# Differential Equations

## First Order Equations # 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