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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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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