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hw1Sols - HOMEWORK 1 MATH 265 L Keshet Due in class on...

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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 ( π/ 2) = 0. (d) t 2 y + 2 ty = t 3 + 1 with y (1) = 1. Solution to Problem 1: (a) The integrating factor is μ ( 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 μ ( 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 μ ( 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
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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). For part (b) the equation can be written as y = y - ty = y (1 - t y ). First we see that y always has to be positive for real values, so only the first two quadrants are relevant. If t < 0 then (1 - t y ) is positive. This means arrows have positive slope in the 2nd quadrant. On the curve (1 - t
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