example of isogonal trajectory

Determine the curves which intersect the origin-centered circles at an angle of ${45}^{\circ }$.

The differential equation of the circles  $x^2+y^2=R^2$  is  $2xdx+2ydy=0$,  i.e.

$$\frac{x}{y}+\frac{dy}{dx}=\mathrm{\hspace{0.33em}0}.$$

Thus, by the model (2) of the parent entry (http://planetmath.org/IsogonalTrajectory), the differential equation of the isogonal trajectory reads

${\displaystyle \frac{x}{y}}+{\displaystyle \frac{y^{\prime }-\tan \frac{\pi }{4}}{1+y^{\prime }\tan \frac{\pi }{4}}}=\mathrm{\hspace{0.33em}0},$(1)

which can be rewritten as

$$y^{\prime }=\frac{y-x}{y+x}=\frac{\frac{y}{x}-1}{\frac{y}{x}+1}.$$

Here, one may take  $\frac{y}{x}:=t$  as a new variable (see ODE types reductible to the variables separable case), when

$$y=xt,y^{\prime }=\frac{dy}{dx}=t+x\frac{dt}{dx},$$

and in the resulting equation

$$t+x\frac{dt}{dx}=\frac{t-1}{t+1}$$

one can separate the variables (http://planetmath.org/SeparationOfVariables):

$$\frac{1+t}{1+t^2}dt=-\frac{dx}{x}$$

Multiplying here by 2 and integrating then give

$$2\arctan t+\ln (1+t^2)=-2\ln x+\ln C^2\equiv -\ln \frac{x^2}{C^2},$$

or equivalently

$$\ln \frac{x^2+x^2{t}^2}{C^2}=-2\arctan t.$$

This is

$$\ln \frac{\sqrt{x^2+y^2}}{C}=-\arctan \frac{y}{x},$$

i.e.

$$\sqrt{x^2+y^2}=C{e}^{-\arctan \frac{y}{x}}.$$

Expressing this in the polar coordinates $r,\phi$ gives the family of the integral curves of the equation (1) in the form

$$r=C{e}^{-\phi }.$$

Consequently, the family of the isogonal trajectories consists of logarithmic spirals.