91Ó°ÊÓ

Polar coordinates offer a unique way to describe any point in a plane. Instead of using the traditional Cartesian system of
setting 'x' and 'y' values, polar coordinates work with a radius and an angle. In this system:

• The radius, denoted as 'r', represents the distance from the point to the origin.
• The angle, written as 'θ', measures the direction of the point from the positive x-axis. It's typically expressed in radians.

Using polar coordinates can simplify the process of solving complex differential equations, especially with curves
that maintain a circular symmetry. In this problem, converting Cartesian elements using the relations
\( x = r \cos \theta \) and \( y = r \sin \theta \) helps in dealing with the given differential equation,
as the intricate terms often become more manageable.

Integration plays a pivotal role in solving differential equations. Here, we employed a basic integration technique
on the equation \( r \, dr + d\theta = 0 \) derived from the transformation in polar coordinates. Integrating involves:

• Identifying separable parts so you can integrate each independently.
• Performing the integration to solve for constants, such as the arbitrary constant 'C' here.

For \( r \, dr \), the integration gives \( \frac{r^2}{2} \), demonstrating how areas under curves for such simple
forms are computed. Meanwhile, integrating \( d\theta \) produces \( \theta \). Together, these contribute to
solving the given differential equation effectively, allowing us to reach the desired expression \( \frac{r^2}{2} + \theta = C \). The use of such integration ensures that the various relationships between variables are correctly captured.
In this scenario, integration seamlessly transitions between variable relationships presented through differential equations.

Cartesian coordinates, involving the familiar 'x' and 'y' axes, typically describe the position of points in a
2D space. They provide a straightforward way to set up and analyze equations and geometry. In the given exercise,
we started with a differential equation in Cartesian coordinates and later transformed it using polar coordinates
to simplify the computation.
After solving the equation in polar terms, reverting back to Cartesian coordinates is crucial to
presenting the solution in a conventional manner. This involves substituting back \( r = \sqrt{x^2 + y^2} \) and
\( \theta = \arctan \left( \frac{y}{x} \right) \). The transformation back ensures that the solution resonates with the provided options and
completes the link between the polar and Cartesian systems, which is key in determining the final answer for our problem:
\( y = x \tan \left( C - \frac{x^2 + y^2}{2} \right) \). This final solution is presented in the Cartesian form aligning with
option (C), verifying the correctness of this approach.