Question

x(y − z)p + y(z − x)q = z(x − y)

Answer 1

$\displaystyle x(y - z)p + y(z - x)q = z(x -y) \\ \frac{\mathrm{d}x}{x(y - z)} = \frac{\mathrm{d}y}{y(z - x)} = \frac{\mathrm{d}z}{z(x -y)} \\ \textsf{Choosing}\,\, \left(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\right)\,\, \textsf{as multipliers} \\ \frac{\mathrm{d}x}{x} + \frac{\mathrm{d}y}{y} + \frac{\mathrm{d}z}{z} = 0 \\ \int \frac{\mathrm{d}x}{x} + \int \frac{\mathrm{d}y}{y} + \int \frac{\mathrm{d}z}{z} = 0 \\ \ln(x) + \ln(y) + \ln(z) = C \\ \ln(xyz) = C \\ xyz = C \\ \textsf{Choosing}\,\, (1, 1, 1) \,\, \textsf{as multipliers} \\ \int\mathrm{d}x + \int\mathrm{d}y + \int\mathrm{d}z = 0\\ x + y + z = C \\ \therefore \textsf{The solution to the PDE is}\,\,\, \phi(x + y + z, xyz) = 0.$