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Charpit's method for a PDE

Short answer

Using Charpit’s equations from F=p²x+q²y−z=0, we get dp/(p−1)=−dx/(2x) → p=1+a/√x. Then substituting and integrating dz = p dx+q dy gives the complete integral z = (√x + a)² + (√y + b)² with arbitrary constants a,b.

Apply Charpit's method to solve the PDE p²x + q²y = z, where the symbols have their usual meanings.

For F(x, y, z, p, q) = 0, Charpit's auxiliary equations are

dp/(Fₓ + pF_z) = dq/(F_y + qF_z) = dz/(pF_p + qF_q) = dx/(−F_p) = dy/(−F_q)

Here F = p²x + q²y − z. So Fₓ = p², F_y = q², F_z = −1, F_p = 2px, F_q = 2qy.

Find a relation. Take dp/(Fₓ + pF_z) = dx/(−F_p):

dp/(p² − p) = −dx/(2px) → dp/(p − 1) = −dx/(2x)

Integrating: ln(p − 1) = −½ ln x + ln a, so p − 1 = a/√x, that is

p = 1 + a/√x

Find q. Put p in the PDE: q²y = z − p²x = z − (√x + a)², so

q = √( [z − (√x + a)²] / y )

Integrate dz = p dx + q dy. Let w = z − (√x + a)². Then dw = dz − (1 + a/√x) dx = dz − p dx = q dy:

dw = √(w/y) dy       →   dw/√w = dy/√y
2√w = 2√y + 2b       →   √w = √y + b

Complete integral

z = (√x + a)² + (√y + b)²

where a and b are arbitrary constants.

Check: p = (√x + a)/√x and q = (√y + b)/√y, so p²x + q²y = (√x + a)² + (√y + b)² = z.

✓ Answer checked by Claude Opus 5.5 (high effort)

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