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Unformatted text preview: Chapter 4
(b) The auxiliary equation to the diﬀerential equation obtained in (a) is 3r 2 + 8r − 3 = 0, which has the roots r= −8 ± 64 − 4(3)(−3) −8 ± 10 = 6 6 ⇒ r = −3, 1 . 3 This yields a general solution y (t) = c1 et/3 + c2 e−3t . (c) Since x = et , we can express y (t) as a function of x by writing y = c1 et/3 + c2 e−3t = c1 et
1/3 + c2 et −3 = c1 x1/3 + c2 x−3 . 41. This equation is a CauchyEuler equation. The substitution x = et leads to the equation (22) with a = 1, b = 2, and c = −6. Thus we have a dy d2 y + (b − a) + cy = 0 2 dt dt ⇒ d2 y dy + − 6y = 0. dt2 dt The auxiliary equation, r 2 + r − 6 = 0, has the roots r = −3 and r = 2. Therefore, a general solution can be written as y = c1 e−3t + c2 e2t = c1 et 43. The substitution x = et yields the equation dy d2 y + (9 − 1) + 17y = 0 dt2 dt ⇒ dy d2 y + 8 + 17y = 0. dt2 dt
−3 + c2 et 2 = c1 x−3 + c2 x2 . Solving the characteristic equation, r 2 + 8r + 17 = 0, we get √ −8 ± 64 − 68 = −4 ± i. r= 2 Thus, the roots are complex with α = −4, β = 1, and a general solution, as a function of t, is given by y (t) = c1 e−4t cos t + c2 e−4t sin t. Now we make the back substitution. Since x = et , we have t = ln x and so y = et 188
−4 (c1 cos t + c2 sin t) = x−4 [c1 cos(ln x) + c2 sin(ln x)] . ...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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