Differential Equations

First Order Equations

1st Order Differential Equations Logo

Using an Integrating Factor

Consider a differential equation of type

\[{P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy }={ 0,}\]

where \(P\left( {x,y} \right)\) and \(Q\left( {x,y} \right)\) 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 \({\large\frac{{\partial \mu }}{{\partial y}}\normalsize} = 0,\) 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 \({\large\frac{{\partial \mu }}{{\partial x}}\normalsize} = 0,\) 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:

\[{z = \frac{x}{y},\;\;\;}\kern-0.3pt{z = xy,\;\;\;}\kern0pt{z = {x^2} + {y^2},\;\;\;}\kern0pt{z = x + y,}\]

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 3

Solve the differential equation \(\left( {x{y^2} – 2{y^3}} \right)dx \) \(+ \left( {3 – 2x{y^2}} \right)dy \) \(= 0.\)

Example 4

Solve the equation \(\left( {xy + 1} \right)dx \) \(+ {x^2}dy \) \(= 0.\)

Example 5

Solve the equation \(ydx +\) \(\left( {{x^2} + {y^2} – x} \right)dy \) \(= 0\) using the integrating factor \(\mu \left( {x,y} \right) =\) \( {x^2} + {y^2}.\)

Example 1.

Solve the equation \(\left( {1 + {y^2}} \right)dx +\) \( xydy = 0.\)

Solution.

First we test this differential equation for exactness:

\[
{{\frac{{\partial Q}}{{\partial x}} }={ \frac{\partial }{{\partial x}}\left( {xy} \right) }={ y,\;\;}}\kern0pt
{{\frac{{\partial P}}{{\partial y}} }={ \frac{\partial }{{\partial y}}\left( {1 + {y^2}} \right) }={ 2y.}}
\]

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

\[{\frac{{{x^2}}}{2} + \frac{{{x^2}{y^2}}}{2} }+{ C }={ 0.}\]

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Problem 1
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Problems 2-5