Differential Equations

1st Order Equations

Using an Integrating Factor

Page 1
Problem 1
Page 2
Problems 2-5

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 on problem description to see 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 it can be seen, 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

\[{\frac{{{x^2}}}{2} + \frac{{{x^2}{y^2}}}{2} }+{ C }={ 0.}\]
Page 1
Problem 1
Page 2
Problems 2-5