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# Answer to Question #223859 in Chemical Engineering for Lokika

Question #223859

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

1
2021-08-23T04:48:39-0400

"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."

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