Page 92 - Engineering Mathematics Workbook_Final
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Differential Equations & Partial Differential Equations

                                                                             1
            45.    The solution of the differential                      (a)  ( xe +  4x  − 4x  −
                               2
                             d y                                             8         xe       ) 1
                                          x
                   equation       −  y =  e  satisfying
                             dx 2
                                                                             1
                                   dy        3                           (b)  ( xe −  4x  − 4x  +
                           0
                    y ( ) 0 =  and    ( ) 0 =   is                                     xe        ) 1
                                   dx        2                               8
                                         x                                   1     4x  − 4x  1   
                                            x
                       y x =
                   (a)  ( ) sinh x +      e                              (c)     e −  xe    +      
                                        2                                    4                2 
                                           x                                                    
                                              x
                   (b)  ( ) x =  y  x cosh x +  e                        (d)   1   xe −  4x  e − 4x  +  1 
                                           2                                 4               2      
                                        x
                       y x =
                                            x
                   (c)  ( ) sinh x −      e                                                    [JAM CA 2011]
                                        2
                                                                  48.    A general solution of the differential
                                            x
                       y x =
                                               x
                   (d)  ( ) 2 cosh x −  x     e                                    d y      d y
                                                                                     3
                                                                                              2
                                            2                            equation       − 3      +  4y =  is
                                                                                                         0
                                                                                   dx 3     dx 2
                                         [JAM CA 2010]
                                                                                                     2x
                                                                                    x
                                                                                           2x
                                                                         (a)  y =  c e +  c e +  c xe
            46.    The solution of the differential                               1      2       3
                               3
                             d y      dy                                            −
                                                                                                      2x
                                                                                     x
                                                                                             2x
                   equation       −  9    =  cos x is                    (b)  y =  c e + c e +   c xe
                             dx 3     dx                                          1       2       3
                                                                                    −
                                                                                     x
                                                                                                      2x
                                                                                              −
                                                                                               x
                                                  1                      (c)  y =  c e +  c xe +   c e
                                                                                  1
                                                                                          2
                                                                                                    3
                   (a)  ( ) x = C e + C e − 3x  + C +  sin x
                                 3x
                       y
                                               3
                                       2
                               1
                                                  10
                                                                                    −
                                                                                                   4x
                                                                                             x
                                                                                     x
                                                                         (d)  y =  c e + c e +  c e
                                                  1                               1       2      3
                                 3x
                       y
                   (b)  ( ) x =  C e + C e − 3x  + C −  sin x
                                               3
                                1
                                       2
                                                  10                                           [JAM CA 2011]

                                                                                          y
                                                                              y
                                                                  49.    Let  ( ) x  and  ( ) x  be twice
                                                  1                            1           2
                                 3x
                       y
                   (c)  ( ) x = C e + C e − 3x  + C +  cos x
                                               3
                               1
                                       2
                                                  10                     differentiable functions on a interval I
                                                                         satisfying the differential equations
                                                  1
                       y
                   (d)  ( ) x =  C e + C e − 3x  + C −  cos x             dy
                                 3x
                                                                                           x
                                1
                                               3
                                       2
                                                  10                        1  −  y −  y =  e  and
                                                                          dx     1    2
                                         [JAM CA 2010]                     dy     dy
                                                                                                        y
                                                                                               0
                                                                         2    1  +  2  −  6y = . Then  ( ) x
            47.    A particular integral of the                             dx    dx       1             1
                   differential equation                                 is
                     2
                    d y  −  16y =  4sinh 2x is
                                        2
                    dx 2


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