Answer to Question #205782 in Differential Equations for kaviya

Question

Form the PDE by eliminating the arbitrary function from 𝑓 ( 𝑥−𝑦 𝑦−𝑧 , 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥) = 0.

Answer 1

$𝑓 ( \frac{x-y}{y-z}, 𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥) = 0\\ let\space u=\frac{x-y}{y-z}, v=𝑥𝑦 + 𝑦𝑧 + 𝑧𝑥\\ Therefore, f(u,v)=0\\ u_x=\frac{1}{y-z}, \\u_y=\frac{-y+z-xq-yq}{(y-z)^2},\\ v_x=x+p,\\v_y=x+q\\ Now,\\ f_x=f_u.u_x+f_v.v_x=0\\ 0=f_u.(\frac{1}{y-z}) +f_v.(x+p)\\ f_u.(\frac{1}{y-z})=-f_v.(x+p) -----(1)\\ f_y=f_u.u_y+f_v.v_y=0\\ 0=f_u.(\frac{-y+z-xq-yq}{(y-z)^2}) +f_v.(x+q)\\ f_u.(\frac{-y+z-xq-yq}{(y-z)^2})=-f_v.(x+q)--(2) \\ \text{Divide equation 1 by 2, we get,}\\ \frac{\frac{1}{y-z}}{\frac{-y+z-xq-yq}{(y-z)^2}}=\frac{x+p}{x+q}\\ \implies \frac{y-z}{-y+z-xq-yq }=\frac{x+p}{x+q}\\ \text{This is the required solution.}$