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1.1 Terminology and Separable Equations   5


                                                                            2


                                                                          1.5


                                                                            1


                                                                          0.5


                                                                            0
                                                                                                  x
                                                           –0.6  –0.4  –0.2   0   0.2   0.4   0.6
                                                                         –0.5


                                                                           –1

                                                         FIGURE 1.1 Some integral curves from Example 1.1.

                                                                      1              1
                                                             y(x) =      ,  y(x) =       ,
                                                                   e −x  − 3      e −x  + 3
                                                                      1                  1
                                                             y(x) =      ,   and  y(x) =    = e x
                                                                   e −x  − 6            e −x
                                        are particular solutions corresponding to k =±3,6, and 0. Particular solutions are also called
                                        integral curves of the differential equation. Graphs of these integral curves are shown in
                                        Figure 1.1.


                                 EXAMPLE 1.2
                                         2
                                        x y = 1 + y is separable, since we can write

                                                                         1       1
                                                                            dy =   dx
                                                                       1 + y     x  2
                                        if y  =−1 and x  = 0. Integrate to obtain
                                                                                  1
                                                                      ln|1 + y|=− + k
                                                                                  x
                                        with k an arbitrary constant. This equation implicitly defines the solution. For a given k,wehave
                                        an equation for the solution corresponding to that k, but not yet an explicit expression for this
                                        solution. In this example, we can explicitly solve for y(x). First, take the exponential of both
                                        sides of the equation to get
                                                                            k −1/x
                                                                    |1 + y|= e e  = ae −1/x ,
                                                                k
                                        where we have written a = e . Since k can be any number, a can be any positive number.
                                        Eliminate the absolute value symbol by writing
                                                                    1 + y =±ae −1/x  = be  −1/x ,
                                        where the constant b =±a can be any nonzero number. Then

                                                                        y =−1 + be  −1/x
                                        with b  = 0.




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                                    October 14, 2010  14:9   THM/NEIL   Page-5          27410_01_ch01_p01-42
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