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Unformatted text preview: Exercises 6.1
(b) To ﬁnd the solution that satisﬁes the initial conditions, we must diﬀerentiate the general solution y (x) twice with respect to x. Thus, we have y (x) = C1 ex − C2 e−x cos 2x − 2C2 e−x sin 2x − C3 e−x sin 2x + 2C3 e−x cos 2x + 2x = C1 ex + (−C2 + 2C3 ) e−x cos 2x + (−2C2 − C3 ) e−x sin 2x + 2x , y (x) = C1 ex + (C2 − 2C3 ) e−x cos 2x − 2 (−C2 + 2C3 ) e−x sin 2x − (−2C2 − C3 ) e−x sin 2x + 2 (−2C2 − C3 ) e−x cos 2x + 2 = C1 ex + (−3C2 − 4C3 ) e−x cos 2x + (4C2 − 3C3 ) e−x sin 2x + 2 . Plugging the initial conditions into these formulas, yields the equations y (0) = C1 + C2 = −1, y (0) = C1 − C2 + 2C3 = 1, y (0) = C1 − 3C2 − 4C3 + 2 = −3. By solving these equations simultaneously, we obtain C1 = −1, C2 = 0, and C3 = 1. Therefore, the solution to the initial value problem is given by y (x) = −ex + e−x sin 2x + x2 . 21. In the standard form, given equation becomes y+ 1 1 3 − ln x y − 3y= . 2 x x x3 Since its coeﬃcients are continuous on (0, ∞), we can apply Theorems 2 and 4 to conclude that a general solution to the corresponding homogeneous equation is yh (x) = C1 x + C2 x ln x + C3 x(ln x)2 and a general solution to the given nonhomogeneous equation is y (x) = yp (x) + yh (x) = ln x + C1 x + C2 x ln x + C3 x(ln x)2 . 345 ...
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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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