nagle_differential_equations_ISM_Part11

nagle_differential_equations_ISM_Part11 - Chapter 2 32 In...

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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 (2) e 2 x dx = 1 + C 1 e - 2 x . Since y 1 (0) = 0, we get 1 + C 1 e 0 = 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 ( - 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 , 0 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 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 ( x ) dx = e R P ( x ) dx d dx 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 μQ dx + C + μ - 1 μQ = - μ - 1 P μQ dx + C + Q , and so dy dx + Py = - μ - 1 P μQ dx + C + Q + P μ - 1 μQ dx + C = Q .

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