Trajectories of Differential Equation Systems with Complex Eigenvalues In an autonomous system of 1st order differential equations in order to find the trajectories one must solve
$$\frac{dy}{dx}=\frac{Ax+By}{Cx+Dy}$$
In the case of complex eigenvalues my notes say that the trajectories are found using
$$\ln (k\cdot x)=\frac{1}{2}\ln[(\omega-a)^{2}+b^{2}]-\frac{a}{b}\arctan\frac{\omega-a}{b}$$
where
$$\omega=\frac{Cx+iDx}{x}$$
I can't find how this relation occurs. How the $\arctan$ term came to being?
 A: $$\frac{dy}{dx}=\frac{Ax+By}{Cx+Dy}=\frac{A+By/x}{C+Dy/x}$$
Now substitute $y=xz$ and $y'=z+xz'$
$$z+xz'=\frac{A+Bz}{C+Dz}$$
$$xz'=\frac{A+Bz}{C+Dz}-z$$
This ODE is seprable.
Edit: $\arctan(x) = \frac{1}{2i} \log(\frac{x-i}{x+i})$.
A: If $y=vx$ then
$$\frac{dy}{dx}=x\frac{dv}{dx}+v=\frac{Ax+Bvx}{Cx+Dvx}=\frac{A+Bv}{C+Dv}$$
$$x\frac{dv}{dx}=\frac{A+Bv}{C+Dv}-v=\frac{A+Bv-Cv-Dv^2}{C+Dv}=-\frac{v^2-\frac{B-C}Dv-\frac DA}{v+\frac CD}$$
$$\frac{v+\frac CD}{v^2-\frac{B-C}Dv-\frac DA}dv=\frac{\left(v-\frac{B-C}{2D}\right)+\frac{B+C}{2D}}{\left(v-\frac{B-C}{2D}\right)^2-\left(\frac{B-C}{2D}\right)^2-\frac DA}dv=-\frac{dx}x$$
Then we can use the formula
$$\int\frac{dx}{x^2+b^2}=\frac1b\tan^{-1}\frac xb+C$$
And the fact that the eigenvalue is complex so
$$-\left(\frac{B-C}{2D}\right)^2-\frac DA=b^2$$
And let
$$\frac{B-C}{2D}=a$$
To get
$$\frac12\ln\left((v-a)^2+b^2\right)+\frac{B+C}{2Db}\tan^{-1}\frac{v-a}{b}=-\ln x+C_1=-\ln(kx)$$
So there are similarities with what you've got, but in general
$$\frac{B+C}{2D}\ne-a=\frac{C-B}{2D}$$
And here I have $v=\frac yx$ instead of $\omega$ as you have and also $-\ln(kx)$ rather than $\ln(kx)$ but I hope this can be viewed as progress in any case.
