Question

sketch the trajectory corresponding to the solution satisfying the specified initial conditions, and indicate the direction of motion for increasing t. $$ d x / d t=-x, \quad d y / d t=2 y ; \quad x(0)=4, \quad y(0)=2 \quad \text { and } \quad x(0)=4, \quad y(0)=0 $$

Short answer

Short Answer: The trajectories for the given system of ODEs with initial conditions are as follows: For the first particular solution (x(0)=4, y(0)=2), the trajectory is a curve starting at point (4,2) and moving towards the positive y-axis while approaching the x-axis. For the second particular solution (x(0)=4, y(0)=0), the trajectory is a horizontal line at y=0, with the motion moving from right to left along the x-axis as time increases.

Step-by-step solution

Find the general solution

To find the general solution of the given ODEs, we will solve each equation separately. For the first ODE, we have \(dx/dt = -x\). This is a first-order linear differential equation, which can be written as \(dx/x = -dt\). Integrating both sides, we get: $$ \int \frac{dx}{x} = \int -dt \\ \ln|x| = -t + C_1 \\ x(t) = e^{-t+C_1} = e^{-t}e^{C_1} \\ $$ Let \(k_1 = e^{C_1}\), then \(x(t) = k_1e^{-t}\). Now, for the second ODE, we have \(dy/dt = 2y\). This is also a first-order linear differential equation, and we proceed similarly: $$ \int \frac{dy}{y} = \int 2dt \\ \ln|y| = 2t + C_2 \\ y(t) = e^{2t+C_2} = e^{2t}e^{C_2} \\ $$ Let \(k_2 = e^{C_2}\), then \(y(t) = k_2e^{2t}\). The general solution is: $$ x(t) = k_1e^{-t} \\ y(t) = k_2e^{2t} $$

Determine the particular solutions using the initial conditions

Now, we determine the particular solutions by satisfying the given initial conditions. We'll consider two cases: Case 1: Initial conditions x(0) = 4, y(0) = 2 Implement the initial conditions in the general solution: $$ x(0) = k_1e^{-0} \rightarrow 4 = k_1 \\ y(0) = k_2e^{2 \cdot 0} \rightarrow 2 = k_2 \\ $$ Thus, for the first initial conditions, the particular solution is: $$ x_1(t) = 4e^{-t} \\ y_1(t) = 2e^{2t} $$ Case 2: Initial conditions x(0) = 4, y(0) = 0 Implement the initial conditions in the general solution: $$ x(0) = k_1e^{-0} \rightarrow 4 = k_1 \\ y(0) = k_2e^{2 \cdot 0} \rightarrow 0 = k_2 \\ $$ Thus, for the second initial conditions, the particular solution is: $$ x_2(t) = 4e^{-t} \\ y_2(t) = 0 $$

Sketch the trajectories and indicate the direction of motion

To sketch the trajectories, we will plot the two particular solutions we found on a Cartesian plane. Indicate the direction of motion for increasing t with arrows. For the first particular solution: As \(t\) increases, \(x_1(t) = 4e^{-t}\) decreases while \(y_1(t) = 2e^{2t}\) increases. The trajectory will be a curve that starts at the initial point (4,2) and goes towards the positive y-axis while approaching the x-axis. For the second particular solution: Since \(y_2(t) = 0\), the trajectory will be a horizontal line at y=0. The point (4,0) represents x(0) and as t increases, x decreases while y remains constant at 0. The trajectory moves from the right to left along the x-axis.

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They're an essential part of modeling the way systems change over time, whether it be in physics, engineering, economics, or biology. In essence, they are the language in which the laws of nature are expressed.

When dealing with first-order linear differential equations, like those in our example, the equation involves the first derivative of the function and can be typically written in the form \( \frac{dx}{dt} = f(x) \). The solution of such an equation gives us a function that describes how a certain quantity changes over time. In our exercise, the two separate differential equations \( dx/dt = -x \) and \( dy/dt = 2y \) each describe an exponential rate of change: one decreasing and the other increasing exponentially.

Initial Conditions

Initial conditions specify the starting point or state of the system described by a differential equation at a particular time. They are crucial for determining the exact solution, which satisfies both the differential equation and the initial conditions. In the context of our exercise, we have the initial conditions \( x(0) = 4 \) and \( y(0) = 2 \) for the first scenario, and \( x(0) = 4 \) and \( y(0) = 0 \) for the second. These conditions allow us to find constants \( k_1 \) and \( k_2 \) in the general solution, thus, yielding the specific trajectories that the system will follow over time.

Integration of Differential Equations

Integrating a differential equation involves finding the antiderivative, or integral, of both sides of the equation, which provides a function that describes the behavior of the system. The process of integration moves us from the rate of change (the derivative) to the actual quantity that changes over time (the function itself).

In our step-by-step solution, both equations are integrated to find \( \ln|x| \) and \( \ln|y| \) before solving for \( x(t) \) and \( y(t) \). Integration constants \( C_1 \) and \( C_2 \) are found using initial conditions and are critical for determining the particular trajectory of the system. This step converts our knowledge of the system's behavior (its rate of change) into an explicit expression of the system's state at any time \( t \).

Trajectory Sketching

Trajectory sketching is the process of graphically representing the solutions of a system of differential equations. It allows us to visualize the system's behavior and understand how different initial conditions can lead to distinct outcomes. The trajectory represents the path a system follows in the state space, with time as an implicit parameter affecting the motion.

In our exercise, one trajectory begins at (4,2) and moves towards the y-axis as time increases, while the other starts at (4,0) and moves left along the x-axis. To indicate the direction of motion for increasing time on our sketch, we use arrows. The sketch provides a tangible depiction of the abstract math, illustrating how the solutions evolve and making it easier to grasp the dynamic nature of the system.