pp2a_5 - Math 293 Practice Prelim 2 Fall 2007 5. Solution:...

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Unformatted text preview: Math 293 Practice Prelim 2 Fall 2007 5. Solution: At first, we need write down the characteristic equation: r 5 - r 3 + 2r 2 + 2r = 0 . (1) This is a polynomial of degree 5, we don't have a formula to get its roots immediately, we need factor it: r 5 - r 3 + 2r 2 + 2r = r(r 4 - r 2 + 2r + 2) = r[r 2 (r 2 - 1) + 2(r + 1)] = r[r 2 (r - 1)(r + 1) + 2(r + 1)] = r(r + 1)[r 2 (r - 1) + 2] . The polynomial r 2 (r - 1) + 2 = r 3 - r 2 + 2 is of degree 3, there is a formula to get its roots, but we don't use it here. Notice that r = -1 is a root of r 3 - r 2 + 2 = 0, so we can divide the polynomial r 3 - r 2 + 2 by r + 1: r 3 - r 2 + 2 = (r + 1)(r 2 - 2r + 2) . Finally, we obtain r 5 - r 3 + 2r 2 + 2r = r(r + 1)2 (r 2 - 2r + 2) = r(r + 1)2 [(r - 1)2 + 1] . The roots of the characteristic equation (1) are: r = 0, -1, -1, 1 i . So the general solution of the differential equation y (5) = y (3) - 2y - 2y is y(x) = c1 + (c2 + c3 x)e-x + (c4 cos x + c5 sin x)ex . 1 ...
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