I1, I2 - Integrating factors | AQA A-Level Further Mathematics

An equation can tell you how a quantity changes without giving the quantity itself. I1 introduces a multiplier that turns a first-order linear differential equation into a single product derivative; I2 uses it to find a whole family of solutions. You will recognise the right form, prepare the equation and explain why the method works.

Recognising a linear first-order equation

How can we find all the functions satisfying $y'+2y=6x$ , when neither term on the left can simply be ignored? Here $y'=\mathrm{d}y/\mathrm{d}x$ : the unknown is a function, not a number.

Integrating term by term would give $y+2\int y\,\mathrm{d}x=3x^2+C$ , which still contains an unknown integral. Instead, we will make the left side the derivative of a product.

Linear first-order differential equation

For an unknown function $y(x)$ , a linear first-order differential equation can be written

\frac{\mathrm{d}y}{\mathrm{d}x}+P(x)y=Q(x)

on an interval where its coefficients are defined. The highest derivative is the first derivative, and $y$ and $y'$ occur only to the first power, without being multiplied together. The coefficients $P$ and $Q$ depend only on the independent variable $x$ .

For $y'+2y=6x$ , $P(x)=2$ and $Q(x)=6x$ . “Linear” describes dependence on $y$ , not on $x$ : $y'+x^2y=\sin x$ is still linear. A missing $y$ -term means $P=0$ , and a zero right side means $Q=0$ ; both are allowed.

By contrast, $y'+y^2=x$ is nonlinear because of $y^2$ , and $yy'+y=x$ is nonlinear because the unknown multiplies its derivative. The equation $y''+y=x$ is linear but second order. These labels matter because a method has conditions, not just a familiar-looking symbol.

Key idea

Look at how the unknown function and its derivatives appear. A coefficient such as $x^2$ is allowed; a coefficient involving the unknown $y$ is not part of this linear form.

Try that distinction before doing any integration.

Making the derivative coefficient one

The coefficient used in an integrating factor is the coefficient of $y$ after the coefficient of $y'$ has been made one.

If the equation is $a(x)y'+b(x)y=c(x)$ , divide every term by $a(x)$ . This is valid only where $a(x)\ne0$ . Work on a connected interval avoiding any excluded point; a solution on one side of such a point does not automatically extend through it.

Worked example 1: a constant coefficient to divide out

Question. Put $2y'+4y=6x$ in standard linear form and identify $P$ and $Q$ .

Plan. Divide every term by the coefficient of $y'$ .

2y'+4y=6x \quad\Longrightarrow\quad y'+2y=3x

The divisor is the nonzero constant $2$ , so it introduces no excluded value of $x$ .

Answer. $P(x)=2$ and $Q(x)=3x$ , for all real $x$ .

Check. Multiplying the standard form by $2$ recovers all three original terms.

Now change only the derivative coefficient from $2$ to $2x$ . Division changes both the coefficient of $y$ and the allowed interval.

Worked example 2: a variable coefficient to divide out

Question. Put $2xy'+4y=6x$ in standard linear form, stating its domain restriction.

Plan. Divide by $2x$ , keeping the restriction next to the result.

\begin{aligned} 2xy'+4y&=6x\\ y'+\frac{2}{x}y&=3 && (x\ne0) \end{aligned}

Answer. $P(x)=2/x$ , $Q(x)=3$ . Use the standard form on $x>0$ or on $x<0$ , separately.

Check. Multiplication by $2x$ returns the original equation when $x\ne0$ . At $x=0$ , the original equation instead requires $4y(0)=0$ ; division has not analysed whether any solution can pass through that point.

The original equation may contain information at an excluded point, but it needs separate examination there.

Watch out

Wrong: reading $P=4$ directly from $2xy'+4y=6x$ . After normalising, $P=2/x$ , not $4$ . Divide the right side as well.

Prepare a new equation in the same way.

Deriving the integrating factor

For our opening equation, multiplying by $e^{2x}$ creates $e^{2x}y'+2e^{2x}y$ . The product rule says this is exactly $(e^{2x}y)'$ . The multiplier has made the extra product-rule term useful.

For a general multiplier $\mu(x)$ , compare

\mu y'+\mu Py=\mu Q, \qquad (\mu y)'=\mu y'+\mu' y

We need $\mu'=P\mu$ . A convenient nonzero solution is obtained from $\mu'/\mu=P$ .

Integrating factor

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

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

Then $\mu'=P\mu$ , so

\frac{\mathrm{d}}{\mathrm{d}x}(\mu y)=\mu Q

Not in the formula booklet: learn it.

Here the integral means any one antiderivative. Adding a constant would only multiply $\mu$ by a nonzero constant; this scales both sides and gives the same solutions. We therefore omit that unnecessary constant when finding the factor.

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Mechanics

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Discrete Mathematics

I1, I2 - Integrating factors

Recognising a linear first-order equation

Making the derivative coefficient one

Deriving the integrating factor

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