# How can I derive differential equations for polar coordinates based on these equations?

A textbook I am using on my own to study differential equations contains a problem: given the two differential equations for $x,y$ below and a real value of $t$, derive the differential equations for the functions $r(t)$ and $θ(t)$, where $x=rcosθ$ and $y = rsinθ$, and $F$ is a function defined in a later part of the problem:

$$\frac{dx}{dt}=-xF\left(\sqrt{x^2 + y^2}\right)-y$$

$$\frac{dy}{dt}=-yF\left(\sqrt{x^2 + y^2}\right)+x$$

By using the chain rule formula and plugging in the values $x=r\cosθ, y=r\sinθ$ into the original equations, I am able to derive:

$$\frac{dx}{dt}=\frac{\partial x}{\partial r}\frac{dr}{dt}+\frac{\partial x}{\partial θ}\frac{dθ}{dt}=cosθ\frac{dr}{dt}-r\sinθ\frac{d\theta}{dt}=-rF(r)\cosθ-r\sinθ$$

$$\frac{dy}{dt}=\frac{\partial y}{\partial r}\frac{dr}{dt}+\frac{\partial y}{\partial θ}\frac{dθ}{dt}=\sinθ\frac{dr}{dt}+r\cosθ\frac{d\theta}{dt}=-rF(r)\sinθ+r\cosθ$$

However, I don't know how to proceed from here. The solution given in the book skips a lot of steps and I can't follow how the author derives the equations. In the book the author jumps from the above equations to:

$$(\cos^2θ+\sin^2θ)\frac{dr}{dt}=-r(\cos^2θ+\sin^2θ)F(r)$$

$$r(\sin^2θ+\cos^2θ)\frac{dθ}{dt}=r(\cos^2θ+\sin^2θ)$$

The above equations pretty much solve themselves, revealing:

$$\frac{dr}{dt}=-rF(r)$$ $$\frac{dθ}{dt}=1$$

But I have no idea how the author got there. Where did the $(\sin^2θ+\cos^2θ)$ in the third step come from?

The equations \begin{align} \cos\theta\frac{dr}{dt}-r\sinθ\frac{d\theta}{dt}&=-rF(r)\cos\theta-r\sin\theta\\ \sin\theta\frac{dr}{dt}+r\cosθ\frac{d\theta}{dt}&=-rF(r)\sin\theta+r\cos\theta \end{align} are a linear system of two equations in the two unknowns $dr/dt$ and $d\theta/dt$. Solve it and you get the desired result. For instance, to obtain $dr/dt$ multiply the first equation by $\cos\theta$, the second by $\sin\theta$ and add them.