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# As in Exercises 3.6, for some problems you | Ch 3.7 - 20E

ISBN: 9780321747730 43

## Solution for problem 20E Chapter 3.7

Fundamentals of Differential Equations | 8th Edition

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Problem 20E

Problem

As in Exercises 3.6, for some problems you will find it essential to have a calculator or computer available.† For Problems 1–17, note whether or not  is bounded.

Chemical Reactions. The reaction between nitrous oxide and oxygen to form nitrogen dioxide is given by the balanced chemical equation 2NO + O2 = 2NO2. At high temperatures the dependence of the rate of this reaction on the concentrations of NO, O2, and NO2 is complicated. However, at 25ºC the rate at which NO2 is formed obeys the law of mass action and is given by the rate equation

Where x (t) denotes the concentration of NO2 at time t, k is the rate constant, α is the initial concentration of NO, and β is the initial concentration of O2. At 25ºC, the constant k is 7.13 X 103 (liter)2/(mole)2(second). Let α = 0.0010 mole/L, β = 0.0041 mole/L, And x (0) = 0 mole/L. Use the fourth-order Runge– Kutta algorithm to approximate x (10). For a tolerance of E = 0.000001, use a stopping procedure based on the relative error.

Step-by-Step Solution:

Step 1 :

In this problem we have to find the value of  using fourth order R-K method.

Given that

Where ,  and

Hence equation become

At  then find x(10).

Step 2 :

Hence

Step 3 :

Put the values  and

Step 4 of 6

Step 5 of 6

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As in Exercises 3.6, for some problems you | Ch 3.7 - 20E

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