Question

Eliminate the arbitrary function from 2=xy+f(x^2+y^2) to obtain the partial differencial equation.

Answer 1

$2=xy+f(x^2+y^2) \iff z =(x+y)f(x^2 - y^2)$

From $z =(x+y)f(x^2 - y^2)$ means that the solution of the partial differential equation is given by an arbitrary function $F( \frac{z}{(x+y)} , x^2-y^2 )=0$ of class C₂ with respect to variables x , y. Take $u = \frac{z}{(x+y)} , v = x^2 - y^2$ , so obtain $F( u,v ) =0$ .

Differentiating respect to x and y and taking $p = z_x , q = z_y$ gives

$F_uu_x + F_vv_x =( \frac{F_u(p(x+y) -z)}{(x+y)^2) +F_v(2x)}) =0$

$F_uu_y + F_vv_y = \frac{F_u( ( q(x+y) -z)}{(x+y)^2)} + F_v(-2y) =0$ .Eliminating $F_u$ and $F_v$ from the above equation gives $\frac{-2y( (p(x+y) -z)}{(x+y)^2)} -\frac{2x(q(x+y) -z)}{(x+y)^2)} =0$ .

Assuming x + y #0 leads to $yp(x+y) +xq(x+y) + (x+y)z =0$ .Follows the required partial differential equation : $yp + xq = -z.$