Exact Differential Equations 
                    
                   The following type of first order differential equations that we’ll be looking at is exact 
                   differential equations.  What’s exact differential equation? 
                                                           
                    Property:                    or                                                                     
                   (1) 
                   is an exact differential equation if and only if 
                                                                                      
                                                                          
                   For exact differential equation (1), there exists a function        such that                  
                                                                                                                    
                                                                                                                              
                                                                                                                         
                   and                 . Thus,                                                                                         
                                                                                                                              
                   Thus, differential equation (1) becomes 
                                                                        
                                                                                     
                                                                        
                   So,            is an implicit solution of differential equation (1). 
                   Let’s look at the detail procedure to find the solution for exact differential equation from the 
                   following examples. 
                   Example 1: Solve the differential equation: 
                                                                                           
                                                                                  
                                          yy2
                    Let M  ycosx2xe ,N sinxx e 1
                                      yy
                    M cosx2xe ,N cosx2xe
                      yx
                    MN
                      yx
                    There exists a function , such that   MN,          .
                                                              xy
                                        y
                     ycosx 2xe               (a)
                     x                                                        
                                  2 y
                      sinxx e 1              (b)
                     y
                    
                    Integrating (a) with respect to x, we have
                                       yy2
                    ycosx 2xe dx ysinxx e h(y)
                        
                    Take dirivative with respect to y, we have
                    sinxx2ey h'(y)
                     y
                           From (b), we have  sinx x2ey 1, so 
                                                          y
                                       22yy
                           sin x  x e h'(y)  sin x x e 1,
                           hy'(   ) 1
                           h()y  yc                                                       
                           Thus,   ysin x x2ey  yc
                            ysin x x2ey  yc  0 is the solution to 
                                              yy2
                            ycosx2xe (sinxx e 1)y'0
                           Example 2: Solve the differential equation: 
                                                                                                           
                           Transfer the original equation to 
                                22
                           (xy 2)dx(x y3)dy 0
                                            22
                           Let M  xy 2,N  x y3
                           M 2xy,N                2xy
                               yx
                           MN
                               yx
                            There exists a function , such that   MN,                                .
                                                                                      xy
                                         2
                           
                              xy           2           (a)
                            x
                                      2
                              x y 3                    (b)
                            y
                           
                           Integrating (a) with respect to x, we have
                             (xy2 2)dx  1 x2y2 2xh(y)
                                                       2
                           Take dirivative with respect to y, we have
                                     2
                           y x y         h'(y)
                                                                 2
                           From (b), we have y xy 3, so 
                             22
                            x yh'(y)  x y3,
                           hy'(   ) 3
                           h(y)  3yc
                                           1 22
                           Thus,   2 x y 2x3yc                                                           
                            1 22
                            2 x y 2x3yc0 is the solution to 
                               22
                            xy 2(3x y)y'
                           Example 3: Solve the differential equation:  
                            
                                                                                                               
                                                                                                          
                                                                                   
                                                           
                           with initial value:         . 
                           Transfer the original equation to 
                              2ty
                           (         2t)dt (ln(t2 1)2)dy  0
                             t2 1
                           Let M  2ty 2t,N ln(t2 1)2
                                         t2 1
                                       22tt
                           MN,
                               yt
                                      22
                                     tt11
                           MN
                               yx
                            There exists a function , such that   MN,                                .
                                                                                      xy
                                        2ty
                             2ta( )
                                 x
                                      t2 1
                           
                                          2
                           
                             y ln(tb1)2                      ( )
                           
                           Integrating (a) with respect to t, we have
                                       2ty                         22
                             ( 2            2t)dt  yln(t 1)t h(y)
                                  t 1
                           Take dirivative with respect to y, we have
                            ln(t2 1)h'(y)
                              y
                           From (b), we have   ln(t2 1)2, so 
                                                          y
                                 22
                           ln(t    1)h'(y) ln(t 1)2,
                           hy'(   ) 2
                           h(y)  2yc
                                                   22
                           Thus,   yln(t 1)t 2yc
                                   22
                            yln(t 1)t 2yc0 is the solution to                                            
                           ( 2ty 2t)dt (ln(t2 1)2)dy  0
                             t2 1
                               y(1)  0
                                        22
                           0ln(1 1)1 20cc0 1
                                      22
                           yln(t 1)t 2y10 is the solution to 
                                   22
                            yln(t 1)        t  2yc0 with initial value y(1)  0.
                           Remark: This equation is first order linear, you can use first order linear equation method to 
                           solve it too. 
                           Integrating Factor: Sometimes it is possible to convert a differential equation that is not exact 
                           into an exact equation by multiplying the equation by a suitable integrating factor.   
                                            ()x
                           Multiply                 both sides to  
                           so that the resulting equation                                                   
                                                                                                                                                   
                                                                               ()x                             ()x                
                                                                                                                                   
                        is exact, then we call                   is an integrating factor. 
                                                          ()x
                        Example 4: Verify that                            is an integrating factor of the equation 
                                                              ()xx
                           (2)                                                               22  
                                                                                   (3xy y )(x xy)y'0
                        Multiplying                      to both sides of (2), we have 
                                            ()xx
                                                                                (3x2y  xy2)(x3  x2y)y'  0  
                         Let M 3x2yxy2,N x3x2y,  we have          
                                     22
                          M 3x 2xy,N 3x 2xy
                             yx
                         MN                                                
                               yx
                        Differential Equation (2) is exact. 
                         Example 5: Given                           is an integrating factor of the equation 
                                                        ()xx
                           (2)                                                               22  
                                                                                   (3xy y )(x xy)y'0
                        Find the solution to (2). 
                        Multiplying                      to both sides of (2), we have 
                                            ()xx
                         (3)                                                       (3x2y  xy2)(x3  x2y)y'  0 
                        (3) is an exact equation, there exists a function  , such that  
                                                                2        2                3      2
                                               M3,x yxy                    Nx x y
                                                                                                     
                                                 xy
                          3x2yxy2dx x3y1x2y2 h(y)
                                                            2
                                 32
                        y  x x yh'(y)
                                    32
                           y x       x y
                        x3x2yh'(y) x3x2y                                                                 
                         h'(y)  c
                        x3y1x2y2c
                                         2
                        x3y1x2y2 0 is the solution to (3xy y2)(x2  xy)y' 0.
                                   2