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Unformatted text preview: HOMEWORK 1: MATH 265, L Keshet Due in class on September 22, 2010 Problem 1: In each case, solve for y ( t ): (a) y + 3 y = 2 e t/ 2 with y (0) = 1. (b) y - 4 y = t with y (1) = 0. (c) ty + 2 y = sin t with y ( / 2) = 0. (d) t 2 y + 2 ty = t 3 + 1 with y (1) = 1. Solution to Problem 1: (a) The integrating factor is ( t ) = e 3 t . The solution is y ( t ) = 4 7 e 7 t/ 2 + Ce 3 t . Using the initial condition, the constant is found and the solution is y ( t ) = 4 7 e 7 t/ 2 + 3 7 e 3 t . (b) The integrating factor is ( t ) = e 4 t . Integration by parts is needed to find that the solution is y ( t ) =- 1 16- 1 4 + Ce 4 t Using the initial condition, we find the constant. The result is y ( t ) =- 1 16- 1 4 + 5 16 e 4 e 4 t (c) The integrating factor is ( t ) = t 2 ( typo corrected ). Integration by parts is needed to find that the solution is y ( t ) = sin( t ) t 2- cos( t ) t + C t 2 The constant is found to be C =- 1 ( typo corrected ). (d) The LHS is actually simply d ( t 2 y ( t )) /dt so we do not need an integrating factor. We can simply integrate both sides. We get y ( t ) = 1 t + t 2 4 + C 1 t 2 . Using the initial condition, we find that the constant is C =- 1 / 4. Problem 2: Draw the direction fields for each of the below and sketch the solution curve corresponding to the indicated initial condition. (a) y = y- 2 cos( t ) + 1, y (0) = 0 (b) y = y- ty , y (0) = 2 Solution to Problem 2: For part (a) we note that the directions of tangent lines to solution curves will all be flat (have slope zero) whenever 0 = y- 2 cos( t ) + 1. We can sketch this set of points. It may be easiest to first rewrite it as y = 2 cos( t )- 1, as we see that it is a cosine of amplitude 2 shifted down the y axis by 1 (dotted line in Fig 1(a)). All along this locus, we sketch horizontal dashes. This locus separates portion of the yt plane for which solutions are increasing (i.e. where y ( t ) > 0) from the portion 1 where solutions are decreasing ( y ( t ) < 0). We can go further by noting that the slope of tangent lines will have a value M all along the locus M = y- 2 cos( t ) + 1. We can similarly sketch such sets of points ( y = 2 cos( t ) + M- 1) and assign to them directions of slope M for various values of M to get a fairly good picture. The (computer generated) direction field, together with one solution curve satisfying the initial condition is shown in Fig 1(a)....
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