4.10c - The integrating factor method | OCR A-Level Further Mathematics A

Some first order equations become integrable when you multiply by the right function. You will build that function from the product rule, recognise when the method applies, and use initial conditions without losing the interval on which your solution is valid.

Building an integrating factor

How could you integrate $y'+2y=x$ when neither $y$ nor $y'$ is known? Multiplying by $e^{2x}$ makes the left-hand side a familiar derivative:

e^{2x}y'+2e^{2x}y=\frac{d}{dx}(e^{2x}y)

The obstruction has become a product-rule calculation in reverse.

You already know

The product rule gives $\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ . A general solution needs an arbitrary constant; an initial condition selects its value.

The method works for a linear first order equation: $y$ and $y'$ occur only to the first power, and their coefficients depend on the independent variable alone. Thus $y'+xy=x^2$ is linear, whereas $y'+y^2=x$ is not. The equation $y'+y=0$ is linear too: a zero right-hand side causes no difficulty.

Integrating factor

An integrating factor for $y'+P(x)y=Q(x)$ is a non-zero function $\mu(x)$ which makes the multiplied left-hand side the derivative of $\mu(x)y$ .

To find it, compare the product rule with the expression you want:

\begin{aligned} \frac{d}{dx}(\mu y)&=\mu y'+\mu' y &&\text{product rule}\\ \mu y'+\mu Py&=\mu y'+\mu' y &&\text{require matching terms}\\ \mu'&=P\mu &&\text{compare coefficients}\\ \mu&=e^{\int P(x)\,dx} &&\text{solve for a non-zero factor} \end{aligned}

Choose an interval on which $P$ and $Q$ are continuous. The exponential factor is positive there. Its integral does not need an added constant: that would merely multiply the factor by a non-zero constant, and multiplying the entire equation by an extra constant changes none of its solutions.

Integrating factor method

For $y'+P(x)y=Q(x)$ , with continuous $P$ and $Q$ on the interval considered, take

\mu=e^{\int P(x)\,dx}

Then

\frac{d}{dx}(\mu y)=\mu Q \quad\Rightarrow\quad \mu y=\int\mu Q\,dx+C

Any non-zero constant multiple of $\mu$ also works.
Not in the formula booklet: learn it.

The surprising part is that the integrating factor depends only on $P$ , not on $Q$ . Once the coefficient of $y$ is fixed, the same factor handles constant, polynomial, exponential or trigonometric right-hand sides, provided you can do the resulting integral.

Key idea

The integrating factor is chosen to complete a product derivative, not to cancel the $y$ term.

Check whether the product rule really matches every term before integrating.

Constant coefficient linear equations

Write the equation with coefficient one on $y'$ , identify $P$ , and multiply every term by $\mu$ . Only then replace the left-hand side by its product derivative.

Worked example 1: a constant forcing term

Question. Find the general solution of $y'+2y=6$ .

Plan. Use $P=2$ , multiply by $e^{2x}$ , integrate and divide by the factor.

\begin{aligned} \mu&=e^{\int2\,dx}=e^{2x} &&\text{find the factor}\\ e^{2x}y'+2e^{2x}y&=6e^{2x} &&\text{multiply every term}\\ \frac{d}{dx}(e^{2x}y)&=6e^{2x} &&\text{product rule}\\ e^{2x}y&=3e^{2x}+C &&\text{integrate}\\ y&=3+Ce^{-2x} &&\text{divide by }e^{2x} \end{aligned}

Why this step? Integration finds $e^{2x}y$ , so dividing by the factor is essential to obtain $y$ .

Answer. $y=3+Ce^{-2x}$ , for real $x$ .

Check. $y'+2y=-2Ce^{-2x}+6+2Ce^{-2x}=6$ .

Change only the right-hand side from $6$ to $6x$ . The integrating factor stays the same, but the integral now needs integration by parts.

Worked example 2: a linear forcing term

Question. Find the general solution of $y'+2y=6x$ .

Plan. Use the same factor and integrate $6xe^{2x}$ by parts.

\begin{aligned} \mu&=e^{2x} && P=2\\ \frac{d}{dx}(e^{2x}y)&=6xe^{2x} &&\text{multiply and collect}\\ e^{2x}y&=3xe^{2x}-\int3e^{2x}\,dx+C &&\text{integration by parts}\\ &=3xe^{2x}-\frac32e^{2x}+C &&\text{integrate}\\ y&=3x-\frac32+Ce^{-2x} &&\text{divide by the factor} \end{aligned}

Why this step? Choose the polynomial to differentiate in integration by parts, because its degree drops.

Answer. $y=3x-\dfrac32+Ce^{-2x}$ , for real $x$ .

Check. $y'=3-2Ce^{-2x}$ , so $y'+2y=3-2Ce^{-2x}+6x-3+2Ce^{-2x}=6x$ .

The factor has done its job before integration begins; do not change it to suit the right-hand side.

Watch out

Wrong: $\int e^{2x}y'\,dx=e^{2x}y$ . This omits the extra product-rule term. Right: combine $e^{2x}y'+2e^{2x}y$ into $\dfrac{d}{dx}(e^{2x}y)$ first.

A new coefficient on the forcing term changes the integral, not the factor.

Normalising variable coefficients

The coefficient used in the integrating factor is the coefficient of $y$ after dividing by the coefficient of $y'$ . For $xy'+2y=x^3$ , it is $2/x$ , not $2$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

4.10c - The integrating factor method

Building an integrating factor

Constant coefficient linear equations

Normalising variable coefficients

Continue this lesson.

Learn faster with Nova