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Lagrange's method for a PDE

Short answer

The auxiliary equations dx/(y−z)=dy/(x−y)=dz/(z−x) give d(x+y+z)=0 so x+y+z = constant. Using multipliers (2x−y−z,2z−x−y,2y−x−z) yields an exact differential, integrating to 2x²−y²−z²−2xy−2xz+4yz = constant. General solution: φ(x+y+z, 2x²−y²−z²−2xy−2xz+4yz)=0.

Solve the partial differential equation (y − z)p + (x − y)q = (z − x) using Lagrange's method.

Reading of the question. The equation is solved exactly as printed: (y − z)p + (x − y)q = z − x.

The auxiliary equations are dx/(y − z) = dy/(x − y) = dz/(z − x).

First integral. Add the numerators and denominators: the sum of the denominators is (y − z) + (x − y) + (z − x) = 0, so dx + dy + dz = 0 and

x + y + z = a

Second integral. Use the multipliers l = 2x − y − z, m = 2z − x − y, n = 2y − x − z. Then

l(y − z) + m(x − y) + n(z − x) = 0

(expanding, every term cancels). So l dx + m dy + n dz = 0, which is exact:

(2x − y − z) dx + (2z − x − y) dy + (2y − x − z) dz = 0

Integrating (it is the differential of ½[2x² − y² − z² − 2xy − 2xz + 4yz]):

2x² − y² − z² − 2xy − 2xz + 4yz = b

General solution

φ( x + y + z, 2x² − y² − z² − 2xy − 2xz + 4yz ) = 0

where φ is an arbitrary function. (Verified by direct substitution: both functions satisfy (y − z)uₓ + (x − y)u_y + (z − x)u_z = 0.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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