Answer to Question #141784 in Differential Equations for Vivek Kumar

Question

Solve -dx/x(x+y) = dy/y(x+y) = dz/(x-y)(2x+2y+z)

Answer 1

We have $\frac{dx}{-x(x+y)} = \frac{dy}{y(x+y)} = \frac{dz}{(x-y)(2x+2y+z)}$ ______________(1)

Now from first two part of equation (1),

$\frac{dx}{-x(x+y)} = \frac{dy}{y(x+y)}$

$\implies \frac{dx}{-x} = \frac{dy}{y}$

Now by integration, we get $-\ln(x) = \ln(y) - \ln(c_1) \implies c_1 = xy$ ________(2)

Now also $\frac{dx}{-x(x+y)} = \frac{dy}{y(x+y)} = \frac{dz}{(x-y)(2x+2y+z)} = \frac{2dx+2dy+dz}{(x-y)z}$

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

$\implies \frac{dz}{2x+2y+z} = \frac{2dx+2dy+dz}{z}$

$\implies zdz = (2x+2y+z) (2dx+2dy+dz)$

Hence by integration, we get

$\frac{z^2}{2} = \frac{(2x+y+z)^2}{2} + \frac{c_2}{2}$

$\implies c_2 = z^2- (2x+y+z)^2$ ______________(3)

Hence, solution is :

$c_2 = f(c_1)$

$\implies z^2- (2x+y+z)^2 = f(xy)$ .