for X. For convenience, continue to write k for the constant in the exponent. 150 – X 5 e180kt 4 %3D 120 – 2X 150 – X = (120 – 2X) e180kt 150 – X = 150e180kt_ Xe180kt 5X e180kt – X = 150e180kt – 150 :)- 5180kt - e180kt - 1) (e180kt – 1) X = 180kt - 1

We have found the constants for the equation relating X to t. Now we can solve for X. For convenience, continue to write k for the constant in the exponent. 150 – X 5 e180kt 4 %3D 120 – 2X 150 – X = (120 – 2X) e180kt 150 – X = 150e180kt_ Xe180kt 5X e180kt – X = 150e180kt – 150 :)- 5180kt - e180kt - 1) (e180kt – 1) X = 180kt - 1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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We have found the constants for the equation relating X to t. Now we can solve for X. For convenience, continue to write k for the constant in the exponent. 150 – X 5 e180kt 4 120 – 2X () 150 – X = (120 – 2X) 5 180kt Xe180kt 2 150 – X = 150e180kt 5X e180kt – X = 150e180kt - 150 180kt - 1 180kt (e 180kt 1) X = 180kt - 1
Transcribed Image Text:We have found the constants for the equation relating X to t. Now we can solve for X. For convenience, continue to write k for the constant in the exponent. 150 – X 5 e180kt 4 120 – 2X () 150 – X = (120 – 2X) 5 180kt Xe180kt 2 150 – X = 150e180kt 5X e180kt – X = 150e180kt - 150 180kt - 1 180kt (e 180kt 1) X = 180kt - 1
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