Question

Charpit’s method,xpq + yq

2 = 1

Answer 1

let S(x, y, z, p, q) = xpq + yq² - 1.

Then S_x = pq, S_y = q², S_z = 0.

$\frac{dp}{S_x+pS_z} = \frac{dq}{S_y+qS_z}$

$\frac{dp}{pq} - \frac{dq}{q^2} = 0$

$\frac{qdp - pdq}{pq^2} = \frac{1}{p}d(\frac{p}{q}) = 0$

p/q = const = a

p = aq.

Putting this into the equation S=0, we have:

q²(ax+y) = 1

$q= \frac{\pm 1}{\sqrt{ax+y}}$

$p = aq = \frac{\pm a}{\sqrt{ax+y}}$

Putting these two formulas in the equation dz = pdx + qdy, we have:

$dz = \pm\frac{adx + dy}{\sqrt{ax+y}} = \pm d(2\sqrt{ax+y})$

Integrating this equation, we finally get:

$z = \pm 2\sqrt{ax+y} + b$

Answer. $z = \pm 2\sqrt{ax+y} + b$