Charpit's Method: Pick One Integral, Then Audit Your Answer

Charpit's method finds a complete integral of F(x, y, z, p, q) = 0, where p = ∂z/∂x and q = ∂z/∂y, when F is nonlinear in p and q. As of November 2026, the official CSIR NET Mathematical Sciences syllabus (Parts B and C, Unit 3) lists "Lagrange and Charpit methods for solving first order PDEs, Cauchy problem for first order PDEs" (CSIR-UGC NET syllabus, CSIR HRDG).

For GATE, Section 9 of the 2027 Mathematics (MA) syllabus lists the "method of characteristics for first order linear and quasilinear partial differential equations" (GATE 2027 MA syllabus, IIT Madras). It does not name Charpit's method. If GATE is your only target, spend your time elsewhere. If you are sitting CSIR NET, read on. Our page on what CSIR NET is covers the exam routes.

Write the PDE as F = 0, find five partial derivatives, form the auxiliary equations, pick one integral, solve for p and q, then integrate. The steps below follow the lecture notes in NPTEL's (National Programme on Technology Enhanced Learning) first-order PDE module and Attila Máté's Brooklyn College note.

Some books write the chain with every sign flipped, as dx/−F_p and dp/(F_x + pF_z). The ratios are identical, so both forms give the same integrals. Keep one convention for the whole paper.

Choose by what F is missing. Look at which of x, y and z the PDE contains. Four patterns, plus a general case, cover the standard types:

The last row is the only one that needs ingenuity. In the NPTEL example p²x + q²y = z, combining the first two ratios and the last two gives p²x = a·q²y. The complete integral is [(1 + a)z]^½ = (ax)^½ + y^½ + b.

Each example below is checked by substituting the answer back into its PDE; the residual (left side minus right side) simplifies to 0.

Example 1: pq = xy. Here F = pq − xy, so F_p = q, F_q = p, F_x = −y, F_y = −x, F_z = 0. The chain is dx/q = dy/p = dz/2pq = dp/y = dq/x. Take dx/q = dp/y and use q = xy/p. This gives x dp = p dx, so p = ax. Then q = y/a, and dz = ax dx + (y/a) dy gives z = ax²/2 + y²/(2a) + b.

Residual: with p = ax and q = y/a, pq − xy = xy − xy = 0.

Example 2: z = pq. Here F = z − pq. The chain contains dp/(−p) = dq/(−q), so p = aq. Solving with the PDE gives q = √(z/a) and p = √(az). Then dz/√z = √a dx + dy/√a. Rename the constant (a² for a) to get 4z = (ax + y/a + b)².

A second integral, q = x + a, also comes from the chain, from dx/(−q) = dq/(−q). It gives p = z/(x + a), and then z = (x + a)(y + b). Both answers satisfy z = pq. Both also pass the independence test in the checklist below: ∂(z, p)/∂(a, b) equals y + b for the first and −z for the second, so neither is identically zero.

Residual: for z = (x + a)(y + b), p = y + b and q = x + a, so z − pq = 0. For 4z = (ax + y/a + b)², pq = (ax + y/a + b)²/4 = z, so z − pq = 0.

Example 3: (p² + q²)y = qz. Here F_x = 0, F_y = p² + q², F_z = −q, F_p = 2py, F_q = 2qy − z. The chain gives dp/(pq) = dq/(−p²), so p dp + q dq = 0 and p² + q² = a². The PDE then gives q = a²y/z, and p = (a/z)√(z² − a²y²). Substituting into dz = p dx + q dy gives z dz − a²y dy = a√(z² − a²y²) dx. The left side is ½ d(z² − a²y²), so √(z² − a²y²) = ax + b. The complete integral is z² = a²y² + (ax + b)².

Residual: with q = a²y/z and p = a(ax + b)/z, (p² + q²)y − qz = a²y[(ax + b)² + a²y²]/z² − a²y = a²y − a²y = 0.

Your answer can differ from the key because you and the book chose different integrals of the auxiliary equations; both can be correct (Example 2). The errors that do cost marks are the three-constant slip and wrongly concluding that no solution fits the boundary data.

The three-constant slip. For p² − q² = 3u (a Math StackExchange question from 2023), the auxiliary equations give p − 3x/2 = a and q + 3y/2 = b. Integrating dz = p dx + q dy then yields u = 3x²/4 + ax − 3y²/4 + by + c. Substituting this three-constant family back into the PDE leaves a residual of a² − b² − 3c. So c is not free: c = (a² − b²)/3. You have two constants, which is correct, but a third constant is a signal that you skipped the substitution.

Expecting the complete integral to fit your boundary data. In that same problem, no member of the family fits u(x, 0) = −x²/4, because 3x²/4 + ax + (a² − b²)/3 cannot equal −x²/4 for all x. The conclusion "there is no solution" is wrong: u = −(x − 2y)²/4 satisfies both the PDE and the condition. A complete integral is a two-constant family of solutions. It need not contain the solution that meets your initial curve. Initial-value problems are handled by the characteristic-strip method, which is the Cauchy problem item in the same syllabus line.

The chain that is hard to integrate. For z² = pqxy (the problem in this Math StackExchange thread), the chain is awkward to integrate, but a short route avoids it. Try z = b·xᵐ·yⁿ. Then p = mz/x, q = nz/y, so pqxy = mn·z². The PDE holds when mn = 1, giving z = b·xᵐ·y^(1/m). This is a complete integral, with two constants m and b.

Run these five checks on every answer. Each one has a pass and a fail.

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