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math 310 a5 Sol - MATH 310-3 Differential Equations...

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MATH 310-3 Fall 2011 Differential Equations Homework Set 5 Some Solutions The following solutions are not necessarily complete, but should highlight the main points for most questions. If you have any further questions regarding the problems and solutions, feel free to discuss them on WebCT. Solutions marked by * are taken from the publisher’s solutions for the textbook by Boyce and DiPrima, 9th edition (Wiley). Section 2.4: * Section 2.4 # 6 : [Recall that cot t = 1 / tan t = cos t/ sin t , so the discontinuities of cot t occur where sin t = 0, that is, for t = , where k Z (that is, k is an integer). Thus the possible intervals of existence are (0 , 1), (1 , π ), and ( kπ, ( k +1) π ) for k 6 = 0. To find the interval of existence for the given initial value problem, we need to determine which of these intervals contains the initial condition at t = 2; the answer is (1 , π ).] * Section 2.4 # 12 : [That is, the hypotheses of the existence-uniqueness theorem are satisfied everywhere except for t = , k Z and y = - 1.] * Section 2.4 # 15 : [For y 0 6 = 0, the interval of existence is ( - 1 2 y - 2 0 , ).] Section 2.4 # 21 : As derived in class (see also Section 2.4 Example 3), the initial value problem y 0 = y 1 / 3 , y (0) = 0 has infinitely many solutions, namely y ( t ) 0 and, for each t 0 0, y ( t ) = ( 0 , 0 t < t 0 , ± 2 3 ( t - t 0 ) 3 / 2 , t t 0 . 1. To get y (1) = 1, we need a positive solution (use “+”) with 1 = 2 3 (1 - t 0 ) 3 / 2 or 2 3 t 0 = 2 3 - 1, which is not possible for t 0 0. Hence there is no solution passing through the point (1 , 1). 2. To get y (2) = 1, we need 1 = 2 3 (2 - t 0 ) 3 / 2 or 2 3 t 0 = 4 3 - 1 = 1 3 , so t 0 = 1 / 2. 3. Considering all (positive and negative) solutions, we have y (2) = ± 2 3 (2 - t 0 ) 3 / 2 for 0 t 0 2 and y (2) = 0 for t 0 > 2. The largest possible values (in absolute value) occur for t 0 = 0, for which | y (2) | = (4 / 3) 3 / 2 . Hence the set of possible values is y (2) [ - (4 / 3) 3 / 2 , (4 / 3) 3 / 2 ].
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(The instructor’s solutions manual provided by the publisher actually contains a solution for this problem, but it is incorrect. . . ) Section 2.4 # 22 : We consider the initial value problem y 0 = - t + ( t 2 + 4 y ) 1 / 2 2 f ( t, y ) , y (2) = - 1 . (a) We verify solutions by substitution. The function y 2 ( t ) = - t 2 / 4 clearly satisfies the differential equation (since the square root vanishes for y 2 , we immediately have y 0 2 = - t/ 2 = f ( t, y 2 )) and y 2 (2) = - 1, and is thus a solution for all t . For y 1 ( t ) = 1 - t , the IC y 1 (2) = - 1 is also satisfied, but we need to be a bit more careful with the DE: we have t 2 + 4 y 1 1 / 2 = t 2 + 4(1 - t ) 1 / 2 = ( t - 2) 2 1 / 2 = | t - 2 | , so f ( t, y 1 ) = - t + t 2 + 4 y 1 1 / 2 / 2 = ( - t + | t - 2 | ) / 2 = - 1 = y 0 for t 2, while f ( t, y 1 ) = 1 - t for t < 2. Hence the function y 1 ( t ) is a solution of the IVP only for t 2. (b) The existence-uniqueness theorem guarantees a unique solution only whenever both f ( t, y ) and ∂f ( t, y ) /∂y = ( t 2 +4 y ) - 1 / 2 are both continuous (which requires t 2 +4 y > 0). In this case, at the initial point ( t, y ) = (2 , - 1) we have t 2 + 4 y = 0, and ∂f/∂y is not continuous; hence the theorem is not applicable and there is no contradiction.
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