Clairaut's Equation for CSIR NET: Solve It, Then Check It

Clairaut's equation is a first-order differential equation y = x·p + f(p), where p = dy/dx and f is continuously differentiable. It has a general solution made of straight lines and a singular solution, a solution that no choice of the constant C produces. The singular solution is the envelope of those lines, the curve that touches each of them.

Alexis Clairaut introduced the equation in 1734, and it is a special case of the Lagrange equation (Wikipedia). To solve it, differentiate both sides with respect to x. The p terms cancel and leave

[x + f′(p)] · dp/dx = 0.

Each factor gives its own solution:

• dp/dx = 0 means p = C, so y = Cx + f(C).
• x + f′(p) = 0 gives the parametric curve x = −f′(p), y = f(p) − p·f′(p), where p is the parameter (MathWorld).

A concrete case: for y = xp + p², the lines are y = Cx + C². The singular curve is y = −x²/4. The line with constant C touches that parabola at x = −2C, where both give y = −C².

CSIR NET names singular solutions of first order ODEs, and in our judgement Clairaut's equation is the most common example of that item. GATE 2027 MA does not name singular solutions. Neither official syllabus uses the word "Clairaut".

Sources: CSIR Human Resource Development Group, Mathematical Sciences syllabus; GATE 2027 Mathematics syllabus, IIT Madras.

Our recommendation: for CSIR NET, learn this routine because Clairaut's equation is the cleanest worked case of a singular solution. The same item can also be tested on other first-order equations through the discriminant method, so practise that as well. For GATE, learn it as part of first order ODEs, but give it less time than items the syllabus names outright. The GATE MA syllabus page lists every section, and our CSIR NET Mathematical Sciences syllabus page covers the rest of the CSIR NET units. Check the current official notification before you plan, because syllabus documents get revised.

Six checks, run in order, take you from the raw question to a verified pair of answers. Each has a pass and a fail, so you can mark your own working.

Is the equation in Clairaut form once you solve for y?

Pass: after you isolate y, the equation reads y = x·p + f(p), with exactly p multiplying x. Fail: the coefficient of x is a function g(p) different from p, which makes it a Lagrange equation (y = x·g(p) + f(p) with g(p) ≠ p) and needs another method.

An equation that looks different can still pass. (y − xp)² = 1 + p² becomes y = xp ± √(1 + p²), and each sign is a separate Clairaut equation.

Is f linear in p?

Pass: f′(p) is not constant, so a singular curve exists. Fail: f′(p) is constant, so stop.

If f(p) = ap + b, every line y = Cx + aC + b passes through the single point (−a, b). There is no envelope. For y = xp + 3p + 2, all lines pass through (−3, 2).

Have you written both solutions?

Pass: your answer has the line family y = Cx + f(C) and the pair x = −f′(p), y = f(p) − p·f′(p). Fail: you stop after the first. Both come from the same differentiated equation, so a single answer leaves half the work undone.

Do you really have to eliminate p?

Pass: the answer is in x and y alone when the question asks for that and solving for p gives a closed form (for example f = p²), or it is left parametric when solving for p needs a root of degree 3 or higher or a non-elementary inverse. Fail: the answer is parametric when the question asks for an equation in x and y and p can be eliminated in one or two steps, or the answer contains an unsolved elimination.

Our recommendation is to eliminate only when f is simple. For f = p², eliminating gives y = −x²/4 in two lines. For f = √(1 + 2p⁴), leave it parametric.

Does your singular solution satisfy the original equation?

Pass: substitute y(x) and p = y′(x) into y = xp + f(p), and the left side minus the right side is zero for every x in the domain. Fail: any nonzero remainder.

Take y = 2 + ln x for f = 1 − ln p. Here p = 1/x, so xp = 1 and f(p) = 1 + ln x, which sum to 2 + ln x. That is a pass, for x > 0. Differentiating can introduce extraneous solutions, so this check is not optional (Vladimir Dobrushkin's Mathematica tutorial, Brown University).

Can you say why the singular solution is not in the family?

Pass: you state that p is constant on every line y = Cx + f(C) but varies along the singular curve. Fail: you call it "the case C = something".

For y = −x²/4, p = −x/2, which changes with x. No single C can reproduce it, although each line touches it at one point.

Six of the seven equations below have a singular solution; the seventh, with linear f, has none. These seven are our selection, chosen to vary the form of f. We checked each on 12 November 2026 using SymPy 1.14.0, a computer algebra library. For each f, we built x = −f′(p), y = f(p) − p·f′(p) and confirmed that the curve satisfies y = xp + f(p) and that dy/dx = p along it. This is a check of seven equations, not a proof for every f.

The same method covers (y − xp)² = 1 + p². Its lines are y = Cx ± √(1 + C²), and its singular solution is the circle x² + y² = 1. The upper semicircle satisfies the equation with the plus sign and the lower semicircle with the minus sign.

You can start free: our free mathematics study content includes demo classes, notes and previous-year questions, and our guide to a CSIR NET test series sets out what to check before you pay for tests.

Our Differential Equations live course is taught in Hindi and English by Ms. Eshita, who teaches Applied Mathematics at SBTech Math. As of November 2026 the batch, which began on 5 October 2026, costs ₹1,999, down from ₹2,499, on our course page.

Before you buy any course, take the CSIR line "singular solutions of first order ODEs" and ask whether the topic list includes it. Send that line to us through our contact form, which we answer within 24 hours, or email sbtechmath@gmail.com, and ask how a new joiner can start the batch. We will tell you plainly whether our course fits your gap.

For the partial differential equation (PDE) side of the same syllabus unit, see our page on Charpit's method.