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*) = *x* ^{2} *y* + 6 *x* – *y* ^{3}, 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*) = *y* ^{2} – 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*) = *y* ^{2} – 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 *xy* ^{2}, – *x* ^{2}, and *y*, so

(Note that the common term *xy* ^{2} 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* – *f* ^{2} 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

The differential equation is exact because

Integrating *M* with respect to *x* gives

and integrating *N* with respect to *y* yields

Therefore, the function *f*( *x,y*) whose total differential is the left‐hand side of the given differential equation is

and the general solution is

The particular solution specified by the IVP must have *y* = 3 when *x* = 0; this condition determines the value of the constant *c*:

Thus, the solution of the IVP is