Solving First-Order Linear Differential Equations using Integrating Factors

The shape to spot. A differential equation is first-order linear when it can be rearranged into the standard form

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

where $P$ and $Q$ are functions of $x$ only (either may be constant). The hallmark is that $y$ and $\frac{dy}{dx}$ each appear only to the first power, never multiplied together and never inside a function like $y^2$ or $\sin y$.

Why an integrating factor helps. The left side $\frac{dy}{dx} + P y$ is almost the derivative of a product. The product rule gives

$\frac{d}{dx}\!\left(\mu y\right) = \mu\frac{dy}{dx} + \frac{d\mu}{dx}\,y.$

If we can find a function $\mu(x)$ — the integrating factor — for which $\frac{d\mu}{dx} = \mu P$, then multiplying the whole equation by $\mu$ turns the left side into a single exact derivative $\frac{d}{dx}(\mu y)$, which we can integrate in one step.

Get to standard form first. The coefficient of $\frac{dy}{dx}$ must be $1$ before you read off $P$. So $x\frac{dy}{dx} + 2y = x^3$ must become

$\frac{dy}{dx} + \frac{2}{x}\,y = x^2 \quad\Rightarrow\quad P(x) = \frac{2}{x}, \; Q(x) = x^2.$

Cambridge tip. Before doing anything else, write the equation in standard form with the coefficient of $\frac{dy}{dx}$ equal to $1$. Mis-reading $P$ from an un-normalised equation is the single most common opening error.

Once the equation is in standard form, the method is always the same four moves.

1. Form the integrating factor. $\mu = e^{\int P\,dx}$ You do not add a constant of integration inside this exponent — any constant $k$ would give $e^{k}\mu$, a constant multiple that cancels out of both sides later. So just take one antiderivative of $P$.

2. Multiply the whole equation by $\mu$. $\mu\frac{dy}{dx} + \mu P\,y = \mu Q.$

3. Recognise the collapsed left side. Because $\frac{d\mu}{dx} = \mu P$ by construction, the left side is exactly the derivative of the product $\mu y$: $\frac{d}{dx}\!\left(\mu y\right) = \mu Q.$ This is the heart of the method — confirm the product-rule expansion mentally each time.

4. Integrate both sides once with respect to $x$ (and never forget the $+C$): $\mu y = \int \mu Q\,dx + C \quad\Rightarrow\quad y = \frac{1}{\mu}\left(\int \mu Q\,dx + C\right).$

A constant-$P$ example. For $\frac{dy}{dx} + 3y = e^{x}$, $P = 3$ so $\mu = e^{\int 3\,dx} = e^{3x}$. Then $\frac{d}{dx}(e^{3x}y) = e^{3x}e^{x} = e^{4x}$, giving $e^{3x}y = \tfrac{1}{4}e^{4x} + C$, i.e. $y = \tfrac{1}{4}e^{x} + Ce^{-3x}$.

A $\mu = x^n$ example. When $P = \frac{n}{x}$, $\int \frac{n}{x}\,dx = n\ln x = \ln x^{n}$, so $\mu = e^{\ln x^{n}} = x^{n}$. This $\mu = x$ or $x^2$ shortcut appears constantly.

Cambridge tip. When you write $\frac{d}{dx}(\mu y) = \mu Q$, simplify $\mu Q$ before integrating. A tidy integrand such as $e^{4x}$ or $x^4$ is far less error-prone than the unsimplified product.

General vs particular solution. Integrating once introduces one arbitrary constant $C$ — this is the general solution, a whole family of curves. A question that supplies an initial or boundary condition such as "$y = 2$ when $x = 0$" (written $y(0) = 2$) wants the single particular solution through that point.

Applying the condition. Substitute the given $x$ and $y$ into the general solution and solve for $C$. For example, if $y = \tfrac{1}{4}e^{x} + Ce^{-3x}$ and $y(0) = 2$: $2 = \tfrac{1}{4}e^{0} + Ce^{0} = \tfrac14 + C \quad\Rightarrow\quad C = \tfrac{7}{4}.$ So $y = \tfrac14 e^{x} + \tfrac74 e^{-3x}$. Substitute the condition only after you have the full general solution including $C$ — never midway.

Always check by differentiating back. Because the integrating-factor method is not in MF19 and the algebra is fiddly, you should differentiate your final $y$ and confirm it satisfies the original ODE. For $y = \tfrac14 e^{x} + \tfrac74 e^{-3x}$: $\frac{dy}{dx} = \tfrac14 e^{x} - \tfrac{21}{4} e^{-3x}, \qquad \frac{dy}{dx} + 3y = \tfrac14 e^{x} - \tfrac{21}{4}e^{-3x} + \tfrac34 e^{x} + \tfrac{21}{4}e^{-3x} = e^{x}.\;\checkmark$ The $e^{-3x}$ terms cancel and we recover $Q(x) = e^{x}$, so the solution is correct.

Numerical accuracy. Where a condition forces a non-exact value (e.g. $C = 1.73\ldots$), give the final answer to 3 significant figures unless an exact form is requested.

Cambridge tip. Spending 20 seconds differentiating your answer back catches sign slips and dropped constants before they cost A marks — examiners reward a verified answer with confidence.

Most exam ODEs reduce to a handful of integrating-factor shapes. Recognising them saves time.

Constant $P = k$. $\int k\,dx = kx \quad\Rightarrow\quad \mu = e^{kx}.$ This is the workhorse for equations modelling growth/decay, e.g. $\frac{dy}{dx} - 2y = \ldots$ gives $\mu = e^{-2x}$.

$P = \dfrac{n}{x}$ (power-of-$x$ factor). $\int \frac{n}{x}\,dx = n\ln x = \ln x^{n} \quad\Rightarrow\quad \mu = x^{n}.$ So $\frac{dy}{dx} + \frac{1}{x}y = \ldots$ has $\mu = x$; $\frac{dy}{dx} + \frac{3}{x}y = \ldots$ has $\mu = x^{3}$. Watch the sign: $P = -\frac{2}{x}$ gives $\mu = x^{-2} = \frac{1}{x^2}$.

$P = \tan x$. $\int \tan x\,dx = \ln(\sec x) \quad\Rightarrow\quad \mu = \sec x.$ Trig integrating factors are common when $Q(x)$ is a trig function; the resulting $\int \mu Q\,dx$ often simplifies neatly via a standard integral in MF19.

A note on $e^{\ln(\cdot)}$. Whenever $\int P\,dx$ produces a logarithm, simplify $e^{\ln u} = u$ immediately. Leaving $\mu = e^{\ln x^2}$ unsimplified is a frequent source of downstream errors.

Cambridge tip. Keep a mental table — $P = k \to e^{kx}$, $P = \frac{n}{x} \to x^n$, $P = \tan x \to \sec x$ — so you spot the integrating factor instantly rather than re-integrating under time pressure.