Page 107 - Engineering Mathematics Workbook_Final
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Differential Equations & Partial Differential Equations

                                                                                        +
            137.  If  y −  1  x   0 then the solution of the            (c) cos y =  (x c  )sin x
                   differential equation
                                                                                        +
                                                                         (d) sec y =  (x c  )cos x
                         1
                    y 1 ( y +  ) y =  ( x x +  ) y  with
                            2
                    y ( ) 0 =  is                                 141.  The general solution of the
                                                                         differential equation

                                                      x
                                                  +
                                              −
                   (a) 1 x e−  +  − x    (b) 1 x e                        dy  +  x sin2y =  x 3 cos y  is
                                                                                                 2
                                                                          dx
                   (c) 1 x e+  +  −  x    (d) 1 x e+  +  x
                                                                                      1                 −  2
                                                                                           2
                                                                                                   c e
                         dy           2log x                             (a) Tan y =    ( x −  ) 1 +    x
            138.  If  x 2   +  2xy =          and                                     2
                         dx             x
                                   y
                           0
                    y ( ) 1 =  then  ( ) e =                             (b) Tan y =  1 ( x −   ) 2 +   − x 2
                                                                                           2
                                                                                                   c e
                                                                                      2
                   (a) e                 (b) 1
                                                                                         2
                                                                                                 c e
                                                                         (c) Tan y =  ( x −  ) 1 +   − x 2
                       1                     1
                   (c)                   (d)
                       e                     e 2                                      1                  2
                                                                                          2
                                                                         (d) Cot y =    ( x −  ) 1 +   x
                                                                                                   c e
                      dy                                                              2
            139.  If     +  2 Tan x =  y  Sin x  and
                      dx                                          142.  The general solution of
                                                                                   dx
                    y            = 0 then the maximum value        ( x y +  3  2  xy )  =  1 is
                       3                                                            dy
                   of ‘y’ is                                                 − 1
                                                                                    2
                                                                                        2 c e
                                                                         (a)     = x − +        − x 2  /2
                       1                     1                                y
                   (a)                   (b)
                       4                     8
                                                                                             
                                                                                         +
                                                                                   2
                                                                         (b)   1  =  x +  2 c e − x 2 /2
                                             1                               y
                   (c) 2                 (d)
                                             2
                                                                             1     2            x 2 /2
                                                                                       2 c e
            140.  The general solution of the                            (c)   y  =  x + + 
                   differential equation
                    dy  +  Tan x   Tan y =  Cos x   Sec y              (d)   1  =  x + 1 c e  − x 2  /2
                                                                                            
                                                                                   2
                                                                                        +
                    dx                                                       y
                   is
                                                                  143.  The differential equation
                   (a) 2sin y = (x c+ −  sincos x  )sec x                ( xy  3  +  y cos  ) x dx +


                   (b) sin y =  (x c+  )cos x                            ( x y +  2  2   sin  ) x dy =  0  is exact for





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