Exact Differential Equations

Consider the equation

f(x,y)  =  C

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.

Example

Solve

4xy + 1 + (2x2 + cos y)y'  =  0

Solution

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

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

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

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

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  =
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 mx  =  0 and

myM + mMy  =  mNx

Solving for my, we get

Nx - My

my  =                      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

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