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- 2081 Chaitra (new course) · 4 marks
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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