CP 9.1 - The integrating factor method | Edexcel A-Level Further Mathematics

A carefully chosen multiplier turns a first-order linear differential equation into one product derivative. You will recognise when the integrating factor method applies, put the equation into the right form, and use constant, algebraic and trigonometric factors to obtain its solution.

Recognising the form and finding the factor

How would you solve $y'+2y=2x$ ? Integrating $y'$ is easy, but the extra $2y$ is an unknown function too. Can one multiplier make both left-hand terms integrate together? The product rule supplies exactly that multiplier.

You already know

The product rule gives $(uv)'=uv'+u'v$ . Separating variables works when you can rearrange an equation into $g(y)\,\mathrm{d}y=f(x)\,\mathrm{d}x$ . The equation $y'+2y=2x$ does not separate in that way, so it needs a different method.

A first-order linear equation has the form $y'+P(x)y=Q(x)$ : the unknown function and its first derivative occur to the first power, with coefficients depending only on $x$ . Thus $y'+x^2y=\sin x$ is linear; $y'+y^2=x$ is not. Divide by the coefficient of $y'$ before identifying $P$ , on an interval where that coefficient is non-zero.

Multiply the standard form by a function $\mu$ . Its left side becomes $\mu y'+\mu Py$ . This is $(\mu y)'$ precisely when $\mu'=P\mu$ . Dividing by $\mu$ and integrating suggests $\ln\mu=\int P\,\mathrm{d}x$ .

Integrating factor

For $y'+P(x)y=Q(x)$ , on an interval where the coefficients are continuous, an integrating factor is

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

It gives

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

You may quote the factor without proof. Choose one antiderivative when calculating it; no arbitrary constant is needed inside the factor. Not in the formula booklet: learn it.

The factor comes from the coefficient of $y$ after normalising the derivative. Keep the sign of that coefficient.

Worked example 1: a constant coefficient

Question. Put $2y'+6y=4x$ into standard linear form and find an integrating factor.

Plan. Divide every term by $2$ , identify $P$ , and exponentiate its integral.

\begin{aligned} y'+3y&=2x&&\text{divide through by }2\\ P(x)&=3&&\text{coefficient of }y\\ \mu&=\mathrm{e}^{\int3\,\mathrm{d}x}=\mathrm{e}^{3x}&&\text{integrate then exponentiate} \end{aligned}

Why this step? Using $6$ before dividing would give $\mathrm{e}^{6x}$ , which would not combine the normalised left side into one product derivative.

Answer. Standard form is $y'+3y=2x$ ; an integrating factor is $\mathrm{e}^{3x}$ .

Check. $(\mathrm{e}^{3x}y)'=\mathrm{e}^{3x}y'+3\mathrm{e}^{3x}y$ , exactly the multiplied left side.

The second example changes the derivative coefficient to a variable. Dividing now makes the coefficient of $y$ a reciprocal function.

Worked example 2: normalising a variable coefficient

Question. For $x>0$ , put $xy'+3y=4x$ into standard form and find an integrating factor.

Plan. Divide by $x$ , then use the logarithmic antiderivative of $3/x$ .

\begin{aligned} y'+\frac3x y&=4&&\text{divide every term by }x\\ \mu&=\mathrm{e}^{\int(3/x)\,\mathrm{d}x}=\mathrm{e}^{3\ln x}&&x>0\\ &=x^3&&\text{inverse exponential and logarithm} \end{aligned}

Why this step? $P(x)=3/x$ , rather than $3$ . Its integral is $3\ln x$ , so the factor is a power of $x$ , not $\mathrm{e}^{3x}$ .

Answer. $y'+\frac3x y=4$ , with integrating factor $x^3$ .

Check. Multiplying by $x^3$ gives $x^3y'+3x^2y=(x^3y)'$ .

There are infinitely many valid integrating factors: any non-zero constant multiple of $\mu$ works too. For example, $7\mathrm{e}^{3x}$ gives seven times the same product derivative. That constant disappears into the final arbitrary constant, leaving exactly the same solution family.

Watch out

Wrong: read $P=3$ directly from $xy'+3y=4x$ . Right: first divide by $x$ , obtaining $P=3/x$ . State the interval when dividing by a variable expression.

Change the coefficient while keeping the normalisation step visible.

Turning the equation into a product derivative

Once you have the factor, multiply the whole equation by it, integrate both sides, and divide to make $y$ the subject.

The left side is no longer two separate integration problems: it is one derivative. Write that derivative explicitly, so the structure remains visible even if the right-hand integration needs work.

Worked example 3: the opening equation

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

Plan. Multiply by $\mathrm{e}^{2x}$ , integrate the product derivative, and use integration by parts on the right.

\begin{aligned} \mu&=\mathrm{e}^{\int2\,\mathrm{d}x}=\mathrm{e}^{2x}\\ (\mathrm{e}^{2x}y)'&=2x\mathrm{e}^{2x}&&\text{multiply every term}\\ \mathrm{e}^{2x}y&=\int2x\mathrm{e}^{2x}\,\mathrm{d}x&&\text{integrate the derivative}\\ &=x\mathrm{e}^{2x}-\frac12\mathrm{e}^{2x}+C&&\text{integration by parts}\\ y&=x-\frac12+C\mathrm{e}^{-2x}&&\text{divide by the factor} \end{aligned}

Why this step? $\int(\mathrm{e}^{2x}y)'\,\mathrm{d}x=\mathrm{e}^{2x}y$ , so you never need to integrate the unknown $y$ separately.

Answer. $y=x-\frac12+C\mathrm{e}^{-2x}$ , where $C$ is arbitrary.

Check. $y'=1-2C\mathrm{e}^{-2x}$ , so $y'+2y=1-2C\mathrm{e}^{-2x}+2x-1+2C\mathrm{e}^{-2x}=2x$ .

The second example changes only the forcing term, from $2x$ to $\mathrm{e}^x$ . The coefficient of $y$ , and therefore the integrating factor, stays the same.

Worked example 4: exponential forcing

Question. Find the general solution of $y'+2y=\mathrm{e}^x$ .

Plan. Use $\mathrm{e}^{2x}$ again, combine the exponential factors and integrate.

\begin{aligned} (\mathrm{e}^{2x}y)'&=\mathrm{e}^{2x}\mathrm{e}^x=\mathrm{e}^{3x}&&\text{multiply through}\\ \mathrm{e}^{2x}y&=\frac13\mathrm{e}^{3x}+C&&\text{integrate}\\ y&=\frac13\mathrm{e}^x+C\mathrm{e}^{-2x}&&\text{divide by }\mathrm{e}^{2x} \end{aligned}

Why this step? The integrating factor depends on $P(x)$ , not on $Q(x)$ . Changing the right side changes the integration, not the factor.

Answer. $y=\frac13\mathrm{e}^x+C\mathrm{e}^{-2x}$ .

Check. $y'+2y=(\frac13+\frac23)\mathrm{e}^x+(-2C+2C)\mathrm{e}^{-2x}=\mathrm{e}^x$ .

Keep the arbitrary constant until information about one particular curve is supplied. The next lesson uses conditions to choose it.

Watch out

Wrong: multiply the left side by $\mathrm{e}^{2x}$ but leave the right side unchanged. Right: $(\mathrm{e}^{2x}y)'=\mathrm{e}^{2x}Q(x)$ . The factor multiplies every term.

Use the same factor with a different exponential on the right.

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CP 9.1 - The integrating factor method

Recognising the form and finding the factor

Turning the equation into a product derivative

Continue this lesson.

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