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Unformatted text preview: Chapter 2 32. In the given equation, P ( x ) = 2, which implies that μ ( x ) = e 2 x . Following guidelines, first we solve the equation on [0 , 3]. On this interval, Q ( x ) ≡ 2. Therefore, y 1 ( x ) = e 2 x Z (2) e 2 x dx = 1 + C 1 e 2 x . Since y 1 (0) = 0, we get 1 + C 1 e = 0 ⇒ C 1 = 1 ⇒ y 1 ( x ) = 1 e 2 x . For x > 3, Q ( x ) = 2 and so y 2 ( x ) = e 2 x Z ( 2) e 2 x dx = 1 + C 2 e 2 x . We now choose C 2 so that y 2 (3) = y 1 (3) = 1 e 6 ⇒ 1 + C 2 e 6 = 1 e 6 ⇒ C 2 = 2 e 6 1 . Therefore, y 2 ( x ) = 1 + (2 e 6 1) e 2 x , and the continuous solution to the given initial value problem on [0 , ∞ ) is y ( x ) = 1 e 2 x , ≤ x ≤ 3 , 1 + (2 e 6 1) e 2 x , x > 3 . The graph of this function is shown in Fig. 2–C , page 72. 34. (a) Since P ( x ) is continuous on ( a,b ), its antiderivatives given by R P ( x ) dx are con tinuously differentiable, and therefore continuous, functions on ( a,b ). Since the function e x is continuous on (∞ , ∞ ), composite functions μ ( x ) = e R P ( x ) dx are continuous on ( a,b ). The range of the exponential function is (0 , ∞ ). This implies that μ ( x ) is positive with any choice of the integration constant. Using the chain rule, we conclude that dμ ( x ) dx = e R P ( x ) dx d dx Z P ( x ) dx = μ ( x ) P ( x ) for any x on ( a,b ). (b) Differentiating (8), we apply the product rule and obtain dy dx = μ 2 μ Z μQdx + C + μ 1 μQ = μ 1 P Z μQdx + C + Q, and so dy dx + Py = μ 1 P Z μQdx + C +...
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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