Elementary Differential Equations and Boundary Value Problems 9th Edition

Published by Wiley
ISBN 10: 0-47038-334-8
ISBN 13: 978-0-47038-334-6

Chapter 1 - Introduction - 1.1 Some Basic Mathematical Models; Direction Fields - Problems - Page 8: 21


(a) $\frac{dy}{dt} = 3 - 3\times10^{-4}y$. (Equivalently, $dq/dt = 300(10^{-2} - q10^{-6})$, where $q$ is in grams and $t$ is in hrs.) (b) $y\rightarrow 10^{4}$. No.

Work Step by Step

Stating the problem: A pond initially contains 1,000,000 gal. of water and an unknown amount of undesirable chemical. Water containing 0.01 g of this chemical per gal. flows into this pond at a rate of 300 gal/hr. The mixture flows out at the same rate, so the amount of water in the pond remains constant. Assume that the chemical is uniformly distributed throughout the pond. (a) Write a differential equation for the amount of chemical in the pond at any time. (b) How much of the chemical will be in the pond after a very long time? Does this limiting amount depend upon the amount that was present initially? Solution: We let $y=y(t)$ denote the amount in grams of chemical in the pond at time $t$. We next isolate the following facts: Rate of input into the pond: 300 gal./hr. Concentration in the pond at $t_0$ is $y_0\cdot 10^{-6}$ g/gal. At a given time, $t_1$, 300 gal/hr with $10^{-2}$g/gal is inputting into the pond, which translates to 3g/hr. # of g in the pond: $y(t)$. Simultaneously, the number of grams of chemical outputting from the pond is $y(t_1)\cdot10^{-6}\cdot 300$g/hr. At the next instant, say $t_2$, we may write an equation expressing the amount of grams of chemical in the pond, letting $\Delta t = t_2 - t_1$, by summing the amount of chemical in the pond at $t_1$ plus the amount imputed over the duration of time,$\Delta t$, minus the amount outputted over the same time duration, $\Delta t$. That is, $y(t_2) = y(t_1) +( 300\cdot 10^{-2} \cdot \Delta t) - (y(t_1)\times 10^{-6}\cdot 300 \cdot \Delta t)$. Letting $\Delta y = y(t_2)-y(t_1)$, we deduce $\Delta y = (3 - 3\times 10^{-4}y)\Delta t$, or, $\frac{\Delta y}{ \Delta t} = 3-3\times 10^{-4}y$. Letting $\Delta t \rightarrow 0$, we find. $\frac{dy}{dt} = 3 - 3\times 10^{-4}y$. (b) To find the amount of unknown chemical in the pond after a very long time, we can use the method of drawing (or imagining) an direction field. We would then find that the value of $y$ (the amount of chemical in the pond in grams) stabilizes, i.e. has $0$ change with time, by setting $0 = 3 - 3\times 10^{-4}y$. I.e., $0= 1 - 10^{-4}y$. This implies $y= 10^{4}$g. This does not vary with different amounts initially in the pond.
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