Question

Solve (X^2-y^2-Z^2)p+2xyq=2xz

Answer 1

Solution:-

By arranging we get

$(y^2+z^2-x^2)p-2xyq+2xz=0\\ \frac{dx}{y^2+z^2-x^2}=\frac{dy}{-2xy}=\frac{dz}{-2xz}\\ \frac{dy}{-2xy}=\frac{dz}{-2xz}\\ \frac{dy}{y}=\frac{dz}{z}\\ \log y=\log z+\log C_1\\ C_1=\frac{y}{z}\\ \text{each fraction}=\frac{xdx+ydy+zdz}{x(y^2+z^2-x^2)-2xy^2-2xz^2}=\\ =\frac{xdx+ydy+zdz}{-x(y^2+z^2+x^2)}\\ \frac{xdx+ydy+zdz}{-x(y^2+z^2+x^2)}=\frac{dy}{-2xy}\\ \frac{2xdx+2ydy+2zdz}{y^2+z^2+x^2}=\frac{dy}{y}\\ \log (y^2+z^2+x^2)=\log y+\log C_2\\ C_2=\frac{y^2+z^2+x^2}{y}\\ f(C_1, C_2)=0\\$

$f(\frac{y}{2}, \frac{y^2+z^2+x^2}{y})=0$