By S.I. Hayek

ISBN-10: 0824704665

ISBN-13: 9780824704667

Hayek (Pennsylvania nation college) offers tools of utilized arithmetic which are rather suited to the applying of arithmetic to actual difficulties in technological know-how and engineering. The textbook is meant for a three-semester graduate direction series.

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**Extra info for Advanced mathematical methods in science and engineering**

**Sample text**

2 Solution Curves of the Differential Equation y = 2x © 2010 by Taylor and Francis Group, LLC 18 Ordinary Differential Equations Integrating the second-order differential equation y = 2x+1 twice, we find y = x2 + x + c1 and (2) y(x) = 1 3 1 2 x + x + c1 x + c2 . 3 2 Hence, (2) is a two-parameter family of solutions of the DE y = 2x + 1. When specifying the number of parameters in a family of functions, we must be careful to avoid calling every constant a parameter. For example, the family of functions y = c1ex+c2 contains two constants c1 and c2 , but this family is not a two-parameter family, since y = c1 ex+c2 = c1ex ec2 = kex .

If 5% of a radioactive substance decomposes in 50 years, what percentage will be present at the end of 500 years? 1000 years? What is the halflife of the substance? 2. If the half-life of a radioactive substance is 1800 years, what percentage is present at the end of 100 years? In how many years does only 10% of the substance remain? 3. If 100 grams of a radioactive substance is present 1 year after the substance was produced and 75 grams is present 2 years after the substance was produced, how much was produced and what is the half-life of the substance?

The original curve y(x) = x3 −4x+1 passes through the point (2, 1). Requiring y(x) = x3 − 4x + C to satisfy the condition y(2) = 1, we see that C must satisfy the equation y(2) = 1 = 23 − 4(2) + C. Hence, C = 1 and we recover the original function. Proceeding one step further, we differentiate y(x) = x3 − 4x + 1 twice and find y (x) = 6x. Integrating this differential equation twice, we obtain the two-parameter family of solutions y(x) = x3 + Ax + B where A and B are arbitrary constants. To recover the original function, in this case, we must specify two additional conditions which will require us to choose A = −4 and B = 1.

### Advanced mathematical methods in science and engineering by S.I. Hayek

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