Answer to Question #168833 in Differential Equations for Hiralal

Question

P^2+px+q=z

Answer 1

$p^2 + px + q = z\\ p^2 + px + q -z =0 \cdot \cdot \cdot \cdot (*)$

Set

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

The Chapit auxiliary equation are

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

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

$dy = \dfrac{dq}{q} \\ q = ae^y$ where a is an arbitrary constant.

Substituting the value of q into (*) , we have

$p^2 +px + ae^y -z = 0$

$p = \dfrac{-x \pm \sqrt{x^2 - 4(ae^y-z)}}{2}$

Taking positive sign only we have that

$p = \dfrac{-x + \sqrt{x^2 - 4(ae^y-z)}}{2}$

Substituting the value of p and q into

$dz = pdx + qdy$

We have

$dz = \bigg( \dfrac{-x + \sqrt{x^2 - 4(ae^y-z)}}{2} \bigg)dx + ae^ydy$

By integrating, we have

$z = \dfrac{-x^2}{4} + \dfrac{ x}{4}\sqrt{x^2 - 4(ae^y-z)} + (z-ae^y)ln\mid x + \sqrt{x^2 - 4(ae^y-z) }\mid + ae^y + b$

where b is an arbitrary constant