Integrating Factors

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 yN 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


   


is not exact, since


   

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