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P2_130_1

# P2_130_1 - Potential problems for midterm I 3.1.18 Find a...

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Potential problems for midterm I 3.1.18 Find a differential equation which has general solution y ( t ) = C 1 e - t/ 2 + C 2 e - 2 t . A second-order, linear, homogeneous differential equation with constant coefficients will have the above as a general solution if the roots of the characteristic equation are - 1 / 2 and - 2. So the characteristic equation is ( r + 1 / 2)( r + 2) = r 2 + (5 / 2) r + 1 = 0. So the differential equation 2 y 00 + 5 y 0 + 2 = 0 works. 3.2.11 Without solving the equation, determine the longest interval in which the solution to the initial value problem ( x - 3) y 00 + xy 0 + (ln | x | ) y = 0 , y (1) = 0 , y 0 (1) = 1 is guaranteed to exist. In order to use theorem 3.2.1, we need to divide the equation by ( x - 3), giving us y 00 + x x - 3 y 0 + ln | x | x - 3 y = 0 . The function x x - 3 is discontinuous at three and the function ln | x | x - 3 is discontinuous at zero and three. So the possible intervals where the solution is defined are ( -∞ , 0), (0 , 3), and (3 , ). The initial point t 0 = 1 is in the second of these, so there is a unique twice differentiable solution on the interval (0 , 3). 3.2.17 If the Wronskian of f and g is 3 e 4 t and f ( t ) = e 2 t , what is g ( t )? The Wronskian of f and g is fg 0 - f 0 g , so we get the differential equation (in g ), e 2 t g 0 - 2 e 2 tg = 3 e 4 t or g 0 - 2 g = 3 e 2 t . We solve this using first-order techniques. The natural integrating factor is μ ( t ) = e - 2 t so , multiplying this through and recognizing the product rule on the left, we get ( e - 2 t g ) 0 = 3. So g ( t ) = 3 te 2 t + Ce 2 t for any constant C will work.

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