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)ex + (c4 cos x + c5 sin x)ex . 1 ...
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This note was uploaded on 03/27/2008 for the course MATH 2930 taught by Professor Terrell,r during the Fall '07 term at Cornell.
 Fall '07
 TERRELL,R
 Math, Differential Equations, Equations

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