Question

Solve. (X²-yz)p +(y²-zx)q =z²-xy

Answer 1

Answer on Question #43673 - Math - Differential Calculus|Equation

Problem. Solve. $(X^2 - yz)p + (y^2 - zx)q = z^2 - xy$

Remark.

We suppose that $p = \frac{\partial z}{\partial x}$ and $q = \frac{\partial z}{\partial y}$.

We need to solve

$$(x^2 - yz) \frac{\partial z}{\partial x} + (y^2 - zx) \frac{\partial z}{\partial y} = z^2 - xy.$$

Solution.

The auxiliary equations are

$$\frac{dx}{x^2 - yz} = \frac{dy}{y^2 - zx} = \frac{dz}{z^2 - xy}.$$

Hence

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

or

$$\frac{d(x - y)}{(x - y)(x + y + z)} = \frac{d(y - z)}{(y - z)(x + y + z)} = \frac{d(z - x)}{(z - x)(x + y + z)}.$$

Therefore

$$\frac{d(x - y)}{x - y} = \frac{d(y - z)}{y - z} = \frac{d(z - x)}{z - x}$$

The solutions of the equations are

$$\ln |x - y| = \ln |y - z| + \ln C_1,$$ $$\ln |y - z| = \ln |z - x| + \ln C_2$$

or

$$\frac{x - y}{y - z} = C_1,$$ $$\frac{y - z}{z - x} = C_2.$$

So the general solution of the equation is

$$\phi\left(\frac{x - y}{y - z}, \frac{y - z}{z - x}\right) = 0,$$

where $\phi$ is a differentiable arbitrary function.

Answer: The general solution of equation is

$$\phi\left(\frac{x - y}{y - z}, \frac{y - z}{z - x}\right) = 0,$$

where $\phi$ is a differentiable arbitrary function.

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