Answer to Question #171132 in Differential Equations for Akash

Question

Solve z=px+qy-log(pq)

Answer 1

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

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

$\dfrac{dx}{\frac{1}{p} -x}= \dfrac{dy}{\frac{1}{q} -y} = \dfrac{dz}{2-px - qy} = \dfrac{dp}{0} = \dfrac{dq}{0}$

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

So, $z = ax + by - log(ab)$