Question

(a) Find an equation of the form \(H(x, y)=c\) satisfied by the trajectories. (b) Plot several level curves of the function \(H\). These are trajectories of the given system. Indicate the direction of motion on each trajectory. $$ d x / d t=y, \quad d y / d t=2 x+y $$

Short answer

Answer: The equation representing the trajectories of the given system of differential equations is \(H(x, y) = x - (2x^2 \cdot \frac{x - C_1}{y} + C_2) = c\), where \(c\) is a constant. In order to determine the direction of motion for each trajectory, we need to examine the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) along the level curves and calculate the slope \(\frac{dy}{dt}/\frac{dx}{dt}\).

Step-by-step solution

Integrate the first equation

We will first integrate the equation \(\frac{dx}{dt} = y\) with respect to time \(t\): $$ \int dx = \int y dt \Longrightarrow x = yt + C_1, $$ where \(C_1\) is the constant of integration.

Solve for \(y\) from the first equation

Rearrange the equation to get \(y\) as a function of \(x\) and \(t\): $$ y = \frac{x - C_1}{t}. $$

Integrate the second equation with substitution

Now integrate the equation \(\frac{dy}{dt} = 2x + y\) with respect to time \(t\). We will substitute the expression for \(y\) obtained in Step 2: $$ \frac{d}{dt} \left(\frac{x - C_1}{t}\right) = 2x + \frac{x - C_1}{t}. $$ To solve this, we will integrate both sides with respect to \(t\) and simplify the result.

Integrate and simplify

Integrate both sides: $$ \int \frac{d}{dt} \left(\frac{x - C_1}{t}\right) dt = \int \left(2x + \frac{x - C_1}{t}\right) dt. $$ This simplifies to: $$ x - C_1 = 2x^2t + C_2, $$ where \(C_2\) is another constant of integration.

Solve for the function \(H(x, y)\)

Now, we can rewrite the equation as $$ H(x, y) = x - (2x^2t + C_2) = c, $$ where \(c\) is a constant. Since \(y = \frac{x - C_1}{t}\), we can substitute this into the equation for \(H(x, y)\): $$ H(x, y) = x - (2x^2 \cdot \frac{x - C_1}{y} + C_2) = c. $$

Plot level curves and indicate direction of motion

In order to find the level curves of the given function, we set \(H(x, y) = c\) for different values of \(c\). After finding these level curves, we plot them on the \(xy\)-plane, indicating the direction of motion at each point by examining the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). The direction will be determined by the slope \(\frac{dy}{dt}/\frac{dx}{dt}\).

Key concepts

Level Curves

In understanding differential equations and their solutions, one crucial concept is that of level curves. Level curves are lines or curves on a graph that represent points where a function takes on a certain constant value. In the context of our exercise, the function was represented by the equation \(H(x, y) = c\), with \(c\) being a constant.

When we plot the level curves for various values of \(c\), what we are effectively doing is mapping out a 'contour map' of the function \(H(x, y)\) over the \(xy\)-plane. This is analogous to topographical maps that show lines of equal elevation. For a given differential equation, these curves depict possible paths, or 'trajectories,' that the system may follow over time.

By plotting several of these level curves, as suggested in part (b) of the exercise, we gain insight into the behavior of the system. It's important to note that each level curve corresponds to a different trajectory, and these trajectories do not intersect, as they represent different constant values of the function. This information is highly useful in visualizing the movement or flow described by the differential equations.

Trajectory Direction

Determining the trajectory direction is a vital step in analyzing the behavior of a system described by differential equations. Each point on a level curve corresponds to a position in which the system might be found at a certain time, and the direction of the trajectory at that point indicates the system's movement from that point.

Within our exercise, the trajectory direction at each point is ascertained by considering the ratio \(\frac{dy}{dt}/\frac{dx}{dt}\). This ratio essentially represents the slope of the tangent to the curve at that point. Remember, \(\frac{dx}{dt}=y\) and \(\frac{dy}{dt}=2x+y\), so the slope at any given point is driven by the values of \(x\) and \(y\) at that point.

By evaluating the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), and hence the slope, we can determine whether the trajectory is moving upwards, downwards, to the right, or to the left on the level curves. Indicating the direction of motion on the trajectories provides a more comprehensive understanding of the system's dynamics over time. It's akin to knowing not only the possible paths a hiker can take in the mountains (the level curves) but also having arrows on the paths showing which way the hiker is heading.

Constant of Integration

The constant of integration is a fundamental aspect in the field of differential equations. Whenever we integrate a function, there's an infinite number of antiderivatives, as adding any constant to a function doesn't change its derivative. These constants are represented by symbols like \(C_1\) and \(C_2\) in our exercise and are crucial for general solutions.

In the given problem, after integrating both \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we encounter constants of integration. The final form of the function \(H(x, y)\) still contains these constants because the general solution to the differential equations encompasses an entire family of functions that describe all possible trajectories. Without additional conditions, such as initial values or boundary conditions, these constants remain undetermined.

The inclusion of the constants of integration hints at the infinitely many possible solutions (trajectories) that exist for the system. Only by applying specific conditions can we isolate a particular solution that matches the constraints of a real-world scenario. Learning to work with these constants is a crucial skill for any student studying differential equations and is integral to understanding the vast range of potential solutions these equations may provide.