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. $$ d x / d t=2 x^{2} y-3 x^{2}-4 y, \quad d y / d t=-2 x y^{2}+6 x y $$

Short answer

Question: Find an equation \(H(x, y) = c\) that is satisfied by the trajectories, and then plot several level curves of this function, given the following first-order differential equations: \(dx/dt = 2x^{2}y - 3x^{2} - 4y\) and \(dy/dt = -2xy^{2} + 6xy\). Answer: The equation \(H(x, y) = c\) for the trajectories is given by \(H(x, y) = x^{3}y - x^{3} + xy^{3} - 7xy + C = c\). To plot the level curves, set different constant values for \(c\) and use computer software or graphing tools to generate the plots. To indicate the direction of motion on each trajectory, examine the signs of \(dx/dt\) and \(dy/dt\) from the given differential equations.

Step-by-step solution

Find function H(x, y) such that dH/dt = 0 along the trajectories

To find a function \(H(x, y)\) such that \(dH/dt = 0\) along the trajectories, we need to find a function that satisfies: $$ \frac{dH}{dt} = \frac{\partial H}{\partial x}\frac{dx}{dt} + \frac{\partial H}{\partial y}\frac{dy}{dt} = 0. $$ Substitute the given \(dx/dt\) and \(dy/dt\) expressions into the equation: $$ \frac{\partial H}{\partial x}(2x^{2}y - 3x^{2} - 4y) + \frac{\partial H}{\partial y}(-2xy^{2} + 6xy) = 0. $$ Now, we will try to find a function H(x, y) that satisfies this equation.

Integrate the partial derivatives to obtain H(x,y)

Integrate the first term with respect to x and the second term with respect to y: $$ H(x, y) = \int(2x^{2}y - 3x^{2} - 4y)dx + \int(-2xy^{2} + 6xy) dy. $$ Integrating each term, we have: $$ H(x, y) = x^{3}y - x^{3} - 4xy + xy^{3} - 3xy + C, $$ where C is an integration constant. Rewrite the expression: $$ H(x, y) = x^{3}y - x^{3} + xy^{3} - 3xy - 4xy + C = x^{3}y - x^{3} + xy^{3} - 7xy + C. $$ Thus, we get the desired equation \(H(x, y) = c\).

Plot level curves of H(x,y)

To plot several level curves of \(H(x, y)\), we can set different constant values for \(c\) and plot the corresponding curves: $$ H(x, y) = x^{3}y - x^{3} + xy^{3} - 7xy + C = c. $$ For the level curves, we can use computer software or graphing tools to generate the plots for different values of \(c\).

Indicate the direction of motion on the trajectories

To indicate the direction of motion on each trajectory, we can examine the signs of \(dx/dt\) and \(dy/dt\) from the given differential equations. (a) If \(dx/dt>0\), then the particle moves to the right, and if \(dx/dt<0\), it moves to the left. (b) If \(dy/dt>0\), the particle moves up, and if \(dy/dt<0\), it moves down. By examining the signs of \(dx/dt\) and \(dy/dt\), we can determine the direction of motion on each trajectory on the level curves of \(H(x,y)\).

Key concepts

Level Curves

Imagine hiking on a mountain, where each elevation is marked with a specific line on the map. These lines are like 'level curves' in mathematics, representing points of a certain height. In the context of differential equations, level curves, also known as contour lines, represent the solutions of an equation where the function is constant, marked by the value 'c'.

For instance, if you have a function representing the topography of a landscape, the level curves indicate the 'flat' regions at different elevations. When dealing with differential equations, these curves illustrate the trajectories over a graph where the rate of change is zero—meaning there is no 'movement' in the value of the function, making them essential in understanding the behavior of dynamical systems. By plotting several level curves for different values of 'c', one can visualize the landscape of solutions and analyze the nature of the trajectories that the system passes through.

Trajectories in Differential Equations

In our mountain metaphor, a trajectory would be the actual path you trek while ascending or descending, indicating your direction and position over time. Similarly, in differential equations, trajectories represent the paths that solutions follow as they evolve. The exercise prompts us to discern these trajectories by finding level curves for the function 'H(x, y) = c'.

These trajectories are governed by the system of differential equations presented, implying how an object, or a single point in our set of equations, moves over time in a 2D space defined by 'x' and 'y'. When plotting level curves for different constants 'c', we observe that these curves are essentially the snapshots of the trajectory at various stages. By determining the direction of motion, as indicated in steps 3 and 4 of our solution, we decipher the dynamical behavior of the system, which is imperative for predicting future states or understanding the stability of the equilibrium points within the system.

Method of Integration for Solving Equations

The method of integration is like a treasure hunt where you need to find the hidden treasure - in our case, the function 'H(x, y)'. This treasure opens up a world of understanding about the system's behavior. Integration involves summing up tiny parts to figure out the total—here, we're summing up infinitely small changes to find the general form of our function.

To solve our treasure hunt, we solve two separate integrals stemming from the partial derivatives of 'H' with respect to 'x' and 'y'. We then piece these parts together to construct our function 'H(x, y)'. We face an incremental process of discovery that, when pieced together, reveals the larger picture of how the entire system evolves over time. This method is very powerful as it converts a complex system of differential equations into a manageable function that represents the entire spectrum of possible solutions.