Differential Equations: Exact Equations

A first-order differential equation is one containing a first—but no higher—derivative of the unknown function. For virtually every such equation encountered in practice, the general solution will contain one arbitrary constant, that is, one parameter, so a first-order IVP will contain one initial condition. There is no general method that solves every first-order equation, but there are methods to solve particular types.

Given a function f( x, y) of two variables, its total differential df is defined by the equation

Example 1: If f( x, y) = x² y + 6 x – y³, then

The equation f( x, y) = c gives the family of integral curves (that is, the solutions) of the differential equation

Therefore, if a differential equation has the form

for some function f( x, y), then it is automatically of the form df = 0, so the general solution is immediately given by f( x, y) = c. In this case,

is called an exact differential, and the differential equation (*) is called an exact equation. To determine whether a given differential equation

is exact, use the Test for Exactness: A differential equation M dx + N dy = 0 is exact if and only if

Example 2: Is the following differential equation exact?

The function that multiplies the differential dx is denoted M( x, y), so M( x, y) = y² – 2 x; the function that multiplies the differential dy is denoted N( x, y), so N( x, y) = 2 xy + 1. Since

the Test for Exactness says that the given differential equation is indeed exact (since M_y = N_x). This means that there exists a function f( x, y) such that

and once this function f is found, the general solution of the differential equation is simply

(where c is an arbitrary constant).

Once a differential equation M dx + N dy = 0 is determined to be exact, the only task remaining is to find the function f ( x, y) such that f_x = M and f_y = N. The method is simple: Integrate M with respect to x, integrate N with respect to y, and then “merge” the two resulting expressions to construct the desired function f.

Example 3: Solve the exact differential equation of Example 2:

First, integrate M( x,y) = y² – 2 x with respect to x (and ignore the arbitrary “constant” of integration):

Next, integrate N( x,y) = 2 xy + 1 with respect to y (and again ignore the arbitrary “constant” of integration):

Now, to “merge” these two expressions, write down each term exactly once, even if a particular term appears in both results. Here the two expressions contain the terms xy², – x², and y, so

(Note that the common term xy² is not written twice.) The general solution of the differential equation is f( x,y) = c, which in this case becomes

Example 4: Test the following equation for exactness and solve it if it is exact:

First, bring the dx term over to the left-hand side to write the equation in standard form:

Therefore, M( x,y) = y + cos y – cos x, and N ( x, y) = x – x sin y.

Now, since

the Test for Exactness says that the differential equation is indeed exact (since M_y = N_x). To construct the function f ( x,y) such that f_x = M and f_y N, first integrate M with respect to x:

Then integrate N with respect to y:

Writing all terms that appear in both these resulting expressions- without repeating any common terms–gives the desired function:

The general solution of the given differential equation is therefore

Example 5: Is the following equation exact?

Since

but

it is clear that M_y ≠ N_x, so the Test for Exactness says that this equation is not exact. That is, there is no function f ( x,y) whose derivative with respect to x is M ( x,y) = 3 xy – f² and which at the same time has N ( x,y) = x ( x – y) as its derivative with respect to y.

Example 6: Solve the IVP