Question

Find particular integrals of the following partial differential equation to represent

surfaces passing through the following curve:

(𝑦 – 𝑧)𝑝 + (𝑧 – 𝑥)𝑞 = 𝑥 – 𝑦 ; 𝑧 = 0, 𝑦 = 2𝑥

Answer 1

$\displaystyle (y - z)p + (z - x)q = x - y\\ \frac{\mathrm{d}x}{y - z} = \frac{\mathrm{d}y}{z - x} = \frac{\mathrm{d}x}{x - y}\\ \mathrm{d}x + \mathrm{d}y + \mathrm{d}z = 0\\ \int\mathrm{d}x + \int\mathrm{d}y + \int\mathrm{d}z = 0\\ x + y + z = c_1\\ \frac{\mathrm{d}y}{c_1 -2x - y} = \frac{\mathrm{d}x}{x - y}\\ (x - y)\mathrm{d}y + (2x + y - c_1)\mathrm{d}x = 0\\ x\mathrm{d}y + y\mathrm{d}x - y\mathrm{d}y + 2x\mathrm{d}x - c_1\mathrm{d}x = 0\\ \mathrm{d}(xy) - y\mathrm{d}y + 2x\mathrm{d}x - c_1\mathrm{d}x = 0\\ \int\mathrm{d}(xy) - \int y\mathrm{d}y + \int 2x\mathrm{d}x - \int c_1 \mathrm{d}x= \int 0\\ xy - \frac{y^2}{2} + x^2 - c_1x = c_2\\ 2x^2 - y^2 + 2xy - 2c_1= 2c_2\\ z = 0, y = 2x\\ x = t, y = 2t, z = 0\\ t + 2t + 0 = c_1, c_1 = 3t\\ 2t^2 - 4t^2 + 4t^2 - 2(3t) = 2c_2\\ 2t^2 - 6t = 2c_2 \\ t^2 - 3t = c_2\\ c_2 = t^2 - c_1\\ 2x^2 - y^2 + 2xy - 2c_1= 2(t^2 - c_1)\\ 2x^2 - y^2 + 2xy = 2t^2\\ 2x^2 - y^2 + 2xy = xy\\ 2x^2 - y^2 + xy = 0\\$