Suppose that a curve $\alpha(t)$ is such that the angle $\phi$ between $\alpha'(t)$ and radial line is constant (and distinct from $\pi/2$) then show that $\alpha(t)=(ae^{bt}\cos(t),ae^{bt}\sin(t))$.
I've already proven the following equality for parametric curves:
$$\tan(\phi)=\frac{\frac{dy}{dx}x-y}{x+\frac{dy}{dx}y}$$ where $\frac{dy}{dx}:=\frac{dy/dt}{dx/dt}$
Supposing that $\tan(\phi)$ is a constant I tried to solve the differential equation but I got stuck.
I'm aware that the system is easier in polar coordinates but I'd like to see if there is a solution for this approach.
Assuming the tangent line is not radial (so $\tan(\frac\pi2+\phi)\in\mathbb{R}$ is a constant $k$) $$ x+y\frac{\mathrm{d}y}{\mathrm{d}x}=k\left(y-x\frac{\mathrm{d}y}{\mathrm{d}x}\right) $$ or equivalently $$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{k y-x}{kx+y}. $$ The RHS is homogeneous degree $0$ (i.e., scaling $(x,y)\mapsto(\lambda x,\lambda y)$ leaves the RHS unchanged for all $\lambda$), and we know the method to solve this class of equations is to let $y=zx$ and get $$ z+x\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{k z-1}{k+z} $$ which rearranges to $$ \frac{\mathrm{d}(\log x)}{\mathrm{d}z}= \frac1x\frac{\mathrm{d}x}{\mathrm{d}z}=\frac{z+k}{k z-1-z(z+k)}=-\frac{z+k}{1+z^2}. $$ Integrating gives $x=F(z)$ and hence you can express the solution either as a parametric curve $(x,y)(z)=(F(z),zF(z))$, or as an implicit curve $x=F(y/x)$.
N.B. Of course this only solve it away from points on the $y$-axis, but you could use $y$ instead of $x$ as the local parameter for the curve for points on the $y$-axis and let $x(y)=ys(y)$ for unknown function $s$, etc.