Answer to Question #170019 in Differential Equations for Hiralal

Question

4xyz=pq+2px^2y+2qxy^2

Answer 1

$4xyz = pq + 2px^2y + 2qxy^2\\ f(x,y,z,p,q) = pq + 2px^2y + 2qxy^2 - 4xyz = 0 ... (*)\\ f_x = 4pxy+2qy^2 - 4yz \\ f_y = 2px^2 + 4 qxy - 4xz\\ f_z = -4xy \\ f_p = q+2x^2y\\ f_q =p+2xy^2$

By charpit method,

$\dfrac{dx}{-q-2x^2y} = \dfrac{dy}{-p-2xy^2} = \dfrac{dz}{-2pq-2px^2y - 2qxy^2} = \dfrac{dp}{2qy^2-4yz} = \dfrac{dq}{2px^2-4xz}$

$\dfrac{xdy-ydx}{qy-px} = \dfrac{xdp-ydq}{2xy(qy-px)} \\ xdp -ydq = 0\\ xp = yq \\ p = \frac{y}{x}q$

Substituting into (*), we have

$yq^2 +4x^2y^2q-4x^2yz = 0 \\ q = -2x^2y \pm 2x \sqrt{x^2y^2+z} \\$

Taking positive sign only

$q = -2x^2y + 2x \sqrt{x^2y^2 +z} \\ \therefore p = -2xy^2 + 2y \sqrt{x^2y^2+z}$

Substituting the value of p and q into pdx+qdy = dz and also integrating gives

$(-2xy^2 + 2y \sqrt{x^2y^2+z} )dx + ( -2x^2y+ 2x \sqrt{x^2y^2+z} )dy = dz$

Integrating term by term, we have

$z = -x^2y^2 +xy\sqrt{x^2y^2 +z} + z ln|y(\sqrt{x^2y^2 +z}+xy)|+ a -x^2y^2 +xy\sqrt{x^2y^2 +z} + z ln|x(\sqrt{x^2y^2 +z}+xy)| + b$

i.e

$z = -2x^2y^2 +2xy\sqrt{x^2y^2+ z} + z [ln|x(\sqrt{x^2y^2+ z} + xy)| + ln|y(\sqrt{x^2y^2+ z} + xy)|] + a + b$ where a and b are constants