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142    CHAPTER 5  Approximation of Solutions



                     1. y = y sin(x); y(0) = 1                     4. y = 2 − x; y(0) = 1


                     2. y = x + y; y(1) =−3                        5. y = y − cos(x); y(1) =−2
                                                                             2

                     3. y = 3xy; y(0) = 5                          6. y = x − y ; y(0) = 4

                     5.3         Taylor and Modified Euler Methods
                                 We will develop two other numerical approximation schemes, both of which are (in general)
                                 more accurate than Euler’s method.
                                    Under certain conditions on f and h, we can use Taylor’s theorem with remainder to write
                                                                         1  2      1  3
                                                                                       (3)


                                                  y(x k+1 ) = y(x k ) + hy (x k ) + h y (x k ) + h y (ξ k )
                                                                         2         6
                                 for some ξ k in [x k , x k+1 ]. If the third derivative of y(x) is bounded, we can make the last term in
                                 this sum as small as we like by choosing h to be small enough, leading to the approximation
                                                                             1  2
                                                         y k+1 ≈ y(x k ) + hy (x k ) + h y (x k ).       (5.1)


                                                                             2
                                 Now, y(x)= f (x, y(x)). This suggests that in equation (5.1) we consider f (x k , y k ) as an approx-
                                 imation of y (x k ) if y k is an approximation of y(x k ). This leaves the term y (x k ) in equation (5.1)



                                 to approximate. To do this, differentiate the equation y (x) = f (x, y(x)) with respect to x to get
                                                                ∂ f      ∂ f

                                                         y (x) =  (x, y) +  (x, y)y (x).
                                                                ∂x       ∂y
                                 We are therefore led to approximate
                                                              ∂ f        ∂ f

                                                       y (x k ) ≈  (x k , y k ) +  (x k , y k )y (x k ).
                                                              ∂x         ∂y
                                 Insert these approximations of y (x k ) and y (x k ) into equation (5.1) to get


                                                                 1  2     ∂ f    ∂ f

                                              y k+1 ≈ y k + hf (x k , y k ) + h  (x k , y k ) +  (x k , y k )y (x k ) .
                                                                 2    ∂x         ∂y
                                   The second-order Taylor method consists of using this expression to approximate y(x k+1 )
                                   by y k+1 We can simplify this expression for the approximate value of y k+1 by using the
                                   notation
                                                                f k = f (x k , y k ),
                                                               ∂ f     ∂ f
                                                                  = f x ,  = f y ,
                                                               ∂x      ∂y
                                                         ∂ f           ∂ f
                                                            (x k , y k ) = f xk ,  (x k , y k ) = f yk .
                                                         ∂x            ∂y
                                   With this notation, the second-order Taylor approximation is
                                                                       1  2
                                                         y k+1 ≈ y k + hf k + h ( f xk + f k f yk ).
                                                                       2



                                    The second-order Taylor method is a one-step method because it approximates the solution
                                 value at x k using the approximation made at x k−1 , which is just one step back. Euler’s method is
                                 also one-step.




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