Answer to Question #138622 in Differential Equations for Malakyaseen

Question

Define the type of given partial differential equation and find the solution.
(y^2+z^2-x^2 )p+2xyq+2zx=0

Answer 1

Lagrange linear equation

$(y^2+z^2-x^2 )p+2xyq+2zx=0\\ (y^2 + z^2 - x^2)p + 2xyq =- 2zx$

from general form,

$Pp +Qq =R$

$P = y^2 + z^2 -x^2\\ Q = 2xy\\ R =-2zx$

From the auxiliary equation,

$\dfrac{dx}{P}= \dfrac{dy}{Q}= \dfrac{dz}{R}$

$\dfrac{dx}{y^2 +z^2 -x^2}= \dfrac{dy}{2xy}= \dfrac{dz}{-2zx} \quad------(i)$

Taking the last 2 ratios

$\dfrac{dy}{2xy}= \dfrac{dz}{-2zx}$

$\dfrac{dy}{y}= -\dfrac{dz}{z}$

Integrating;

$\log y = -(\log z) + \log c_1\\ \log y = -\log z + \log c_1\\ \log y +\log z = \log c_1\\ \log yz =\log c_1$

$yz = c_1$

$c_1 = yz \quad------(ii)$

Taking the first 2 ratios

$\dfrac{dx}{y^2 +z^2 -x^2}= \dfrac{dy}{2xy}$

but,

$c_1 = yz\\ \therefore z= \dfrac{c_1}{y}$

substituting the value of z into the first 2 ratios,

we have,

$\dfrac{dx}{y^2 +(\frac{c_1}{y})^2 -x^2}= \dfrac{dy}{2xy}$

$2xy\, dx = (y^2 +(\dfrac{c_1}{y})^2 -x^2)dy$

$2xy\, dx - (y^2 +(\dfrac{c_1}{y})^2 -x^2)dy = 0$

$2xy\, dx - (y^2 +\dfrac{c_1^2}{y^2}-x^2)dy = 0 \quad------(iii)$

Integrating equation (iii)

$\large \int2xy\, dx - \int(y^2 +\dfrac{c_1^2}{y^2}-x^2)dy = 0$

$\large x^2y-\dfrac{y^3}{3}+\dfrac{c_1^2}{y} = c_2$

but, $c_1 = yz$

$\therefore\large x^2y-\dfrac{y^3}{3}+yz^2 = c_2 \quad------(iv)$

$\therefore$ the general solution is;

$\large f(yz,\,\, x^2y- \dfrac{y^3}{3} +yz^2) = 0$