Exact Differential Equations

Consider the equation

        f(x,y)  =  C

Taking the gradient we get

        fx(x,y)i + fy(x,y)j  =  0

We can write this equation in differential form as 

        fx(x,y)dx+ fy(x,y)dy  =  0

Now divide by dx (we are not pretending to be rigorous here) to get

        fx(x,y)+ fy(x,y) dy/dx  =  0

Which is a first order differential equation. The goal of this section is to go backward.  That is if a differential equation if of the form above, we seek the original function f(x,y) (called a potential function).  A differential equation with a potential function is called exact.  If you have had vector calculus, this is the same as finding the potential functions and using the fundamental theorem of line integrals.



        4xy + 1 + (2x2 + cos y)y'  =  0


We seek a function f(x,y)  with 

        fx(x,y)  =  4xy + 1    and     fy(x,y)  =  2x2 + cos y

Integrate the first equation with respect to x to get

        f(x,y)  =  2x2y + x + C(y)    Notice since y is treated as a constant,. we write C(y).

Now take the partial derivative with respect to y to get

        fy(x,y)  =  2x2 + C'(y)

We have two formulae for fy(x,y) so we can set them equal to eachother.

          2x2 + cos y  =  2x2 + C'(y)

That is 

        C'(y)  =  cos y


        C(y)  =  sin y


         f(x,y)  =  2x2y + x + sin y

The solution to the differential equation is

        2x2y + x + sin y  =  C

Does this method always work?  The answer is no.  We can tell if the method works by remembering that for a function with continuous partial derivatives, the mixed partials are order independent.  That is 

        fxy  =  fyx

If we have the differential equation

        M(x,y) + N(x,y)y'  =  0

then we say it is an exact differential equation if 

        My(x,y)  =  Nx(x,y)

Theorem:  Solutions to Exact Differential Equations

Let M, N, My, and Nx be continuous with

          My  =  Nx

Then there is a function f with 

          fx  =  M          and          fy  =  N

such that 

          f(x,y)  =  C

is a solution to the differential equation 

          M(x,y)  + N(x,y)y'  =  0


Solve the differential equation 

        y + (2xy - e-2y)y'  =  0


We have 

        M(x,y)  =  y    and    N(x,y)  =  2xy - e-2y

Now calculate 

        My  =  1    and    Nx  =  2y

Since they are not equal, finding a potential function f is hopeless.  However there is a glimmer of hope if we remember how we solved first order linear differential equations.  We multiplied both sides by an integrating factor m.  We do that here to get

        mM + mNy'  =  0

For this to be exact we must have 

        (mM)y  =   (mN)x  

Using the product rule gives

        myM + mMy  =   mxN + mNx 

We now have a new differential equation that is unfortunately more difficult to solve than the original differential equation.  We simplify the equation by assuming that either m is a function of only x or only y.  If it is a function of only x, then my  =  0 and 

        mMy  =   mxN + mNx         

Solving for mx, we get

                    My - Nx
     mx  =                      

If this is a function of y only, then we will be able to find an integrating factor that involves y only.

If it is a function of only y, then mx  =  0 and 

        myM + mMy  =  mNx         

Solving for my, we get

                    Nx - My
     my  =                      m

If this is a function of y only, then we will be able to find an integrating factor that involves y only.

For our example

                    Nx - My                  2y - 1
     my  =                      m  =                       m  =  (2 - 1/y)m   
                         M                          y 

Separating gives

        dm/m  =  (2 - 1/y) dy

Integrating gives

        ln m  =  2y  -  ln y

        m  =  e2y - ln y  =  y -1e2y

Multiplying both sides of the original differential equation by m gives

        y(y -1e2y) + (y -1e2y)(2xy - e-2y)y'  =  0

        e2y + (2xe2y - 1/y)y'  =  0

Now we see that 

        My  =  2e2y  =    Nx  

Which tells us that the differential equation is exact.  We therefore have

        fx (x,y)  =  e2y 

Integrating with respect to x gives

        f(x,y)  =  xe2y + C(y)

Now taking the partial derivative with respect to y gives

        fy(x,y)  =  2xe2y + C'(y)  =  2xe2y - 1/y

So that 

        C'(y)  =  1/y

Integrating gives

        C(y)  =  ln y

The final solution is 

        xe2y + ln y  =  0



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