Exact Differential Equations                Solving an Exact DE                  Making a DE Exact                  Conclusion
                                                       MATH312
                       Section 2.4: Exact Differential Equations
                                               Prof. Jonathan Duncan
                                                    Walla Walla University
                                                 Spring Quarter, 2008
    Exact Differential Equations                Solving an Exact DE                  Making a DE Exact                  Conclusion
    Outline
           1    Exact Differential Equations
           2    Solving an Exact DE
           3    Making a DE Exact
           4    Conclusion
    Exact Differential Equations                Solving an Exact DE                  Making a DE Exact                  Conclusion
    A Motivating Example
           Our tools so far allow us to solve first-order differential equations
           which are separable and/or linear.
           Example
           Is the following differential equation separable or linear?
                              (tanx −sinx siny)dx +(cosx cosy)dy = 0
           After rewriting as shown, what do you notice?
                                               dy = sinx siny −tanx
                                               dx              cosx cosy
                   The equation is not separable.
                   The equation is not linear.
                   Weneed a new solution method for this DE!
    Exact Differential Equations                Solving an Exact DE                  Making a DE Exact                  Conclusion
    Working Backwards
           Wedevelop our method using Calculus notation.
           Differentials
           Recall that if f (x,y) has continuous first partials on some region of the
           xy-plane, then with z = f (x,y) the differential is:
                                                   dz = ∂f dx + ∂f dy
                                                            ∂x           ∂y
           Why is this of use? Recall our motivating example.
           Example
           Now, to solve
                                 (tanx −sinx siny)dx +(cosx cosy)dy = 0
           we find an f(x,y) for which ∂f = (tanx −sinx siny) and
                                                       ∂x
           ∂f = (cosx cosy), and set f (x,y) = c for any constant c so that dz = 0.
           ∂y