If a differential equation of the form
is not exact as written, then there exists a function μ( x,y) such that the equivalent equation obtained by multiplying both sides of (*) by μ,
is exact. Such a function μ is called an integrating factor of the original equation and is guaranteed to exist if the given differential equation actually has a solution. Integrating factors turn nonexact equations into exact ones. The question is, how do you find an integrating factor? Two special cases will be considered.
- Case 1:
Consider the differential equation M dx + N dy = 0. If this equation is not exact, then M y will not equal N x ; that is, M y – N x ≠ 0. However, if
is a function of x only, let it be denoted by ξ( x). Then
will be an integrating factor of the given differential equation.
- Case 2:
Consider the differential equation M dx + N dy = 0. If this equation is not exact, then M y will not equal N x ; that is, M y – N x ≠ 0. = 0. However, if
is a function of y only, let it be denoted by ψ( y). Then
will be an integrating factor of the given differential equation.
Example 1: The equation
However, note that
is a function of x alone. Therefore, by Case 1,
will be an integrating factor of the differential equation. Multiplying both sides of the given equation by μ = x yields
which is exact because
Solving this equivalent exact equation by the method described in the previous section, M is integrated with respect to x,
and N integrated with respect to y:
(with each “constant” of integration ignored, as usual). These calculations clearly give
as the general solution of the differential equation.
Example 2: The equation
is not exact, since
However, note that
is a function of y alone (Case 2). Denote this function by ψ( y); since
the given differential equation will have
as an integrating factor. Multiplying the differential equation through by μ = (sin y) −1 yields
which is exact because
To solve this exact equation, integrate M with respect to x and integrate N with respect to y, ignoring the “constant” of integration in each case:
These integrations imply that
is the general solution of the differential equation.
Example 3: Solve the IVP
The given differential equation is not exact, since
However, note that
which can be interpreted to be, say, a function of x only; that is, this last equation can be written as ξ( x) ≡ 2. Case 1 then says that
will be an integrating factor. Multiplying both sides of the differential equation by μ( x) = e 2 x yields
which is exact because
Now, since
and
(with the “constant” of integration suppressed in each calculation), the general solution of the differential equation is
The value of the constant c is now determined by applying the initial condition y(0) = 1:
Thus, the particular solution is
which can be expressed explicitly as
Example 4: Given that the nonexact differential equation
has an integrating factor of the form μ( x,y) = x a y b for some positive integers a and b, find the general solution of the equation.
Since there exist positive integers a and b such that x a y b is an integrating factor, multiplying the differential equation through by this expression must yield an exact equation. That is,
is exact for some a and b. Exactness of this equation means
By equating like terms in this last equation, it must be the case that
The simultaneous solution of these equations is a = 3 and b = 1.
Thus the integrating factor x a y b is x 3 y, and the exact equation M dx + N dy = 0 reads
Now, since
and
(ignoring the “constant” of integration in each case), the general solution of the differential equation (*)—and hence the original differential equation—is clearly