Using Laplace transforms calculate the solution of the ordinary differential equation d 2 y d t 2 − 2 d y d t + y = e − t , subject to the initial conditions y(0)=0 and d y d t ( 0 ) = 1. Select the correct answer from the following list. a ) y ( t ) = 3 t e t 2 + e − t 4 − e t 4 b ) y ( t ) = t e t + t 2 e t 2 c ) y ( t ) = t e − t 2 − e − t 4 + e t 4 d ) y ( t ) = t e − t + t 2 e − t 2 e ) y ( t ) = t e − t 2 − e − t 2 + e t 2

Question

Using Laplace transforms calculate the solution of the ordinary differential equation                     d        2            y              d              t        2              −  2            d      y              d      t        +  y  =      e          −      t      , subject to the initial conditions y(0)=0 and             d      y              d      t        (  0  )  =  1. Select the correct answer from the following list.  a  )  y  (  t  )  =            3      t              e        t              2    +            e              −        t              4    −            e      t        4      b  )  y  (  t  )  =  t      e    t    +                    t        2                    e        t              2      c  )  y  (  t  )  =            t              e                  −          t                      2    −            e              −        t              4    +            e      t        4      d  )  y  (  t  )  =  t      e          −      t        +                    t        2                    e                  −          t                      2      e  )  y  (  t  )  =            t              e                  −          t                      2    −            e              −        t              2    +            e      t        2

sweetwisdomgw · Accepted Answer

d        2            y              d              t        2              −  2            d      y              d      t        +  y  =      e          −      t                 y  (  0  )  =  0  ,               y    ′    (  0  )  =  1Auxillary equation and given by      m    2    =  2  m  +  1  =  0  ⇒  (  m  −  1      )    2    =  0  ⇒  m  =  1  ,  1complementary solution is      y    2    (  x  )  =  (      c    1    +      c    2    t  )      e    t   where       c    1   and       c    2   are some constantParticular integral      1                  D        2            −      2      D      +      1            e          −      t        =      1          (      −      1              )        2            −      2      (      −      1      )      +      1            e          −      t        =      1    4        e          −      t      complete solution  (y)  y  (  t  )  =  (      c    1    +      c    2    t  )      e    t    +      1    4        e          −      t        +      1    4        e          −      t      using integral conclusion  y  (  0  )  =      c    1    +      1    4    ⇒      c    1    =  −      1    4          y    ′    (  t  )  =  (      c    1    +      c    2    t  )      e    t    +      e    t    ×      c    2    −      1    4        e          −      t              y    ′    (  0  )  =      c    1        e    0    +      c    2    −      1    4        e          −      0        ⇒      c    1    +      c    2    −      1    4      ⇒  y  (  t  )  =  −      1    4        e    t    +      3    2    t      e          t        +      1    4        e          −      t

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