Answer to Question #138659 in Differential Equations for Mirza

Question

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

Answer 1

Given differential equation is $(y^2+z^2-x^2 )p+2xyq+2zx=0$

Then we can write,

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

Taking last two terms,

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

Integrating both sides, we get

$\int\frac{dy}{y}=\int \frac{dz}{-z}$

$log |y| = -log|z| + logC_1 \implies yz = C_1$ (1)

then $z = \frac{C_1}{y}$

taking first two terms,

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

putting value of z from equation (1),

$\frac{dx}{(y^2+\frac{C_1^2}{y^2}-x^2 )}=\frac{dy}{2xy}$

${(y^2+\frac{C_1^2}{y^2}-x^2 )}{dy}={2xy}{dx}$

${2xy}{dx} - {(y^2+\frac{C_1^2}{y^2}-x^2 )}{dy} = 0$ (2)

This equation is exact differential equation, as

$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = 2x$

where $M ={2xy}, N = - {(y^2+\frac{C_1^2}{y^2}-x^2 )}$

then integrating both sides equation (2),

$\int{2xy}{dx} - \int{(y^2+\frac{C_1^2}{y^2}-x^2 )}{dy} = C_2$

$x^2y-\frac{y^3}{3}+\frac{C_1^2}{y} = C_2$

putting value of $C_1$

$x^2y-\frac{y^3}{3}+yz^2 = C_2$ (3)

Hence equation (1) and (3) are the required solutions of the equation.