Answer to Question #170018 in Differential Equations for Hiralal

Question

z=px+qy+3(pq)^1/3

Answer 1

$z = px + qy +3(pq)^{\frac{1}{3}}\\ f(x,y,z,p,q) = px + qy +3(pq)^{\frac{1}{3}} - z = 0 \\ f_x = p \\ f_y = q \\ f_z = -1\\ f_p = x + p^{-\frac{2}{3}} q^{\frac{1}{3}} \\ f_q = y + p^{\frac{1}{3}} q^{-\frac{2}{3}} \\$

$\dfrac{dx}{-f_p} = \dfrac{dy}{-f_q} = \dfrac{dz}{-pf_p -qf_q} = \dfrac{dp}{f_x +pf_z} = \dfrac{dq}{f_y + qf_z}$

$\dfrac{dx}{-x - p^{-\frac{2}{3}} q^{\frac{1}{3}}} = \dfrac{dy}{- y -p^{\frac{1}{3}} q^{-\frac{2}{3}}} = \dfrac{dz}{-xp-yq-2(pq)^\frac{1}{3}} = \dfrac{dp}{0} = \dfrac{dq}{0}$

By integrating, we have p= a, q = b ; where a and b are arbitrary constants.

So, $z = ax + by +3(ab)^{\frac{1}{3}}$