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Unformatted text preview: M351 Sample Hour ExaminationANSWERS 1. (a) For dx dt 1 t x = t 2 , t &gt; , the associated homogeneous equation is dx dt 1 t x = 0 . (b) The general solution of the homogeneous equation is x ( t ) = C exp Z t 1 s ds = C e ln ( t ) = C t . (c) Since d dt t 3 2 = 3 2 t 2 , substitution in the differential equation yields 3 2 t 2 1 t t 3 2 = 3 2 1 2 t 2 = t 2 . Hence the general solution is given by x ( t ) = C t + t 3 2 . 2. The separable equation dy dx = y 2 y 2 , has constant solution obtained by setting the right hand side equal to zero and solving the quadratic equation y 2 y 2 = 0 which has roots y = 1 and y = 2. These give the constant solutions. Now separate variables to get 1 ( y + 1)( y 2) dy dx = 1 , or, using a partial fraction decomposition 1 3 1 y 2 1 y + 1 dy dx = 1 . Integrating, 1 3 ln y 2 y +1 = t + C or y 2 y +1 1 / 3 = K e t . 3. Consider the initial value problem for the linear differential equation dy dx + p ( x ) y = q ( x ) , y ( x o ) = y o , 1 (a) d...
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This note was uploaded on 12/02/2009 for the course MATH 352 taught by Professor Staff during the Spring '08 term at University of Delaware.
 Spring '08
 Staff

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