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Numerical Methods

                   for the initial value problem                  17.    The root of the equation
                    dy  =  x +  2  y ,  ( ) 0 =  is                      x log x =   4.77  with initial guess
                                2
                                                                              10
                                           1
                                   y
                    dx                                                   x =  6, obtained by using Newton
                                                                           0
                   (a) 1.322             (b) 1.122                       Raphson method after second
                                                                         iteration is
                   (c) 1.222             (d) 1.110
                                                                  18.    Let the following discrete data be
            15.    Consider the system of equations                      obtained from a curve  y =  y ( ) x :

                     5 −  1 1          10 
                                    x
                                                                   x:     0      0.25           0.5
                                  
                                              
                   
                                         
                                    y = 
                     2   4   0         12                                  0.75  1
                                         
                     1   1   5         − 1                         y:     1      0.98960        95890
                                    z
                                  
                                              
                   
                                         
                                  
                                         
                                              
                   
                   Using Jacobi’s method with the initial                       90890          8415
                   guess  , y z =                                      Let S be the solid of revolution
                                 ,
                           x
                                     2,3,0 , the
                            0   0  0
                   solution after two approximations is                  obtained by rotating the above curve
                                                                         about the x-axis between x = 0 and x
                                           
                                     −
                              −
                   (a) 2.64, 1.70, 1.12                                 = 1, and let V denote its volume. The
                                                                         approximate value of V, obtained
                              −
                   (b) 2.64, 1.70,1.12                                                   1
                                                                         using Simpson’s   rule, is
                                   −
                   (c) 2.64,1.70, 1.12                                                   3
                                                                  19.    Using the Gauss-Seidel iteration
                                        
                   (d) 2.64,1.70,1.12                                   method with the initial guess,
                                                                          , y z =                      
                                                                                ,
                                                                          x
                                                                                    3.5,2.25,1.625 , the
                                      1 2 0                               0  0  0
                                                                     solution after two approximations for
            16.    The matrix  A =   1 3 1      can be
                                                                       system of equations 2x −    x =  7,
                                          0 1 3                    − x +  2x −  x =  1,  x +   2x =
                                                                                                 1
                                                                                                      2
                                                                                              −
                   decomposed uniquely into the                             1     2    3         2     3   1 is
                   product A = LU, where L and U are                     (a) [5.3125, 4.4491, 2.1563]
                   lower and upper triangular matrices
                   respectively with L has diagonal                      (b) [5.3125, 4.3125, 2.6563]
                   entries 1. The solution of the system                 (c) [5.3125, 4.4491, 2.6563]
                                    
                                     t
                   LX =   1 2 2  is
                                                                         (d) [5.4991, 4.4491, 2.1563]
                                
                                 t
                   (a) 1 1 1            (b) 1 1 0    t

                                
                                                      
                                 t
                                                       t
                   (c) 0 1 1            (d) 1 0 1

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