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:



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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.