Using an Integrating Factor
Consider a differential equation of type
\[P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0,\]
where P (x, y) and Q (x, y) are functions of two variables x and y continuous in a certain region D. If
\[\frac{{\partial Q}}{{\partial x}} \ne \frac{{\partial P}}{{\partial y}},\]
the equation is not exact. However, we can try to find so-called integrating factor, which is a function \(\mu \left( {x,y} \right)\) such that the equation becomes exact after multiplication by this factor. If so, then the relationship
\[\frac{{\partial \left( {\mu Q\left( {x,y} \right)} \right)}}{{\partial x}} = \frac{{\partial \left( {\mu P\left( {x,y} \right)} \right)}}{{\partial y}} \]
is valid. This condition can be written in the form:
\[Q\frac{{\partial \mu }}{{\partial x}} + \mu \frac{{\partial Q}}{{\partial x}} = P\frac{{\partial \mu }}{{\partial y}} + \mu \frac{{\partial P}}{{\partial y}},\;\; \Rightarrow Q\frac{{\partial \mu }}{{\partial x}} - P\frac{{\partial \mu }}{{\partial y}} = \mu \left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right).\]
The last expression is the partial differential equation of first order that defines the integrating factor \(\mu \left( {x,y} \right).\)
Unfortunately, there is no general method to find the integrating factor. However, one can mention some particular cases for which the partial differential equation can be solved and as a result we can construct the integrating factor.
1. Integrating Factor Depends on the Variable \(x:\) \(\mu = \mu \left( x \right).\)
In this case we have \(\frac{{\partial \mu }}{{\partial y}} = 0,\) so the equation for \(\mu \left( {x,y} \right)\) can be written in the form:
\[\frac{1}{\mu }\frac{{d\mu }}{{dx}} = \frac{1}{Q}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right).\]
The right side of this equation must be a function of only \(x.\) We can find the function \(\mu \left( x \right)\) by integrating the last equation.
2. Integrating Factor Depends on the Variable \(y:\) \(\mu = \mu \left( y \right).\)
Similarly, if \(\frac{{\partial \mu }}{{\partial x}} = 0,\) we get the following ordinary differential equation for the integrating factor \(\mu:\)
\[\frac{1}{\mu }\frac{{d\mu }}{{dy}} = -\frac{1}{P}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right),\]
where the right side depends only on \(y.\) The function \(\mu \left( y \right)\) can be found by integrating the given equation.
3. Integrating Factor Depends on a Certain Combination of the Variables \(x\) and \(y:\) \(\mu = \mu \left( {z\left( {x,y} \right)} \right).\)
The new function \({z\left( {x,y} \right)}\) can be, for example, of the following type:
\[z = \frac{x}{y},\;\; z = xy,\;\; z = {x^2} + {y^2},\;\; z = x + y,\]
and so on.
Here it is important that the integrating factor \(\mu \left( {x,y} \right)\) becomes a function of one variable \(z:\)
\[\mu \left( {x,y} \right) = \mu \left( z \right)\]
and can be found from the differential equation:
\[\frac{1}{\mu }\frac{{d\mu }}{{dz}} = \frac{{\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}}}{{Q\frac{{\partial z}}{{\partial x}} - P\frac{{\partial z}}{{\partial y}}}}.\]
We assume that the right side of the equation depends only on \(z\) and the denominator is not zero.
Below we consider some particular examples of the equation
\[P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0,\]
where we can determine the integrating factor. The general conditions of existence of the integrating factor are derived in the theory of Lie Group.
Solved Problems
Click or tap a problem to see the solution.
Example 1
Solve the equation \[\left( {1 + {y^2}} \right)dx + xydy = 0.\]
Example 2
Solve the differential equation \[\left( {x - \cos y} \right)dx - \sin ydy = 0.\]
Example 1.
Solve the equation \[\left( {1 + {y^2}} \right)dx + xydy = 0.\]
Solution.
First we test this differential equation for exactness:
\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {xy} \right) = y,\;\; \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {1 + {y^2}} \right) = 2y.\]
As one can see, this equation is not exact. We try to find an integrating factor to convert the equation into exact. Calculate the function
\[\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}} = 2y - y = y.\]
One can notice that the expression
\[\frac{1}{Q}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = \frac{1}{{xy}} \cdot y = \frac{1}{x}\]
depends only on the variable \(x.\) Hence, the integrating factor will also depend only on \(x:\) \(\mu = \mu \left( x \right).\) We can get it from the equation
\[\frac{1}{\mu }\frac{{d\mu }}{{dx}} = \frac{1}{x}.\]
Separating variables and integrating, we obtain:
\[\int {\frac{{d\mu }}{\mu }} = \int {\frac{{dx}}{x}} ,\;\; \Rightarrow \ln \left| \mu \right| = \ln \left| x \right|,\;\; \Rightarrow \mu = \pm x.\]
We choose \(\mu = x.\) Multiplying the original differential equation by \(\mu = x,\) produces the exact equation:
\[\left( {x + x{y^2}} \right)dx + {x^2}ydy = 0.\]
Indeed, now we have
\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {{x^2}y} \right) = 2xy = \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x + x{y^2}} \right) = 2xy.\]
Solve the resulting equation. The function \(u\left( {x,y} \right)\) can be found from the system of equations:
\[\left\{ \begin{array}{l}
\frac{{\partial u}}{{\partial x}} = x + x{y^2}\\
\frac{{\partial u}}{{\partial y}} = {x^2}y
\end{array} \right..\]
It follows from the first equation that
\[u\left( {x,y} \right) = \int {\left( {x + x{y^2}} \right)dx} = \frac{{{x^2}}}{2} + \frac{{{x^2}{y^2}}}{2} + \varphi \left( y \right).\]
Substitute this in the second equation to determine \(\varphi \left( y \right):\)
\[\frac{{\partial u}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{{x^2}}}{2} + \frac{{{x^2}{y^2}}}{2} + \varphi \left( y \right)} \right] = {x^2}y,\;\; \Rightarrow {x^2}y + \varphi'\left( y \right) = {x^2}y,\;\; \Rightarrow \varphi'\left( y \right) = 0.\]
It follows from here that \(\varphi \left( y \right) = C,\) where \(C\) is a constant.
Thus, the general solution of the original differential equation is given by
\[\frac{{{x^2}}}{2} + \frac{{{x^2}{y^2}}}{2} + C = 0.\]
Example 2.
Solve the differential equation \[\left( {x - \cos y} \right)dx - \sin ydy = 0.\]
Solution.
If we test this equation for exactness, we find that
\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( { - \sin y} \right) = 0,\;\; \frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {x - \cos y} \right) = \sin y.\]
Hence, this equation is not exact. We try to construct an integrating factor. Notice that
\[\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}} = \sin y,\]
and the expression
\[\frac{1}{Q}\left( {\frac{{\partial P}}{{\partial y}} - \frac{{\partial Q}}{{\partial x}}} \right) = \frac{{\sin y}}{{\left( { - \sin y} \right)}} = - 1\]
is constant.
Hence, we can find the integrating factor as a function \(\mu \left( x \right)\) by solving the following equation:
\[\frac{1}{\mu }\frac{{d\mu }}{{dx}} = - 1,\;\; \Rightarrow \int {\frac{{d\mu }}{\mu }} = - \int {dx} ,\;\; \Rightarrow \ln \left| \mu \right| = - x,\;\; \Rightarrow \mu = {e^{ \pm x}}.\]
We choose the function \(\mu = {e^{ - x}}\) and make sure that the equation becomes exact after multiplication by \(\mu = {e^{ - x}}:\)
\[{e^{ - x}}\left( {x - \cos y} \right)dx - {e^{ - x}}\sin ydy = 0,\]
\[\frac{{\partial Q}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( { - {e^{ - x}}\sin y} \right) = {e^{ - x}}\sin y = \frac{{\partial P}}{{\partial y}} = \frac{\partial }{\partial }\left( {{e^{ - x}}\left( {x - \cos y} \right)} \right) = {e^{ - x}}\sin y.\]
Its general solution can be found from the system of equations:
\[\left\{ \begin{array}{l}
\frac{{\partial u}}{{\partial x}} = {e^{ - x}}\left( {x - \cos y} \right)\\
\frac{{\partial u}}{{\partial y}} = - {e^{ - x}}\sin y
\end{array} \right..\]
Here it is more convenient to integrate the second equation with respect to \(y:\)
\[u\left( {x,y} \right) = \int {\left( { - {e^{ - x}}\sin y} \right)dy} = {e^{ - x}}\cos y + \psi \left( x \right).\]
Substituting this in the first equation, we have
\[\frac{{\partial u}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{ - x}}\cos y + \psi \left( x \right)} \right] = {e^{ - x}}\left( {x - \cos y} \right),\;\; \Rightarrow - \cancel{{e^{ - x}}\cos y} + \psi'\left( x \right) = x{e^{ - x}} - \cancel{{e^{ - x}}\cos y},\;\; \Rightarrow \psi'\left( x \right) = x{e^{ - x}}.\]
Integrating by parts gives:
\[\psi \left( x \right) = \int {x{e^{ - x}}dx}
= \left[ {\begin{array}{*{20}{l}}
{u = x}\\
{v' = {e^{ - x}}}\\
{u' = 1}\\
{v = - {e^{ - x}}}
\end{array}} \right]
= - x{e^{ - x}} - \int {\left( { - {e^{ - x}}} \right)dx}
= - x{e^{ - x}} + \int {{e^{ - x}}dx}
= - x{e^{ - x}} - {e^{ - x}}.\]
Thus, the general solution of the equation is given by
\[{e^{ - x}}\cos y - x{e^{ - x}} - {e^{ - x}} = C\;\; \text{or}\;\; {e^{ - x}}\left( {\cos y - x - 1} \right) = C,\]
where \(C\) is an arbitrary real number.
See more problems on Page 2.
