Show that the system $$ \frac{d x}{d t}=-2 y, \quad \frac{d y}{d t}=x-5, $$ has closed elliptical orbits. (Hint: Consider first transforming \(x\) to \(x-5\) and \(y\) to \(y / 2\) and then using a polar transformation.)

To better understand elliptical orbits, we first transform the system's coordinates from Cartesian to polar. Polar coordinates describe a point in terms of its distance from a reference point (called the radius, \( r \)) and the angle formed with a reference direction (called the angle, \( \theta \)).
This transformation simplifies the analysis of problems where symmetry and rotation are involved. By converting the variables \( x \) and \( y \) into polar coordinates, we can leverage trigonometric relationships to further simplify the differential equations. For our problem, the new transformed coordinates become \( x' = r \cos(\theta) \) and \( y' = r \sin(\theta) \).
This helps us deal with the motion and behavior of the system more intuitively.

A key element to recognize in this problem is the presence of closed orbits. Closed orbits imply that the motion of the system repeats itself in a cyclical pattern, forming closed loops.
For the system \( \frac{dx}{dt} = -2y \) and \( \frac{dy}{dt} = x - 5 \), transforming to a new set of variables allows us to see these closed orbits more clearly. The transformation \( x' = x - 5 \) and \( y' = \frac{y}{2} \) is used to shift and scale the original variables.
Through this transformation, it becomes easier to analyze the phase space of the system and identify the elliptical paths that represent the closed orbits.

Transforming a differential system is often necessary to make complicated systems more tractable. For our system, we start by redefining \( x \) and \( y \) as \( x' = x - 5 \) and \( y' = \frac{y}{2} \).
Next, we substitute these new variables into the original differential equations. This gives us a transformed system: \( \frac{dx'}{dt} = -4y' \) and \( \frac{dy'}{dt} = x' \).
The goal is to simplify the equations, making them easier to solve or analyze. By choosing appropriate transformations, we can often reduce the complexity of the system, revealing underlying structures or symmetries.

The chain rule is essential when dealing with transformations, particularly in differential equations. It relates the rate of change of a function to the rates of change of its variables.
For instance, when we transform our variables to polar coordinates, we need the chain rule to express how derivatives like \( \frac{dx'}{dt} \) and \( \frac{dy'}{dt} \) change according to \( \frac{dr}{dt} \) and \( \frac{d\theta}{dt} \).

• For \( x' \), we have \( \frac{dx'}{dt} = \cos(\theta) \frac{dr}{dt} - r \sin(\theta) \frac{d\theta}{dt} \).
• Similarly, for \( y' \), we have \( \frac{dy'}{dt} = \sin(\theta) \frac{dr}{dt} + r \cos(\theta) \frac{d\theta}{dt} \).

Using the chain rule correctly allows us to connect Cartesian and polar derivatives seamlessly.

Understanding the concept of constant radius in polar coordinates is crucial, especially when dealing with closed orbits. In a system where the radius \( r \) does not change over time \( \left( \frac{dr}{dt} = 0 \right) \), the motion is constrained to a circle in the polar plane.
In our problem, we found that \( \frac{dr}{dt} = 0 \), indicating that \( r \) remains constant. This simplifies the analysis and shows that the closed orbits are circles, which in this transformed context translate to ellipses in the original coordinates.
This insight helps us conclude that the original system features elliptical orbits, providing a clear visualization of the motion dynamics.