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Unformatted text preview: Chapter 6
Equation (6.5) will equal zero and, therefore, the diﬀerential equation will be satisﬁed for r = ±1 and r = 2. Thus, three solutions to the diﬀerential equation are y = x, y = x−1 , and y = x2 . Since these functions are linearly independent, they form a fundamental solution set. (b) Let y (x) = xr . In addition to (6.4), we need the fourth derivative of y (x). y (4) = (y ) = r (r − 1)(r − 2)(r − 3)xr−4 = (r 4 − 6r 3 + 11r 2 − 6r )xr−4 . Thus, if y = xr is a solution to this fourth order CauchyEuler equation, then we must have x4 (r 4 − 6r 3 + 11r 2 − 6r )xr−4 + 6x3 (r 3 − 3r 2 + 2r )xr−3 +2x2 (r 2 − r )xr−2 − 4xrxr−1 + 4xr = 0 ⇒ ⇒ (r 4 − 6r 3 + 11r 2 − 6r )xr + 6(r 3 − 3r 2 + 2r )xr + 2(r 2 − r )xr − 4rxr + 4xr = 0 (r 4 − 5r 2 + 4)xr = 0. (6.6) Therefore, in order for y = xr to be a solution to the equation with x > 0, we must have r 4 − 5r 2 + 4 = 0. Factoring this equation yields r 4 − 5r 2 + 4 = (r 2 − 4)(r 2 − 1) = (r − 2)(r + 2)(r − 1)(r + 1) = 0. Equation (6.6) will be satisﬁed if r = ±1, ±2. Thus, four solutions to the diﬀerential equation are y = x, y = x−1 , y = x2 , and y = x−2 . These functions are linearly independent, and so form a fundamental solution set. (c) Substituting y = xr into this diﬀerential equation yields (r 3 − 3r 2 + 2r )xr − 2(r 2 − r )xr + 13rxr − 13xr = 0 ⇒ (r 3 − 5r 2 + 17r − 13)xr = 0. Thus, in order for y = xr to be a solution to this diﬀerential equation with x > 0, we must have r 3 − 5r 2 + 17r − 13 = 0. By inspection we ﬁnd that r = 1 is a root to this equation. Therefore, we can factor this equation as follows (r − 1)(r 2 − 4r + 13) = 0. 358 ...
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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