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Unformatted text preview: MATH 3103 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 publishers 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 = k , where k Z (that is, k is an integer). Thus the possible intervals of existence are (0 , 1), (1 , ), and ( k, ( k +1) ) 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 , ).] * Section 2.4 # 12 : [That is, the hypotheses of the existenceuniqueness theorem are satisfied everywhere except for t = k , k Z and y = 1.] * Section 2.4 # 15 : [For y 6 = 0, the interval of existence is ( 1 2 y 2 , ).] Section 2.4 # 21 : As derived in class (see also Section 2.4 Example 3), the initial value problem y = y 1 / 3 , y (0) = 0 has infinitely many solutions, namely y ( t ) 0 and, for each t 0, y ( t ) = ( , t < t , 2 3 ( t t ) 3 / 2 , t t . 1. To get y (1) = 1, we need a positive solution (use +) with 1 = 2 3 (1 t ) 3 / 2 or 2 3 t = 2 3 1, which is not possible for t 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 ) 3 / 2 or 2 3 t = 4 3 1 = 1 3 , so t = 1 / 2. 3. Considering all (positive and negative) solutions, we have y (2) = 2 3 (2 t ) 3 / 2 for 0 t 2 and y (2) = 0 for t > 2. The largest possible values (in absolute value) occur for t = 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 ]. (The instructors 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 = 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 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 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....
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This note was uploaded on 01/30/2012 for the course MATH 310 303 taught by Professor Rwk during the Spring '11 term at Simon Fraser.
 Spring '11
 RWK
 Math, Differential Equations, Equations

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