Linear first-order

At a Glance

• Frequency: 6 sub-parts across 5 of 14 years (2015, 2016, 2022, 2023, 2026)
• Priority tier: T3
• Marks (count): 10 (5)
• Average solve time: ~6 min
• Difficulty mix: easy 5
• Section: B | Dominant type: computation

Why This Chapter Matters

Linear first-order ODEs appear in Section B compulsory question Q5 in most years — always for 10 marks, always easy. The integrating-factor technique is a single five-step algorithm that works identically on every question: rearrange, read off $P$ and $Q$, compute $\mu = e^{\int P\,dx}$, multiply through, integrate. The only variety is how the RHS simplifies after multiplying by $\mu$ — each question is designed to collapse to something integrable by inspection. Master one template, collect 10 free marks.

Minimum Theory

Standard form. A linear first-order ODE has the form $\frac{dy}{dx} + P(x)\,y = Q(x).$ If the equation is not in this form, divide by the coefficient of $dy/dx$ first.

Integrating factor. Compute $\mu(x) = \exp\!\int P(x)\,dx.$ Multiplying both sides by $\mu$ converts the left side to an exact derivative: $\frac{d}{dx}(\mu y) = \mu Q.$

General solution. Integrate both sides and divide by $\mu$: $y = \frac{1}{\mu}\!\left[\int \mu Q\,dx + C\right].$

Question Archetypes

linear-first-order (4 question(s); 2015, 2016, 2016, 2023)

Solve a linear first-order ODE via an integrating factor

Recognition Cues

• The equation contains $y$ and $dy/dx$ to the first power only, with no $y^2$ or product $y \cdot dy/dx$.
• It may not look linear at first — watch for a symmetric form $M\,dx + N\,dy = 0$ where treating $y$ as the dependent variable yields a linear equation.
• A 10-mark Section B question asking you to “solve” a first-order ODE, with no hint of exactness, Bernoulli, or Clairaut form, is almost certainly this archetype.

Solution Template

1. Rearrange to standard form: divide by the coefficient of $dy/dx$ to get $y' + P(x)y = Q(x)$.
2. Compute the integrating factor: $\mu = e^{\int P\,dx}$ (no constant of integration needed).
3. Multiply through: the LHS becomes $\dfrac{d}{dx}(\mu y)$.
4. Integrate the RHS: $\mu y = \int \mu Q\,dx + C$.
5. Divide by $\mu$ to get $y$ explicitly; apply any initial condition to find $C$.

Worked Example 1

2015 Paper 1, 2015-P1-Q5a (10 marks)

Solve $x\cos x\,\dfrac{dy}{dx}+y(x\sin x+\cos x)=1$.

Step 1 — Standard form. Divide by $x\cos x$: $\frac{dy}{dx} + \Bigl(\tan x + \frac{1}{x}\Bigr)y = \frac{\sec x}{x}.$ So $P = \tan x + 1/x$ and $Q = \sec x / x$.

Step 2 — Integrating factor. $\int P\,dx = \int\tan x\,dx + \int\frac{dx}{x} = -\ln|\cos x| + \ln|x| = \ln\!\left(\frac{x}{\cos x}\right),$ $\mu = x\sec x.$

Step 3 — Multiply and integrate. The RHS simplifies: $\frac{d}{dx}(x\sec x\cdot y) = x\sec x\cdot\frac{\sec x}{x} = \sec^2 x.$ Integrate: $x\sec x\cdot y = \tan x + C.$

Step 4 — Solve for $y$. $\boxed{y = \frac{\sin x + C\cos x}{x}.}$

Worked Example 2

2016 Paper 1, 2016-P1-Q5c (10 marks)

Solve $\dfrac{dy}{dx} = \dfrac{1}{1+x^2}\bigl(e^{\arctan x}-y\bigr)$.

Step 1 — Standard form. Move $y/(1+x^2)$ to the left: $\frac{dy}{dx} + \frac{1}{1+x^2}\,y = \frac{e^{\arctan x}}{1+x^2}.$ So $P = 1/(1+x^2)$ and $Q = e^{\arctan x}/(1+x^2)$.

Step 2 — Integrating factor. $\mu = e^{\int dx/(1+x^2)} = e^{\arctan x}.$

Step 3 — Multiply and integrate. Substitute $t = \arctan x$: $\frac{d}{dx}(e^{\arctan x}\,y) = \frac{e^{2\arctan x}}{1+x^2},$ $e^{\arctan x}\,y = \int e^{2t}\,dt = \frac{1}{2}e^{2\arctan x}+C.$

Step 4 — Solve for $y$. $\boxed{y = \frac{1}{2}e^{\arctan x} + C\,e^{-\arctan x}.}$

Worked Example 3

2016 Paper 1, 2016-P1-Q6a (10 marks)

Solve $\{y(1-x\tan x)+x^2\cos x\}\,dx - x\,dy = 0$.

Step 1 — Standard form. Rearrange to express $dy/dx$: $\frac{dy}{dx} - \frac{1-x\tan x}{x}\,y = x\cos x.$ So $P = -(1/x) + \tan x$ and $Q = x\cos x$.

Step 2 — Integrating factor. $\int P\,dx = -\ln x - \ln|\cos x| = -\ln(x\cos x),$ $\mu = \frac{1}{x\cos x}.$

Step 3 — Multiply and integrate. The RHS collapses exactly: $\frac{d}{dx}\!\left(\frac{y}{x\cos x}\right) = \frac{1}{x\cos x}\cdot x\cos x = 1.$ Integrate: $\frac{y}{x\cos x} = x + C.$

Step 4 — Solve for $y$. $\boxed{y = x\cos x\,(x+C).}$

Worked Example 4

2023 Paper 1, 2023-P1-Q5a (10 marks)

Solve $\dfrac{dy}{dx}-2xy=2$, $y(0)=1$, in the form $y=e^{x^2}[1+\sqrt{\pi}\,\operatorname{erf}(x)]$.

Step 1 — Standard form. Already in standard form: $P = -2x$, $Q = 2$.

Step 2 — Integrating factor. $\mu = e^{\int(-2x)\,dx} = e^{-x^2}.$

Step 3 — Multiply and integrate. $\frac{d}{dx}(e^{-x^2}y) = 2e^{-x^2}.$ Recall $\operatorname{erf}(x) = \dfrac{2}{\sqrt\pi}\int_0^x e^{-t^2}\,dt$, so $\int 2e^{-x^2}\,dx = \sqrt\pi\,\operatorname{erf}(x) + \text{const}$: $e^{-x^2}y = \sqrt\pi\,\operatorname{erf}(x) + C.$

Step 4 — Apply initial condition. At $x=0$: $1 = 0 + C$, so $C=1$. $\boxed{y = e^{x^2}[1+\sqrt\pi\,\operatorname{erf}(x)].}$

Common Traps

• Failing to rearrange first. In the 2016(Q5c) and 2016(Q6a) problems the ODE is not in standard form — there is a $+y/(1+x^2)$ hiding on the right, or a $-x\,dy$ in a symmetric form. Moving $y$-terms to the left before reading off $P$ and $Q$ is essential.
• Dropping the factor of $\frac{1}{2}$. In 2016(Q5c), integrating $e^{2\arctan x}/(1+x^2)$ yields $\frac{1}{2}e^{2\arctan x}$. The factor $\frac{1}{2}$ is easy to miss when the substitution is done mentally.
• Sign in $\int\tan x\,dx$. The integral is $-\ln|\cos x|$ (not $+\ln|\cos x|$). A sign error here flips the integrating factor entirely: $e^{-\ln(x\cos x)} = 1/(x\cos x)$, not $x\cos x$.
• erf normalisation. $\int e^{-x^2}\,dx = \tfrac{\sqrt\pi}{2}\operatorname{erf}(x)$, so $\int 2e^{-x^2}\,dx = \sqrt\pi\,\operatorname{erf}(x)$. Writing $\int 2e^{-x^2}\,dx = \operatorname{erf}(x)$ drops the $\sqrt\pi$ factor and breaks the initial condition.

linear-solution-form (1 question(s); 2022)

Derive/prove a closed form for the general solution of $y'+Py=Q$

Recognition Cues

• The question says “Show that the general solution of $y'+Py=Q$ can be written as [formula].”
• The given formula involves $Q/P$ as a leading term — it looks structurally different from the standard IF solution but is algebraically equivalent.
• This is a derivation question, not a computation question: start from the standard method and reach the target via integration by parts.

Solution Template

1. Write the standard IF solution: $\mu y = \int \mu Q\,dx + C_0$.
2. Write $Q = P\cdot(Q/P)$ inside the integral; recognise $e^{\int P\,dx}\cdot P\,dx = d(e^{\int P\,dx})$.
3. Integrate by parts: $\int (Q/P)\,d(\mu) = (Q/P)\mu - \int\mu\,d(Q/P)$.
4. Assemble, divide by $\mu$, and rename $C = -C_0$ to match the target sign.

Worked Example

2022 Paper 1, 2022-P1-Q5a (10 marks)

Show that the general solution of $y'+Py=Q$ can be written as $y=\frac{Q}{P}-e^{-\int P\,dx}\!\left\{C+\int e^{\int P\,dx}\,d\!\left(\frac{Q}{P}\right)\right\}.$

Step 1 — Standard solution. The integrating-factor method gives $\mu y = \int\mu Q\,dx + C_0, \qquad \mu = e^{\int P\,dx}.$

Step 2 — Rewrite the integral. Inside the integral write $Q = P\cdot(Q/P)$. Since $d(\mu) = P\mu\,dx$: $\int \mu Q\,dx = \int \frac{Q}{P}\,d(\mu).$

Step 3 — Integrate by parts. With $u = Q/P$ and $dv = d(\mu)$: $\int\frac{Q}{P}\,d(\mu) = \frac{Q}{P}\cdot\mu - \int\mu\,d\!\left(\frac{Q}{P}\right).$

Step 4 — Assemble. Substituting back: $\mu y = \frac{Q}{P}\cdot\mu - \int\mu\,d\!\left(\frac{Q}{P}\right) + C_0.$ Divide by $\mu$ and set $C = -C_0$: $\boxed{y = \frac{Q}{P} - e^{-\int P\,dx}\!\left\{C + \int e^{\int P\,dx}\,d\!\left(\frac{Q}{P}\right)\right\}.}\qquad\blacksquare$

Common Traps

• The key rewrite. Writing $Q = P\cdot(Q/P)$ inside the integral is the non-obvious step that makes integration by parts work. Without it, there is no natural $u$ and $dv$.
• Relabelling the constant. The standard solution has $+C_0$; after the manipulation the constant flips sign, giving $C = -C_0$. Failing to rename produces the wrong sign inside the braces.

Marks-Aware Writing

For a 10-mark computation question: Write all five steps explicitly — standard form (with division shown), integrating factor (with the integral written out), the $d(\mu y)/dx$ step, integration of the RHS, and the final $y$. A correct final answer with two steps missing earns at most 5 marks. Label each step clearly.

For the 2022 proof question: Structure the derivation in numbered steps. The integration-by-parts step must display $u$, $dv$, and the result explicitly — “integrate by parts and simplify” earns no marks. Conclude with $\blacksquare$ and box the target expression.

Practice Set

• 2024-P1-Q5a (10 m) — — Hint: this is an orthogonal-trajectories question; the elimination step produces a linear ODE in $u = r^2$ or similar, solvable by IF.
• 2020-P1-Q5b (10 m) — — Hint: after eliminating the family parameter, the ODE for the orthogonal trajectories is linear in $x^2 + y^2$.