Question

solve x²(y-z)p+y²(z-x)q=z²(x-y)=0

Answer 1

Given the equation:

$$x²(y-z)p+y²(z-x)q-z²(x-y)=0$$

We can rewrite the equation as:

$$x²(y-z)p+y²(z-x)q=z²(x-y)$$

The above equation is of the form:

$$Pp +Qq = R$$

Where:

$$P = x²(y-z)\\ Q=y²(z-x)\\ R= z²(x-y)$$

The auxilliary equation is of the form:

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

Substituting for P, Q and R:

$$\frac{dx}{x²(y-z)}=\frac{dy}{y²(z-x)}=\frac{dz}{z²(x-y)} \qquad \cdots eqn (i)$$

Solvijng the above:

$$\frac{\frac{dx}{x²}}{y-z}=\frac{\frac{dy}{y²}}{z-x}=\frac{\frac{dz}{z²}}{x-y}$$

$$\frac{\frac{dx}{x²}+\frac{dy}{y²}+\frac{dz}{z²}}{y-z+ z-x+x-y}=k \qquad \text{(say)}\\ \frac{\frac{dx}{x²}+\frac{dy}{y²}+\frac{dz}{z²}}{0}=k\\ \implies \frac{dx}{x²}+\frac{dy}{y²}+\frac{dz}{z²}=0$$

Integratiing through:

$$\int \frac{dx}{x²}+\int \frac{dy}{y²}+\int \frac{dz}{z²}=\int 0\\ -\frac{1}{x}-\frac{1}{y}-\frac{1}{z}= c\\ \implies c_1 = -\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \Big)$$

Also,

$$\frac{\frac{dx}{x}}{x(y-z)}=\frac{\frac{dy}{y}}{y(z-x)}=\frac{\frac{dz}{z}}{z(x-y)}\\ \frac{-\frac{dx}{x}}{x(z-y)}=\frac{-\frac{dy}{y}}{y(x-z)}=\frac{-\frac{dz}{z}}{z(y-x)}\\ \frac{-\frac{dx}{x} + (-\frac{dy}{y})+ (-\frac{dz}{z})}{x(z-y)+y(x-z)+z(y-x)} = k \qquad \text{(say)}\\ \frac{-\frac{dx}{x} -\frac{dy}{y}-\frac{dz}{z}}{xz-xy+xy-yz+yz-xy} = k\\ \frac{-\frac{dx}{x} -\frac{dy}{y}-\frac{dz}{z}}{0} = k\\ \implies -\frac{dx}{x} -\frac{dy}{y}-\frac{dz}{z} = 0$$

Integrating through:

$$\int-\frac{dx}{x}\int -\frac{dy}{y}\int-\frac{dz}{z}=\int 0\\ -\int\frac{1}{x}dx-\int \frac{1}{y}dy-\int\frac{1}{z}dz=\int 0\\ - \ln x - \ln y - \ln z = \ln c\\ \ln x^{-1} + \ln y^{-1} + \ln z^{-1} = \ln c\\ \ln \Big(\frac{1}{x}\Big) + \ln \Big(\frac{1}{y}\Big) + \ln \Big(\frac{1}{z}\Big) = \ln c\\ \ln \Big(\frac{1}{xyz} \Big) =\ln c\\ \frac{1}{xyz} = c\\ \implies c_2 = \frac{1}{xyz}$$

Therefore, the general solution is:

$$f(u,v) = \Bigg(-\Big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \Big), \frac{1}{xyz} \Bigg) = 0$$