Solve the equation
$$\frac{dx}{x^2+b^2} =\frac{dy}{xy-bz}=\frac{dz}{xz+by}$$
How to show the curves are conics.
First of all , I need to find its integral curves.
I tried to solve. But I get strange results. So I cannot write there. Sorry. Please show me way. Thanks
I found one solution.
$$\frac{xdy+bdz}{x^2y-bxz+bxz+b^2z}=\frac{dx}{x^2+b^2}$$
Then I get $bz=c_1$
which integrals do you need? these?
$\int \frac{dx}{x^2+b^2} =A=\frac{1}{b} \left(- \frac{i}{2} \log{\left (- i b + x \right )} + \frac{i}{2} \log{\left (i b + x \right )}\right)$
$\int \frac{dy}{xy-bz}=B=\frac{1}{x} \log{\left (- b z + x y \right )}$
$\int\frac{dz}{xz+by}=C=\frac{1}{x} \log{\left (- b y + x z \right )}$