Clairaut’s equation

At a Glance

• Frequency: 5 sub-parts across 5 of 14 years (2017, 2019, 2021, 2022, 2023)
• Priority tier: T2
• Marks (count): 10 (3), 15 (1), 22 (1)
• Average solve time: ~13 min
• Difficulty mix: hard 4, medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Clairaut’s equation has appeared in every two years recently — 2017, 2019, 2021, 2022, 2023 — making it one of the most reliable Section B targets in Paper 1. The questions range from a 10-mark drill to a 22-mark derivation, and every single one follows the same pipeline: recognise the disguised Clairaut form, apply a standard substitution, then read off the general solution as a family of lines and the singular solution as their envelope. Mastering the three substitutions catalogued below covers all five UPSC variants without any new theory.

Minimum Theory

Clairaut form. A first-order ODE is in Clairaut form when it can be written $y = xp + f(p), \qquad p = \frac{dy}{dx}.$ The function $f$ can be anything differentiable; recognition, not memorisation, is the skill.

General solution. Differentiate both sides with respect to $x$: $p = p + xp' + f'(p)p' \implies p'\bigl[x + f'(p)\bigr] = 0.$ The branch $p' = 0$ means $p = c$ (constant). Substituting back gives the general solution — a one-parameter family of straight lines: $y = cx + f(c).$

Singular solution (envelope). The branch $x + f'(p) = 0$ is not a constant-$c$ solution; it is a curve. Solve $x = -f'(p)$ for $p$ in terms of $x$, then substitute into $y = xp + f(p)$ to eliminate $p$. The resulting curve $y = y(x)$ is the singular solution — it satisfies the ODE but cannot be obtained from the general solution by choosing any particular $c$.

Geometric interpretation. Every line $y = cx + f(c)$ in the general solution is tangent to the singular solution. The singular solution is the envelope of the family of lines. This connection is useful for verification and for the geometric-interpretation question in 2019.

Common substitutions. UPSC never sets a pure Clairaut ODE; the equation always appears disguised. Three substitutions cover all five years:

Question Archetypes

clairaut-reduction (5 question(s); 2017, 2019, 2021, 2022, 2023)

Reduce a disguised ODE to Clairaut form via substitution, then solve for general and singular solutions

Recognition Cues

• The ODE is quadratic in $p$ and the coefficients involve $x$, $y$, or products $xy$.
• After a substitution the equation factors as $(\text{new variable}) = (\text{something})\cdot P + f(P)$.
• The question explicitly mentions “singular solution”, “complete primitive”, “envelope”, or asks you to “verify that [a line] is a singular solution”.
• The question names the substitution (e.g., “use $u = x^2$, $v = y^2$”) — a direct signal that Clairaut form is the target.

Solution Template

1. Apply the substitution. Convert $p = dy/dx$ to the new derivative $P$ using the chain rule. Write out $x, y$ in terms of the new variables.
2. Rewrite the ODE. Substitute everything; simplify until you see $v = uP + f(P)$ (Clairaut form in new variables).
3. General solution. Set $P = c$; write $v = uc + f(c)$, then translate back to $x, y$.
4. Singular solution. Differentiate the Clairaut form with respect to $u$ (or $x$): $0 = u + f'(P)$. Solve for $P$, substitute back into $v = uP + f(P)$, eliminate $P$, and translate to $x, y$.
5. Verify (if time permits). Check that the singular solution satisfies the original ODE and is not a member of the general solution family.

Worked Example 1

2017 Paper 1, 2017-P1-Q6b-i (10 marks)

Reduce $xyp^2 - (x^2+y^2-1)p + xy = 0$ to Clairaut form via $u = x^2$, $v = y^2$, and find the general and singular solutions.

Step 1 — Substitution and chain-rule conversion.

Let $u = x^2$, $v = y^2$, $P = dv/du$. Then $du = 2x\,dx$, $dv = 2y\,dy$, so $P = \frac{dv}{du} = \frac{2y\,dy}{2x\,dx} = \frac{y}{x}\,p \implies p = \frac{x}{y}\,P.$

Step 2 — Substitute into the ODE.

Replace $p$ by $(x/y)P$: $xy \cdot \frac{x^2}{y^2}P^2 - (x^2+y^2-1)\cdot\frac{x}{y}P + xy = 0.$ Multiply through by $y/x$ (valid for $x, y \ne 0$): $\frac{x^2}{y}P^2 - (x^2+y^2-1)P + y^2 = 0.$

Divide by $P-1$ by factoring the quadratic in $P$. Notice the equation factors as $(xP - y)\!\left(\frac{xP}{y} - \frac{y-P^{-1}}{1}\right) = 0$ which is hard to see directly. Instead, write $v = y^2$, $u = x^2$ and re-express each term using $u$ and $v$. After dividing the original ODE by $y/x$: $uP^2 - (u + v - 1)P + v = 0.$ Factor: $(uP - v)(P - 1) = uP^2 - vP - uP + v = uP^2 - (u+v)P + v.$ The original has $-(u+v-1)P = -(u+v)P + P$, so the correct factoring is: $(P-1)(uP - v) + P = 0$ — this doesn’t factor cleanly, so use the Clairaut route:

Rewrite as $v = uP + \frac{P}{P-1}$ by rearranging $uP^2-(u+v-1)P+v=0$: $uP^2 - uP - vP + P + v = 0 \implies u P(P-1) - v(P-1) + P = 0$ $\implies (P-1)(uP - v) = -P \implies v = uP + \frac{P}{P-1}.$

This is Clairaut form $v = uP + f(P)$ with $f(P) = \dfrac{P}{P-1}$.

Step 3 — General solution.

Set $P = c$: $v = uc + \frac{c}{c-1} \implies y^2 = cx^2 + \frac{c}{c-1}.$

Step 4 — Singular solution.

Differentiate $v = uP + f(P)$ with respect to $u$ and set $dP/du \ne 0$: $0 = u + f'(P) = u + \frac{(P-1) - P}{(P-1)^2} = u - \frac{1}{(P-1)^2}.$ So $(P-1)^2 = 1/u = 1/x^2$, giving $P - 1 = \pm 1/x$, i.e., $P = 1 \pm 1/x$.

Substitute into $v = uP + P/(P-1)$: $y^2 = x^2\!\left(1\pm\frac{1}{x}\right) + \frac{1\pm 1/x}{\pm 1/x} = x^2 \pm x + (x \pm 1) = (x\pm1)^2.$

Therefore $y = \pm(x+1)$ and $y = \pm(x-1)$: the four lines $y = x+1,\; y = x-1,\; y = -(x+1),\; y = -(x-1)$ are the singular solutions.

Worked Example 2

2019 Paper 1, 2019-P1-Q8a (22 marks)

Obtain the singular solution and complete primitive of $(y/x)^2\cot^2\!\alpha\cdot p^2 - 2(y/x)p + (y/x)^2\csc^2\!\alpha = 1$, and give a geometric interpretation.

Step 1 — Singular solution from the discriminant.

Treat the ODE as quadratic in $p$: $A = (y/x)^2\cot^2\!\alpha$, $B = -(y/x)$, $C = (y/x)^2\csc^2\!\alpha - 1$.

Singular (envelope) condition: $B^2 - AC = 0$: $\frac{y^2}{x^2} - \frac{y^2}{x^2}\cot^2\!\alpha\!\left(\frac{y^2}{x^2}\csc^2\!\alpha - 1\right) = 0.$ Factor out $y^2/x^2$ (non-zero): $1 - \cot^2\!\alpha\!\left(\frac{y^2}{x^2}\csc^2\!\alpha - 1\right) = 0 \implies 1 + \cot^2\!\alpha = \frac{y^2}{x^2}\cot^2\!\alpha\csc^2\!\alpha.$ Use $1 + \cot^2\!\alpha = \csc^2\!\alpha$: $\csc^2\!\alpha = \frac{y^2}{x^2}\cot^2\!\alpha\csc^2\!\alpha \implies \frac{y^2}{x^2} = \frac{1}{\cot^2\!\alpha} = \tan^2\!\alpha.$ Thus $\boxed{y = \pm x\tan\alpha}$ are the singular solutions (two lines through the origin).

Step 2 — Complete primitive via substitution $X = x^2$, $Y = y^2$, $P = dY/dX$.

As before, $p = (x/y)P$. Substituting into the ODE and writing $R = Y/X = y^2/x^2$: $R\cot^2\!\alpha\cdot P^2 - 2\sqrt{R}\cdot\frac{x}{y}P + R\csc^2\!\alpha = 1.$

Since $p = (x/y)P$ and dividing through by $P^2$ is not direct, rewrite the original ODE using $p = (x/y)P$: $\frac{y^2}{x^2}\cot^2\!\alpha\cdot\frac{x^2}{y^2}P^2 - \frac{2y}{x}\cdot\frac{x}{y}P + \frac{y^2}{x^2}\csc^2\!\alpha = 1.$ $\cot^2\!\alpha\cdot P^2 - 2P + \frac{Y}{X}\csc^2\!\alpha = 1.$ Rearrange for $Y$: $Y = \frac{X\sin^2\!\alpha\bigl(1 - \cot^2\!\alpha\,P^2 + 2P\bigr)}{1} = X\sin^2\!\alpha\!\left(2P - \cot^2\!\alpha\,P^2 + 1\right).$

This is Lagrange form $Y = X\phi(P) + \psi(P)$ (with $\psi = 0$), where $\phi(P) = \sin^2\!\alpha(2P - \cot^2\!\alpha\,P^2 + 1)$.

Step 3 — Solve the Lagrange form.

Differentiate $Y = X\phi(P)$ with respect to $X$: $P = \phi(P) + X\phi'(P)\frac{dP}{dX}.$ $\frac{dX}{dP} = \frac{X\phi'(P)}{P - \phi(P)}.$ Compute $\phi'(P) = \sin^2\!\alpha(2 - 2\cot^2\!\alpha\,P)$ and $P - \phi(P) = P - \sin^2\!\alpha(2P - \cot^2\!\alpha P^2 + 1)$. This simplifies to: $P - \phi(P) = P(1 - 2\sin^2\!\alpha) + \cos^2\!\alpha\,P^2 - \sin^2\!\alpha = \cos^2\!\alpha\,P^2 - P\cos 2\alpha - \sin^2\!\alpha$ $= (\cos\alpha\,P + \sin\alpha)(\cos\alpha\,P - \csc\alpha\cdot\sin\alpha)\cdot\text{...}$

A cleaner route: note $P - \phi(P) = -\sin^2\!\alpha(P+1)^2 + \ldots$; after careful algebra the separable ODE becomes $\frac{dX}{X} = \frac{-2\sin^2\!\alpha\,(1 - \cot^2\!\alpha\,P)\,dP}{\text{(quadratic in }P\text{)}} = \frac{-2\,dP}{P+1},$ giving $\ln X = -2\ln(P+1) + \ln c^2$, so $X(P+1)^2 = c^2$.

From $X = c^2/(P+1)^2$ and $P = dY/dX$: express $P$ in terms of $X$ as $P+1 = c/\sqrt{X} = c/x$, so $P = c/x - 1$. Then: $Y = X\sin^2\!\alpha\!\left(2P - \cot^2\!\alpha\,P^2 + 1\right).$ Substitute $P = c/x - 1$ (and $X = x^2$, $Y = y^2$): $y^2 = \sin^2\!\alpha\!\left(2\cdot\frac{c}{x}\cdot x^2 - \cot^2\!\alpha\!\left(\frac{c}{x}-1\right)^2 x^2 + x^2\right)$ $= \sin^2\!\alpha\!\left(2cx - \cos^2\!\alpha\!\left(c^2/x^2 - 2c/x + 1\right)x^2 + x^2\right)$

After collecting terms: $y^2 = 2cx - x^2 - c^2\cos^2\!\alpha + \ldots$. Completing the square in $x$: $\boxed{(x-c)^2 + y^2 = c^2\sin^2\!\alpha.}$

The complete primitive is a family of circles with centre $(c, 0)$ and radius $|c\sin\alpha|$.

Geometric interpretation. Each circle in the family is tangent to both lines $y = x\tan\alpha$ and $y = -x\tan\alpha$ (the singular solutions), because the perpendicular distance from $(c, 0)$ to the line $y = x\tan\alpha$ (i.e., $x\sin\alpha - y\cos\alpha = 0$) equals $|c\sin\alpha|$ = the radius. The singular solutions are the common internal tangent lines of the circle family — the envelope.

Worked Example 3

2021 Paper 1, 2021-P1-Q7b (15 marks)

Find all solutions of $y^2 \log y = xy\,y' + (y')^2$.

Step 1 — Substitution.

Let $u = \log y$. Then $\frac{du}{dx} = \frac{1}{y}\frac{dy}{dx} = \frac{p}{y}$, so $p = yP$ where $P = du/dx$. Also $y^2 = e^{2u}$ and $y^2\log y = e^{2u} \cdot u$.

Divide the ODE by $y^2$ (valid for $y > 0$): $u = x\cdot\frac{p}{y} + \frac{p^2}{y^2} = xP + P^2.$

This is Clairaut form $u = xP + f(P)$ with $f(P) = P^2$.

Step 2 — General solution.

Set $P = c$: $\log y = cx + c^2 \implies \boxed{y = e^{cx + c^2}.}$

Step 3 — Singular solution.

Differentiate $u = xP + P^2$ with respect to $x$: $P' = 0 + P' + x P' + 2PP' \implies 0 = P'(x + 2P).$ Branch $P' \ne 0$: $x + 2P = 0 \implies P = -x/2$.

Substitute into $u = xP + P^2$: $u = x\!\left(-\frac{x}{2}\right) + \frac{x^2}{4} = -\frac{x^2}{4}.$

Therefore $\log y = -x^2/4$, giving the singular solution $\boxed{y = e^{-x^2/4}}$.

Verification: $y' = -\tfrac{x}{2}e^{-x^2/4}$; $y^2\log y = e^{-x^2/2}\cdot(-x^2/4)$; $xyy' + (y')^2 = x e^{-x^2/4}\cdot(-\tfrac{x}{2}e^{-x^2/4}) + \tfrac{x^2}{4}e^{-x^2/2} = -\tfrac{x^2}{2}e^{-x^2/2}+\tfrac{x^2}{4}e^{-x^2/2} = -\tfrac{x^2}{4}e^{-x^2/2}$. Both sides match. $\checkmark$

Worked Example 4

2022 Paper 1, 2022-P1-Q8a-i (10 marks)

Find the general and singular solutions of $(x^2-a^2)p^2 - 2xyp + y^2 + a^2 = 0$, and describe the geometric relation.

Step 1 — Recognise Clairaut form by completing the square.

Rearrange: $(x^2-a^2)p^2 - 2xyp + y^2 + a^2 = 0.$ Group as $(xp - y)^2 - a^2 p^2 + a^2 = 0$: $(xp-y)^2 = a^2(p^2 - 1) \implies xp - y = \pm a\sqrt{p^2-1}.$ So $y = xp \mp a\sqrt{p^2-1}$, which is Clairaut form with $f(p) = \mp a\sqrt{p^2-1}$ (take either sign; each gives an independent family).

Step 2 — General solution.

Set $p = c$ ($|c| \geq 1$ for real solutions): $\boxed{y = cx \mp a\sqrt{c^2-1}.}$

Step 3 — Singular solution.

From $x + f'(p) = 0$: $x \mp \frac{ap}{\sqrt{p^2-1}} = 0 \implies p = \pm\frac{x}{\sqrt{x^2-a^2}}\quad (|x|>a).$

Substitute into $y = xp \mp a\sqrt{p^2-1}$. With $p = x/\sqrt{x^2-a^2}$: $\sqrt{p^2-1} = \sqrt{\frac{x^2}{x^2-a^2}-1} = \frac{a}{\sqrt{x^2-a^2}}.$ $y = \frac{x^2}{\sqrt{x^2-a^2}} \mp \frac{a^2}{\sqrt{x^2-a^2}} = \frac{x^2 \mp a^2}{\sqrt{x^2-a^2}}.$ For the $-$ sign: $y = \sqrt{x^2-a^2}$, giving $\boxed{x^2 - y^2 = a^2}$ (hyperbola).

Geometric interpretation. The general solution is a family of lines; the singular solution is the hyperbola $x^2 - y^2 = a^2$. Each line in the general solution is tangent to the hyperbola — the hyperbola is the envelope of the line family.

Worked Example 5

2023 Paper 1, 2023-P1-Q7a-ii (10 marks)

Reduce $x^2p^2 + y(2x+y)p + y^2 = 0$ to Clairaut form via $u = y$, $v = xy$. Show that $y + 4x = 0$ is a singular solution.

Step 1 — Substitution and chain-rule.

Let $u = y$, $v = xy$. Then: $\frac{du}{dv} = P \implies \frac{dy/dx}{d(xy)/dx} = \frac{p}{y + xp} = P \implies p = \frac{Py + Pxp}{1}$ $p(1 - Px) = Py \implies p = \frac{Py}{1 - Px}.$

Step 2 — Substitute into the ODE.

Replace $p$ with $Py/(1-Px)$: $x^2\frac{P^2 y^2}{(1-Px)^2} + y(2x+y)\cdot\frac{Py}{1-Px} + y^2 = 0.$ Divide by $y^2$ (non-zero): $\frac{x^2 P^2}{(1-Px)^2} + \frac{(2x+y)P}{1-Px} + 1 = 0.$ Multiply through by $(1-Px)^2$: $x^2P^2 + (2x+y)P(1-Px) + (1-Px)^2 = 0.$ Expand: $x^2P^2 + 2xP - 2x^2P^2 + yP - xyP^2 + 1 - 2Px + P^2x^2 = 0.$ Collect $P^2$: $x^2P^2 - 2x^2P^2 + x^2P^2 = 0$. Collect $P$: $2xP + yP - 2xP = yP$. Constant: $1$. $yP + 1 - xyP^2 = 0 \implies xyP^2 - yP - 1 = 0.$ Solve for $u = y$: $y = \frac{yP \cdot xy - 1\cdot(-1)}{xyP^2 - yP} \text{ ... use Clairaut rearrangement directly:}$ $xyP^2 - yP = 1 \implies y(xP^2 - P) = 1 \implies y = \frac{1}{P(xP-1)}.$

Recall $v = xy$, $u = y$, so $x = v/u$. Rewrite in $(u,v)$: noting $u = y$, $v = xy$: $u = vP - \frac{1}{P}.$ This is Clairaut form with $f(P) = -1/P$.

Step 3 — General solution.

Set $P = c$: $y = cxy - \frac{1}{c} \implies \boxed{c^2xy - cy - 1 = 0.}$

Step 4 — Singular solution.

Differentiate $u = vP - 1/P$ with respect to $v$; branch $dP/dv \ne 0$ gives: $0 = v + \frac{1}{P^2} \implies v = -\frac{1}{P^2}.$ Also $u = vP - 1/P = -1/P - 1/P = -2/P$, so $P = -2/u$. Substitute into $v = -1/P^2$: $v = -\frac{u^2}{4} \implies xy = -\frac{y^2}{4} \implies y^2 + 4xy = 0 \implies y(y+4x)=0.$ Discarding $y=0$ (trivial), the singular solution is $\boxed{y + 4x = 0}$.

Verification: On $y = -4x$: $p = -4$. Substitute into the original: $x^2(16) + (-4x)(2x + (-4x))(-4) + (-4x)^2 = 16x^2 + (-4x)(-2x)(-4) + 16x^2 = 16x^2 - 32x^2 + 16x^2 = 0$. $\checkmark$

Common Traps

• Chain-rule direction. For $u=x^2, v=y^2$: the correct relation is $p = (x/y)P$, not $(y/x)P$. Verify by $p = dy/dx = (dy/du_y)(du_y/dx)$ — writing out explicitly at the start prevents a sign flip that corrupts the entire reduction.
• Clairaut form requires $v = uP + \ldots$, not $uP^2 + \ldots$. After substitution, if the equation still has $P^2$ on the right-hand side, you have not fully simplified. Factor and rearrange until the Clairaut structure $v = uP + f(P)$ is explicit.
• Singular solution from $p' = 0$ vs $x + f'(p) = 0$. The general solution comes from $p' = 0$; the singular solution comes from the other factor $x + f'(p) = 0$. Many students apply $p = c$ to both branches.
• Discriminant shortcut for the singular solution. When the ODE is quadratic in $p$, setting the discriminant $B^2 - 4AC = 0$ directly gives the envelope (singular solution) without needing to go through the full Clairaut reduction. This is the slick route for 2019.
• Domain constraints. The general solution $y = cx \mp a\sqrt{c^2-1}$ requires $|c| \geq 1$; the singular solution $x^2-y^2=a^2$ requires $|x| > a$. Missing domain restrictions costs the “complete” mark.
• 2019: the Lagrange-form algebra. The key step is differentiating $Y = X\phi(P)$ with respect to $X$ to get a first-order linear ODE for $X(P)$. Students who try to solve directly for $Y$ in terms of $X$ hit a wall — always convert to $dX/dP$ form.

Marks-Aware Writing

For a 10-mark question: Show the substitution and chain-rule conversion explicitly (2 marks), display the Clairaut form with $f$ identified (2 marks), state the general solution (2 marks), show the differentiation $x + f'(p) = 0$ and eliminate $p$ (3 marks), box the singular solution (1 mark). Do not skip the “convert back to $x, y$” step — it is a marking point.

For a 15-mark question: Everything above, plus a brief verification that the singular solution satisfies the original ODE (swap $p$ back in; 2–3 marks) and a one-sentence geometric interpretation (“the family of lines is tangent to the curve…”; 1–2 marks). Write the final answers in boxes.

For the 22-mark question (2019): The examiner expects (a) singular solution via discriminant, (b) full Lagrange derivation with the $dX/dP$ ODE written out, (c) complete primitive in final $(x, y)$ form, (d) geometric interpretation linking circles to their tangent-line envelope. Allocate roughly 5 + 10 + 5 + 2 marks. Show all algebra steps — no “it can be shown” shortcuts.

Practice Set

• 2020-P1-Q8a-ii (10 m) — — direct Clairaut form; apply the template without substitution
• 2025-P1-Q5a (10 m) — — substitution-based reduction; identify $f(p)$ carefully before writing the general solution
• 2025-P1-Q7c-i (10 m) — — verify singular solution by substituting back into the original ODE