Exact Differential Equations

Consider the equation

        f(x,y)  =  C

Taking the gradient we get

        f_x(x,y)i + f_y(x,y)j  =  0

We can write this equation in differential form as 

        f_x(x,y)dx+ f_y(x,y)dy  =  0

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

        f_x(x,y)+ f_y(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.

Example

Solve

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

Solution

We seek a function f(x,y)  with 

        f_x(x,y)  =  4xy + 1    and     f_y(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

        f_y(x,y)  =  2x² + C'(y)

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

          2x² + cos y  =  2x² + C'(y)

That is 

        C'(y)  =  cos y

or

        C(y)  =  sin y

Hence 

         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 

        f_xy  =  f_yx

If we have the differential equation

        M(x,y) + N(x,y)y'  =  0

then we say it is an exact differential equation if 

        M_y(x,y)  =  N_x(x,y)

Example

Solve the differential equation 

        y + (2xy - e^-2y)y'  =  0

Solution

We have 

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

Now calculate 

        M_y  =  1    and    N_x  =  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

        m_yM + mM_y  =   m_xN + mN_x 

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 m_y  =  0 and 

        mM_y  =   m_xN + mN_x         

Solving for m_x, we get

                    M_y - N_x
        m_x  =                      
                         N

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 m_x  =  0 and 

        m_yM + mM_y  =  mN_x         

Solving for m_y, we get

                    N_x - M_y
        m_y  =                      m
                         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

                    N_x - M_y                  2y - 1
        m_y  =                      m  =                       m  =  (2 - 1/y)m   
                         M                          y 

Separating gives

        dm/m  =  (2 - 1/y) dy

Integrating gives

        ln m  =  2y  -  ln y

        m  =  e^{2y - ln y}  =  y ^-1e^2y

Multiplying both sides of the original differential equation by m gives

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

        e^2y + (2xe^2y - 1/y)y'  =  0

Now we see that 

        M_y  =  2e^2y  =    N_x  

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

        f_x (x,y)  =  e^2y 

Integrating with respect to x gives

        f(x,y)  =  xe^2y + C(y)

Now taking the partial derivative with respect to y gives

        f_y(x,y)  =  2xe^2y + C'(y)  =  2xe^2y - 1/y

So that 

        C'(y)  =  1/y

Integrating gives

        C(y)  =  ln y

The final solution is 

        xe^2y + ln y  =  0

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