Chem Differential Eq HW Solutions Fall 2011 180

Chem Differential Eq HW Solutions Fall 2011 180 - A180...

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Unformatted text preview: A180 Appendix A Ordinary Differential Equations: Review of Concepts and Methods and so π π π 1 )] + sin[2(x − )] + x sin 2x. 2 8 2 4 using the addition formulas for the cosine and sine, we can write y = cos[2(x − cos[2(x − π )] = − cos 2x and 2 sin[2(x − π )] = − sin 2x, 2 and so y = − cos 2x − 1 π1 π sin 2x + x sin 2x = − cos 2x + − + x sin 2x. 8 4 84 73. An antiderivative of g(x) = ea x cos bx is a solution of the differential equation y = ea x cos bx. We assume throughout this exercise that a = 0 and b = 0. For these special cases the integral is clear. To solve the differential equation we used the method of undetermined coefficients. The solution of homogeneous equation y = 0 is y = C . To find a particular solution of y = eax cos bx we try y = eax (A cos bx + B sin bx) y = aeax (A cos bx + B sin bx) + eax (−bA sin bx + bB cos bx) = eax (Aa + bB ) cos bx + eax (aB − bA) sin bx. Plugging into the equation, we find eax (Aa + bB ) cos bx + eax (aB − bA) sin bx = eax cos bx Aa + bB = 1, aB − bA = 0 ⇒ A = so eax cos bx dx = a b , B= 2 ; a2 + b2 a + b2 eax a cos x + b sin bx + C. a2 + b2 ...
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