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Step-by-Step Solution
Step 1: Write the Velocity Components
Given:
$$\vec{v} = k \bigl(y \hat{i} + x \hat{j}\bigr).$$
Here, $k$ is a constant, and $\vec{v}$ is the velocity vector of the particle in the $xy$-plane.
Step 2: Relate Velocity to Time Derivatives of Coordinates
The velocity vector can also be written as
$$\vec{v} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}.$$
Comparing components, we get:
$$\frac{dx}{dt} = k\,y,$$
$$\frac{dy}{dt} = k\,x.$$
Step 3: Find the Slope of the Trajectory
Divide the expressions for $\frac{dy}{dt}$ and $\frac{dx}{dt}$ to eliminate $t$:
$$
\frac{dy/dt}{dx/dt} = \frac{k\,x}{k\,y} \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{x}{y}.
$$
Step 4: Separate Variables and Integrate
Rewrite the differential equation and integrate both sides:
$$
\frac{dy}{dx} = \frac{x}{y} \quad \Longrightarrow \quad y\,dy = x\,dx.
$$
Integrate:
$$
\int y\,dy = \int x\,dx,
$$
which gives
$$
y^2 = x^2 + C,
$$
where $C$ is the constant of integration.
Step 5: Interpret the Final Equation
The general equation of the path of the particle is
$$y^2 = x^2 + \text{constant},$$
which matches the correct answer.
