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Unformatted text preview: ACM 95b/100b Solutions for Problem Set 1 Sean Mauch sean@caltech.edu January 18, 2002 1 Problem 1 Problem. Find the general solution and plot some integral curves for each of the following ODEs. What is the behavior of the solution as t ? 1. ty y = t 2 2. y + y cos t = sin t cos t Solution. 1. We consider the differential equation ty y = t 2 . We can solve this differential equation by inspection. We guess a particular solution of the form y p = at 2 . By substituting into the differential equation we see that y p = t 2 . We see that y h = t is a homogeneous solution. Now we have the general solution of the differential equation. y = t 2 + ct We could also solve this problem by finding an integrating factor. y y t = t I = exp Z 1 t d t = e ln  t  = 1  t  We choose I = 1 /t as the integrating factor. d d t y t = 1 y t = t + c y = t 2 + ct The solution, plotted in Figure 1, is unbounded as t . 2. We consider the differential equation y + y cos t = sin t cos t. 121 1 221.510.5 0.5 1 1.5 2 Figure 1: The integral curves of ty y = t 2 . Again we can solve this differential equation by inspection. We see that y p = sin t 1 is a particular solution and y h = e sin t is a homogeneous solution. The general solution of the differential equation is y = sin t 1 + c e sin t . We could also solve this problem by finding an integrating factor. I = exp Z cos t d t = e sin t d d t ( e sin t y ) = e sin t sin t cos t e sin t y = Z e sin t sin t cos t d t + c e sin t y = Z sin t e x x d x + c e sin t y = [( x 1)e x ] sin t + c y = sin t 1 + c e sin t The solution, plotted in Figure 2, is bounded but does not have a limit as t . 2 Problem 2 Problem. Solve the following linear initial value problems and in each case describe the interval on which the solution is defined....
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This note was uploaded on 01/08/2011 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.
 Winter '09
 NilesA.Pierce

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