Answer to Question #195906 in Differential Equations for Anuradhasingh adau

Question

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

Answer 1

We have 

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

 

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

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

Now also 

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

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

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

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

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

By integration, we get

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

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

Hence, solution is :

$$c_2 = f(c_1)$$

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

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