Unformatted text preview: Exercises 6.2
Similarly, we ﬁnd other two roots. With the initial guess r0 = −1 − 2i, we ﬁnd that r1 = (−1 − 2i) − . . . r6 ≈ −0.499994 − 1.322875i , r7 ≈ −0.500000 − 1.322876i , r8 ≈ −0.500000 − 1.322876i . Therefore, the other two roots are r ≈ −0.5 − 1.323i and r = −0.5 − 1.323i = −0.5 + 1.323i . (−1 − 2i)4 + 2(−1 − 2i)3 + 4(−1 − 2i)2 + 3(−1 − 2i) + 2 4(−1 − 2i)3 + 6(−1 − 2i)2 + 8(−1 − 2i) + 3 ≈ −0.830703 − 1.652798i , Thus, the auxiliary equation has four complex roots, and a general solution to the given diﬀerential equation is given by y (x) ≈ c1 e−0.5x cos(0.866x) + c2 e−0.5x sin(0.866x) + c3 e−0.5x cos(1.323x) + c4 e−0.5x sin(1.323x) . 31. (a) If we let y (x) = xr , then we see that y = rxr−1 , y = r (r − 1)xr−2 = (r 2 − r )xr−2 , y = r (r − 1)(r − 2)xr−3 = (r 3 − 3r 2 + 2r )xr−3 . Thus, if y = xr is a solution to this third order CauchyEuler equation, then we must have x3 (r 3 − 3r 2 + 2r )xr−3 + x2 (r 2 − r )xr−2 − 2xrxr−1 + 2xr = 0 ⇒ ⇒ (r 3 − 3r 2 + 2r )xr + (r 2 − r )xr − 2rxr + 2xr = 0 (r 3 − 2r 2 − r + 2)xr = 0. (6.5) (6.4) Therefore, in order for y = xr to be a solution to the equation with x > 0, we must have r 3 − 2r 2 − r + 2 = 0. Factoring this equation yields r 3 − 2r 2 − r + 2 = (r 3 − 2r 2) − (r − 2) = (r − 2)(r 2 − 1) = (r − 2)(r + 1)(r − 1) = 0. 357 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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