### Integrating Factor

#### 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 $y(x)$:

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

}}

###### 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 $y$:

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

}}

###### 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.