Chem Differential Eq HW Solutions Fall 2011 167

Chem Differential Eq HW Solutions Fall 2011 167 - Section...

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Unformatted text preview: Section A.1 Linear Ordinary Differential Equations A167 Solutions to Exercises A.1 1. We solve the equation y + y = 1 in two different ways. The first method basically rederives formula (2) instead of just appealing to it. Using an integrating factor. In the notation of Theorem 1, we have p(x) = 1 and q(x) = 1. An antiderivative of p(x) is thus 1 · dx = x. The integrating factor is µ(x) = e p(x) dx = ex . Multiplying both sides of the equation by the integrating factor, we obtain the equivalent equation ex [y + y] = ex ; dx [e y] = ex , dx d where we have used the product rule for differentiation to set dx [exy] = ex [y + y]. Integrating both sides of the equation gets rid of the derivative on the left side, and on the right side we obtain ex dx = ex + C . Thus, ex y = ex + C y = 1 + Ce−x, ⇒ where the last equality follows by multiplying by e−x the previous equality. This gives the solution y = 1 + Ce−x up to one arbitrary constant, as expected from the solution of a first order differential equation. Using formula (2). We have, with p(x) = 1, p(x) dx = x (note how we took the constant of integration equal 0): y = e−x C + 1 · ex dx = e−x [C + ex ] = 1 + Ce−x . 5. According to (2), y = ex C + sin xe−x dx . To evaluate the integral, use integration by parts twice sin xe−x dx = = − sin xe−x + e−x cos x dx − sin xe−x + cos x(−e−x ) − So sin xe−x dx = −e−x sin x + cos x sin xe−x dx = 2 e−x sin x dx; 1 − e−x sin x + cos x . 2 1 y = ex C − e−x sin x + cos x 2 = Cex − 1 sin x + cos x . 2 9. We use an integrating factor e p(x) dx =e tan x dx = e− ln(cos x) = 1 = sec x. cos x Then sec xy − sec x tan x y = d [y sec x] = dx y sec x = y = sec x cos x; 1 x+C = x cos x + C cos x. ...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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