Question

z=(x-y)F(x²+y²) obtain the partial differential equation by eliminating the arbiyrary function from the following equation

Answer 1

Let us obtain the partial differential equation by eliminating the arbiyrary function from the following equation $z=(x-y)F(x^2+y^2).$ Since $z_x'=F(x^2+y^2)+(x-y)F'(x^2+y^2)2x$ and $z_y'=-F(x^2+y^2)+(x-y)F'(x^2+y^2)2y,$ we conclude that $yz_x'=yF(x^2+y^2)+2xy(x-y)F'(x^2+y^2)$ and $xz_y'=-xF(x^2+y^2)+2xy(x-y)F'(x^2+y^2).$ After substracting we get the equation $yz_x'-xz_y'=yF(x^2+y^2)+xF(x^2+y^2).$ Then $yz_x'-xz_y'=(x+y)F(x^2+y^2),$ and hence $F(x^2+y^2)=\frac{yz_x'-xz_y'}{x+y}.$ It follows that $z=(x-y)F(x^2+y^2)=(x-y)\frac{yz_x'-xz_y'}{x+y}.$ Consequently, the partial differential equation is the following:

$z(x+y)=(x-y)(yz_x'-xz_y').$