57_pdfsam_math 54 differential equation solutions odd

57_pdfsam_math 54 differential equation solutions odd - x...

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Exercises 2.3 (d) We want the value of the initial point for the solution in part (c) to be the value of the solution found in part (b) at the point x = 2. This value is given by y (2) = 2 1+2 e 2 =1+2 e 2 . Thus the initial point we seek is y (2) = 1 + 2 e 2 . Using this initial point to Fnd the constant C g iveninpart(c)y ie lds 1+2 e 2 = y (2) = 2 3 1 9 + Ce 6 C = 4 9 e 6 +2 e 4 . Thus, the solution of the equation on the interval x> 2isg ivenby y = x 3 1 9 + ± 4 9 e 6 +2 e 4 ² e 3
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Unformatted text preview: x ≥ 0: y = x − 1 + 2 e − x , ≤ x ≤ 2; x 3 − 1 9 + ³ 4 9 e 6 + 2 e 4 ´ e − 3 x , 2 < x. (e) The graph of the solution is given in ±igure B.18 of the answers in the text. 33. (a) Writing the equation in standard form yields dy dx + 2 x y = 3 . Therefore, P ( x ) = 2 /x and µ ( x ) = exp ³Z 2 x dx ´ = exp (2 ln | x | ) = | x | 2 = x 2 . Hence d dx ( x 2 y ) = 3 x 2 ⇒ x 2 y = Z 3 x 2 dx = x 3 + C ⇒ y = x + C x 2 53...
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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.

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