Numeracy, Maths and Statistics - Academic Skills Kit

Integrating Factor

ContentsToggle Main Menu 1 Definition 2 Method 3 Worked Examples 4 Video Example 5 Workbook 6 See Also 7 External Resources

Definition

The integrating factor method is a technique for solving first order ordinary differential equations of the form

$\frac{\mathrm{d} y}{\mathrm{d} x} + f(x)y = g(x),$

where $f(x)$ and $g(x)$ are any two arbitrary functions of $x$ only. Equations of this form are not separable, however we can combine the two terms on the left-hand side into a single derivative by using an integrating factor.

The integrating factor IF is given by integrating $f(x)$ and then exponentiating:

$IF = e^{F(x)},$

where $F(x)$ is defined by $\displaystyle F(x) = \int f(x) \mathrm{d} x$ .

It is important to note that the constant of integration is not included; this method requires that the derivative of $F(x)$ is $f(x)$ , i.e. $F'(x)=f(x)$ , and the constant of integration is not necessary to meet this condition.

Method

Multiplying the original differential equation by the integrating factor $IF= e^{F(x)}$ gives

$e^{F(x)}\frac{\mathrm{d} y}{\mathrm{d} x} + f(x) e^{F(x)}y = g(x) e^{F(x)}.$

Now note that the left hand side is the derivative of the integrating factor multiplied by $y$ :

$\frac{\mathrm{d}}{\mathrm{d} x}\left[ e^{F(x)}y \right] = e^{F(x)}\frac{\mathrm{d} y}{\mathrm{d} x} + f(x) e^{F(x)}y.$

This result comes from the product rule, with the first term on the right-hand side containing the derivative of $y$ and the second term containing the derivate of $e^{F(x)}$ .

The original equation can now be written in the form

$\frac{\mathrm{d}}{\mathrm{d} x} \left[ e^{F(x)}y \right] = g(x) e^{F(x)}.$

This equation can be solved by integrating both sides:

$\int \frac{\mathrm{d}}{\mathrm{d} x} \left[ e^{F(x)}y \right] \mathrm{d} x = \int g(x) e^{F(x)} \mathrm{d} x,$

As integration is the opposite of differentiation, the left-hand side can be simplified:

$\int\frac{\mathrm{d}}{\mathrm{d} x}\left[ e^{F(x)}y\right]\mathrm{d} x = e^{F(x)}y.$

Hence the equation becomes:

$e^{F(x)}y = \int g(x) e^{F(x)} \mathrm{d} x.$

Provided that the integral on the right-hand side is simple enough to compute, this will lead to a solution.

Worked Examples

Example 1

Solve $\dfrac{\mathrm{d} y}{\mathrm{d} x} + 3y = x.$

Solution

The general form of a first order linear ordinary differential equation is:

$\frac{\mathrm{d} y}{\mathrm{d} x} + f(x)y = g(x).$

In the given equation, $f(x) = 3$ and $g(x) = x$ .

Recall that the integrating factor is given by $IF= e^{F(x)}$ , where $\displaystyle F(x)=\int f(x) \mathrm{d} x$ . For this equation, the function $F(x)$ is:

$F(x)=\int3\mathrm{d} x=3x.$

As mentioned above, it is not necessary to include a constant of integration.

The integrating factor for this equation is therefore given by:

$IF = e^{3x}.$

Multiplying both sides of the original equation by the integrating factor gives:

$e^{3x}\frac{\mathrm{d} y}{\mathrm{d} x} + 3 e^{3x}y = x e^{3x}.$

Note: At this point it is advisable to check that the left-hand side is indeed the derivative of the integrating factor multiplied by $y$ :

\[\frac{\mathrm{d </div>{\mathrm{d} x} \left[ e^{3x}y \right] = e^{3x}\frac{\mathrm{d} y}{\mathrm{d} x} + 3 e^{3x}y.\]

The simplified form of the equation is therefore:

$\frac{\mathrm{d}}{\mathrm{d} x} \left[ e^{3x}y \right] = x e^{3x}.$

Integrating both sides gives:

$e^{3x}y = \int x e^{3x} \mathrm{d} x.$

To compute the integral on the right-hand side it is necessary to use integration by parts.

Recall the formula for integration by parts:

$\int uv'dx = uv - \int u'v dx.$

Choose $u=x$ and $v' = e^{3x}$ . Then:

$\begin{align} u&=x, & v&=\dfrac{1}{3} e^{3x}, \\ u'&=1, & v'&= e^{3x}, \end{align}$

and the integral becomes:

$\begin{align} \int x e^{3x}\mathrm{d} x &= x \cdot \frac{1}{3} e^{3x} - \int 1 \cdot \frac{1}{3} e^{3x}\mathrm{d} x, \\ &= \frac{x}{3} e^{3x} - \frac{1}{9} e^{3x} + C. \end{align}$

Note: At this point it is necessary to include the constant of integration.

Substituting this result for the integral back into the equation gives:

$e^{3x}y = \frac{x}{3} e^{3x} - \frac{1}{9} e^{3x} + C.$

Multiplying both sides by $e^{-3x}$ gives the solution for $y(x)$ :

$y = \frac{x}{3} - \frac{1}{9} + C e^{-3x}.$

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Video Example 1

Newcastle University Maths-Aid uses the integrating factor method to find the general solution of $\dfrac{\mathrm{d}y}{\mathrm{d}x}+3y=x$ .

Example 1

Solve $\dfrac{\mathrm{d} y}{\mathrm{d} x} - \dfrac{1}{x}y = x^2.$

Solution

The general form of a first order linear ordinary differential equation is:

$\frac{\mathrm{d} y}{\mathrm{d} x} + f(x)y = g(x).$

In the given equation, $f(x) = -\dfrac{1}{x}$ and $g(x) = x^2$ .

Note: If the function $f(x)$ includes a minus sign it is essential to include this minus sign when computing the integrating factor.

Recall that the integrating factor is given by $IF= e^{F(x)}$ , where $\displaystyle F(x)=\int f(x) \mathrm{d} x$ . For this equation, the function $F(x)$ is:

$F(x)=\int-\frac{1}{x}\mathrm{d} x=-\ln (x) = \ln (x^{-1}).$

As mentioned above, it is not necessary to include a constant of integration.

The integrating factor for this equation is therefore given by:

$IF = e^{\ln(x^{-1})}.$

By the laws of logarithms this simplifies to become:

$IF = \frac{1}{x}.$

Multiplying both sides of the original equation by the integrating factor gives:

$\frac{1}{x} \frac{\mathrm{d} y}{\mathrm{d} x} - \frac{1}{x^2}y = x.$

Note: At this point it is advisable to check that the left-hand side is indeed the derivative of the integrating factor multiplied by $y$ :

\[\frac{\mathrm{d </div>{\mathrm{d} x} \left[ \frac{1}{x}y \right] = \frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x} - \frac{1}{x^2}y.\]

The simplified from of the equation is therefore:

$\frac{\mathrm{d}}{\mathrm{d} x} \left[ \frac{1}{x}y \right] = x.$

Integrating both sides gives:

$\begin{align} \frac{1}{x}y &= \int x \mathrm{d} x \\ &= \frac{1}{2}x^2+C \end{align}$

Multiplying both sides by $x$ gives the solution for $y$ :

$y=\frac{1}{2}x^3+Cx.$

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Video Example 2

Newcastle University Maths-Aid uses the integrating factor method to find the general solution of $\dfrac{\mathrm{d}y}{\mathrm{d}x}-\dfrac{1}{x}y=x^2$ .

Video Example

Prof. Robin Johnson uses the integrating factor method to find the general solution of $x\dfrac{\mathrm{d}y}{\mathrm{d}x}+2y=x^2$ .

Workbook

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

First order differential equations

External Resources

Solving differential equations with integrating factors workbook at mathcentre.
Integrating factors videos at Khan Academy.

More Support

You can get one-to-one support from Maths-Aid.