Pc7–Pc10 - First order equations and integrating factors | OCR A-Level Further Mathematics B (MEI)

An integrating factor turns a first-order linear differential equation into one product derivative that you can integrate. Pc7 distinguishes general and particular solutions; Pc8 identifies the right equation form; Pc9 explains the factor; Pc10 uses it to solve equations and apply conditions. You will normalise carefully, keep the arbitrary constant, and work on an interval where the equation is defined.

Recognising the equation and its solution family

What could you multiply $y'+2y/x=x^2$ , for $x>0$ , by to make its left side a single product derivative?

The product rule is the clue: differentiating $x^2y$ gives $x^2y'+2xy$ . Multiplying the whole equation by $x^2$ would produce exactly that left side. The integrating-factor method finds this multiplier systematically, even when spotting it is difficult.

You already know

The product rule gives $(uv)'=u'v+uv'$ . Integrating a derivative introduces an arbitrary constant. A condition such as “the curve passes through $(1,2)$ ” means $y=2$ when $x=1$ .

Start by making the coefficient of $dy/dx$ equal to $1$ . Read the coefficient of $y$ only after this division.

First-order linear differential equation

An equation is first-order linear when it can be written

\frac{dy}{dx}+P(x)y=Q(x)

where $P$ and $Q$ depend only on $x$ . “First-order” means that no derivative above the first occurs; “linear” means that $y$ and its derivative occur to the first power and are not multiplied together. For example, $xy'+2y=x^3$ is linear on an interval excluding $x=0$ ; $y'+y^2=x$ is not.

Normalising $xy'+2y=x^3$ gives $y'+(2/x)y=x^2$ . Its coefficient is $P(x)=2/x$ , not $2$ . The point $x=0$ cannot be crossed by this normalisation; solve on an appropriate interval instead.

General and particular solutions

A general solution describes the solution family and contains an arbitrary constant for a first-order equation. A particular solution selects a member of that family by fixing the constant, usually from a given condition. For example, $y'=2x$ has general solution $y=x^2+C$ ; through $(0,3)$ , its particular solution is $y=x^2+3$ .

The equation controls the rate, while the condition selects the curve. Do not set the arbitrary constant to zero simply because the question asks for a general solution.

Key idea

Divide to reach $y'+P(x)y=Q(x)$ before finding the factor. A general solution retains its free constant; a particular solution uses a supplied condition to determine it.

This check asks only for recognition and normalisation.

Making a product derivative

Choose a multiplier whose derivative is that multiplier times $P(x)$ ; this is exactly what the product rule needs.

If $I'=PI$ , then $(Iy)'=Iy'+I'y=I(y'+Py)$ . Multiplying the equation by $I$ therefore makes the left side $(Iy)'$ , while the right side becomes $IQ$ . The factor works because of this equality, not because multiplication itself makes a difficult integral easier.

Integrating factor

For $y'+P(x)y=Q(x)$ on an interval where $P$ and $Q$ are continuous, choose

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

Then

\frac{d}{dx}\bigl(I(x)y\bigr)=I(x)Q(x),\qquad I(x)y=\int I(x)Q(x)\,dx+C

Use one convenient antiderivative in the exponent; multiplying an integrating factor by a non-zero constant makes no difference.

Not in the formula booklet: learn it.

An exponential factor never vanishes. After integrating, divide by the same factor to isolate $y$ ; do not leave the answer as a formula for $Iy$ unless that is what the question requests.

Worked example 1: a constant right side

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

Plan. Use $I=e^{2x}$ , identify the product derivative, integrate, then divide by the factor.

\begin{aligned} P(x)&=2,\quad I=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? The factor's derivative is $2e^{2x}$ , so the two left-side terms are exactly the derivative of $e^{2x}y$ .

Answer. $y=3+Ce^{-2x}$ , where $C$ is an arbitrary real constant.

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

The next example changes only the forcing term from $6$ to $6e^x$ . The coefficient of $y$ is unchanged, so the same integrating factor works.

Worked example 2: an exponential right side

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

Plan. Multiply by the same factor, combine the exponential factors on the right and integrate the product derivative.

\begin{aligned} I&=e^{2x} && P(x)=2 \\ e^{2x}y'+2e^{2x}y&=6e^{3x} && \text{multiply through} \\ \frac{d}{dx}(e^{2x}y)&=6e^{3x} && \text{product derivative} \\ e^{2x}y&=2e^{3x}+C && \text{integrate} \\ y&=2e^x+Ce^{-2x} && \text{divide by the factor} \end{aligned}

Why this step? The right side is $e^{2x}(6e^x)=6e^{3x}$ . Multiplying only the left side would change the equation instead of solving it.

Answer. $y=2e^x+Ce^{-2x}$ , with arbitrary real $C$ .

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

Two different solutions of the second equation differ by $(C_1-C_2)e^{-2x}$ . If their constants differ, that difference is never zero, so the solution curves do not cross. One point really does select one member of this family.

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Pc7–Pc10 - First order equations and integrating factors

Recognising the equation and its solution family

Making a product derivative

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