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MTH 2613 Elementary Differential Equations Score _____________ Examination 2 Student Name: _______________________________________________________________________________ This exam is a closed book, open notes exam. You are allowed to use a simple, non-integrating calculator. All work must be shown to receive full credit for the problem. Solutions are to be worked in the space provided. 1. Determine the general solution for the given homogeneous differential equations. 20
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Differential Calculus

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MTH 2613 Elementary Differential Equations Score _____________Examination 2Student Name: _______________________________________________________________________________ This exam is a closed book, open notes exam. You are allowed to use a simple, non-integrating calculator. Allwork must be shown to receive full credit for the problem. Solutions are to be worked in the space provided. 1. Determine the general solution for the given homogeneous differential equations. 20 points a.   6 11 3 0  y y y          2 6 11 3 03 1 2 3 01 3,3 2 m mm mm          13321 2  xx  y x c e c e    b.   4 8 0  y y y       2 4 8 04 16 3224 42 22 m mmii              2 21 2 cos2 sin2  x x  y x c e x c e x      c.   10 25 0  y y y             3 222 10 25 010 25 05 00, 5 mult.2 m m mm m mm mm            5 51 2 3  x x  y x c c e c xe         2. Determine the general solution to the differential equation given by tan  y y x   . 25 points Begin by finding the solution to the homogeneous equation: 21,2 01 0  y ymm i       1 2 cos sin c  y c x c x     Applying the Variation of Parameters method and replacing the arbitrary constants with unknown functions,we see 1 2 cos sin  p  y v x v x   . Differentiating the particular solution and assuming the “ assumption ” step,we see  1 2 1 21 2 1 2 cos sin sin cossin cos cos sin  p p  y v x v x v x v x y v x v x v x v x               Substituting into the given differential equation, we see 1 2 1 2 1 21 2 tansin cos cos sin cos sin tansin cos tan  y y xv x v x v x v x v x v x xv x v x x              Using Cramer ’ s Rule to determine the unknown functions, we see   122 0 sintan coscos sinsin cossincos1cos 1coscos secsin ln sec tan  x x xv dx x x x x x xdx xdx x x x dx x x x        2 cos 0sin tancos sinsin cossin1sincos  x x xv dx x x x x xdx xdx x     Thus,     1 2 cos sin sin ln sec tan cos cos sin cos ln sec tan  p  y v x v x x x x x x x x x x            . And,the general solution is given by the function   1 2 cos sin cos ln sec tan  y x c x c x x x x      .  3. Determine the general solution to the given system of equations. 25 points 2 2 12 3 dx dy x ydt dt dx dy ydt dt          Begin by converting the equation to differential operator notation and isolating the nonhomogeneous terms:              2 2 12 1 3  D x D y D x D y         Eliminating the  x terms and solving for    y t  , we see                                     2 2 12 2 2 1 2 32 2 2 1 1 2 32 1 6  D D x D D y D D D x D D y D D D D D y D D D D y                     Solving the homogeneous problem, we see    2 1 02,1 m mm      21 2 t t c  y c e c e     Using the Method of Undetermined Coefficients to solve for the nonhomogeneous term with  p  y At B   ,we see       3 2 60 3 2 60 3  y y y A At B A B              21 2 3 t t   y t c e c e     Solving the second equation for dxdt  and substituting into the first equation, we can solve for    x t  :   2 21 2 1 221 2 12212 3 223 52 2 t t t t t t  dy x ydt c e c e c e c ec e c e                   Thus, the general solution is given by    21 221 2 3 52 23 t t t t   x t c e c e y t c e c e          4. Find two power series solutions to the differential equation given by   1 0  y x y y      . 30 points Recall, the derivatives for the arbitrary power series are given by       01122 1 nnnnnnnnn  y x c x y x nc x y x n n c x      Substituting into the given differential equation,      2 12 1 02 12 1 1 0 1 1 01 0 n n nn n nn n nn n n nn n n nn n n n n n c x x nc x c xn n c x nc x nc x c x                              Adjusting the summations to be expansions of the function   k   y x x  , we see      2 10 1 0 0 2 1 1 0 k k k k k k k k k k k k  k k c x kc x k c x c x                     Extracting the 0 k   terms and combining the summations, we see                2 10 1 0 02 1 0 2 112 1 0 2 11 2 1 1 02 2 1 1 02 2 1 1 1 0 k k k k k k k k k k k k k k k k k k k k k k k  k k c x kc x k c x c xc c c k k c kc k c c xc c c k k c k c k c x                                             Solving for the coefficients, we see   2 1 02 1 0 2 012! c c cc c c           2 12 1 2 01, 12 k k k k k k  k c c cc c c k k            Expanding the summation, we see                     3 2 1 1 0 1 1 04 3 2 1 0 1 0 1 05 4 3 1 0 1 0 1 0 1 1 1 11: 33 3 2 3!1 1 1 1 12: 3 6 44 4 3! 2! 4!1 1 1 1 13: 6 4 3 18 85 5 4! 3! 5! k c c c c c c c ck c c c c c c c c ck c c c c c c c c c                                 
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