Consider the equations $$ \frac{d x}{d t}=y, \quad \frac{d y}{d t}=-x . $$ (b) By transforming variables, obtain $$ \frac{d r^{2}}{d t}=0, \quad \frac{d \theta}{d t}=-1 . $$ (b) Conclude that there are circular solutions. What is the direction of rotation? Are these cycles stable?

A differential equation is a mathematical equation that relates some function with its derivatives. In this exercise, we explore two first-order differential equations: $$ \frac{d x}{d t}=y, \frac{d y}{d t}=-x.$$ These equations describe how the variables x and y change with respect to time t. These equations are interdependent, meaning the derivative of one variable depends on the value of the other variable.

To analyze these differential equations more easily, we can transform them into another coordinate system, such as polar coordinates. Polar coordinates represent points in terms of their distance (radius) from the origin and their angle from a fixed direction. This transformation can simplify the equations and make the patterns in the solutions more evident.

By converting our equations to polar coordinates, where x = r cos(θ) and y = r sin(θ), we can obtain new expressions that demonstrate the circular nature of the solutions. Transforming our original equations, we get: $$ \frac{d r^{2}}{d t}=0, \frac{d \theta}{d t}=-1.$$ These new equations indicate that:

• The term \frac{d r^2}{d t} = 0 means that r^2 is constant over time, or in other words, r (the radius) does not change. This implies motion along a circle with a fixed radius.
• The term \frac{d \theta}{d t} = -1 indicates that the angle θ is decreasing at a uniform rate, pointing to a clockwise rotation.

This transformation reveals that the solutions to our differential equations are circular, indicating a periodic behavior in the system. Thus, the system exhibits circular motion with a constant radius.

Stability analysis involves determining how a system responds to small disturbances or perturbations. For circular solutions, we want to know if slight changes in radius or angle will cause the system to significantly deviate from its original path.

For our transformed equations:$$ \frac{d r^{2}}{d t}=0 \text{ and } \frac{d \theta}{d t}=-1, $$we see that any small change in the radius r would not change the nature of motion along the circle because \frac{d r^2}{d t} = 0 always holds true. This indicates that the radius remains fixed.

Thus, these circular solutions are stable in the sense that a perturbation in the radius will not grow or diminish over time. However, any deviation in θ simply changes the phase of the rotation but the uniform circular motion in a clockwise direction persists.

In summary, the system is stable because small perturbations in r do not amplify over time, ensuring that the circular motion continues unaltered.