20d_07s_fs

20d_07s_fs - Math 20D Final Exam Solutions 2 2 2 June 12,...

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Math 20D Final Exam Solutions June 12, 2007 1. (a) An integrating factor is e t/ 2 / 2 and so e t 2 / 2 y = i t 3 e t 2 / 2 dt . The integral can be done using integration by parts with u = t or it can be done by substitution with x = t 2 / 2 and then integration by parts. The result is ( t 2 - 2) e t 2 / 2 + C and so y = t 2 - 2 + Ce t 2 / 2 . Using the initial condition, 0 = - 2 + C and so C = 2. (b) Separate variables: e y dy = - t 2 dt . Thus - e y = t 1 + C . (c) Set y = e rt to obtain r 2 - 2 r +2 = 0 and so r = 1 ± - 1. Hence the general solution is y = C 1 e (1+ i ) t + C 2 e (1 i ) t , which can be written y = e t ( D 1 sin t + D 2 cos t ). (d) The homogeneous equation y ′′ - 4 y + y = 0 gives r 2 - 4 r + 4 = 0 and so r = - 2 is a double root. Thus the general solution to the homogeneous equation is y = ( C 1 + C 2 t ) e 2 t . One can use variation of parameters or undetermined coe±cients to ²nd a particular solution. Using the latter, we set y = Ce t and
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This note was uploaded on 06/18/2009 for the course MATH 20D taught by Professor Mohanty during the Fall '06 term at UCSD.

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20d_07s_fs - Math 20D Final Exam Solutions 2 2 2 June 12,...

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