Question

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

Short answer

The trajectory for this motion is described by the equation: $$x = 4 e^{-\frac{2x}{y}}$$ Starting at the point (4,2) and moving towards the origin (0,0) as t increases. The direction of motion is towards the origin, since both x(t) and y(t) decrease with increasing t.

Step-by-step solution

Solve the differential equations independently

We have two independent differential equations: 1. $$\frac{dx}{dt} = -x$$ 2. $$\frac{dy}{dt} = -2y$$ We can solve each of these equations by using the method of separation of variables. For the first equation, we get: $$\frac{dx}{x} = -dt$$ Now, integrate both sides: $$\int\frac{dx}{x} = -\int dt$$ Which gives us: $$\ln|x| = -t + C_1$$ Where C_1 is the integration constant. Solve for x: $$x(t) = e^{-t+C_1} = e^{-t}e^{C_1}$$ Similarly, for the second equation, we get: $$\frac{dy}{y} = -2 dt$$ And after integrating both sides: $$\ln|y| = -2t + C_2$$ Where C_2 is another integration constant. Now, solve for y: $$y(t) = e^{-2t+C_2} = e^{-2t}e^{C_2}$$ Now we have the general solutions for both equations. Next, we will apply the initial conditions to obtain the particular solutions.

Apply the initial conditions

Given the initial conditions, we have: $$x(0)=4 \Rightarrow e^{-0+C_1}e^0 = 4 \Rightarrow e^{C_1}=4$$ And: $$y(0)=2 \Rightarrow e^{-2(0)+C_2}e^0 = 2 \Rightarrow e^{C_2}=2$$ So, the particular solutions for x(t) and y(t) are: $$x(t) = 4e^{-t}$$ $$y(t) = 2e^{-2t}$$

Sketch the trajectory and indicate the direction of motion for increasing t

The equations for x(t) and y(t) describe the trajectory of the motion. From these equations, we can see that as t increases, both x(t) and y(t) decrease: For x(t): $$\frac{d}{dt}x(t) = \frac{d}{dt}(4e^{-t}) = -4e^{-t}=-x(t)<0$$ For y(t): $$\frac{d}{dt}y(t) = \frac{d}{dt}(2e^{-2t}) = -4e^{-2t}=-2y(t)<0$$ To sketch the trajectory, consider the parametric equations: $$x = 4e^{-t}$$ $$y = 2e^{-2t}$$ Here, we can eliminate t by dividing both equations: $$\frac{y}{x} = \frac{2e^{-2t}}{4e^{-t}}$$ Simplifying, we get: $$\frac{y}{x} = \frac{1}{2}e^{-t}$$ Solve for $$e^t$$: $$e^t = \frac{2x}{y}$$ Now substitute this expression back into $$x(t)$$: $$x = 4 e^{-\frac{2x}{y}}$$ So, the trajectory for this motion will be described by: $$x = 4 e^{-\frac{2x}{y}}$$ This is the trajectory of an exponential decay curve starting at the point (4,2) and moving towards the origin (0,0) as t increases. The direction of motion will be towards the origin, as both x(t) and y(t) decrease with increasing t.

Key concepts

Separation of Variables

The method of separation of variables is a common technique used to solve ordinary differential equations (ODEs). It involves rearranging an equation to isolate the variables on different sides of the equation, hence 'separating' them. Let's break down how this works using a simple example.

For the differential equation \(\frac{dx}{dt} = -x\), we can separate the variables by moving all terms involving \(x\) to one side and all terms involving \(t\) to the other side, yielding \(\frac{dx}{x} = -dt\). What follows is the integration of both sides, leading to a function in terms of \(x\) and a function in terms of \(t\), coupled with an integration constant. This process converts the differential equation into an equation that is easier to handle and solve. It is widely used because it simplifies the equation to the point where the solution can be easily obtained by integration.

Integration Constants

When solving differential equations by integrating, we often encounter integration constants such as \(C_1\) and \(C_2\). These constants appear because when we integrate a function, we get a family of functions that differ by a constant. These are the 'plus C' terms you may remember from basic calculus.

In the context of differential equations, these constants play a vital role as they allow us to tailor our general solution to meet specific initial conditions or boundary values. For instance, in the given problem, after integrating, we obtained natural logarithms of the absolute values of \(x\) and \(y\) equal to the product of a negative time and the integration constants. By applying initial conditions, \(x(0) = 4\) and \(y(0) = 2\), we were able to solve for these constants and thus generate the particular solutions tailored to the problem's specifics. The presence of these constants helps transform a general solution into one that fits specific scenarios.

Exponential Decay

Exponential decay refers to a process where a quantity decreases at a rate proportional to its current value, a concept that's beautifully illustrated by the solutions to the given problem's differential equations. The expressions \(x(t) = 4e^{-t}\) and \(y(t) = 2e^{-2t}\) both suggest that as time \(t\) increases, the values of \(x\) and \(y\) diminish exponentially.

The negative exponents indicate the decay nature; the variable being multiplied by \(e\) to the power of a negative constant times \(t\) shows us that as \(t\) becomes larger, the overall expression shrinks rapidly. Exponential decay models are pervasive in natural phenomena, where they describe everything from the cooling of a hot object to radioactive decay. Understanding the behavior of these models is important not only in math but also in various scientific and engineering disciplines.

Parametric Equations

Parametric equations are a set of equations that express a set of quantities as explicit functions of one or more independent variables known as parameters. In the problem given, we have a system of parametric equations with \(t\) as the parameter: \(x = 4e^{-t}\) and \(y = 2e^{-2t}\).

The beauty of parametric equations lies in their ability to represent curves in the plane that may be difficult to describe using a single equation in just \(x\) and \(y\). By eliminating the parameter (in our case, \(t\)), we can find a relationship between \(x\) and \(y\) that defines the trajectory independently of time. This is particularly useful when we wish to visualize the path traced by a moving object, as it accounts for both the horizontal and vertical changes simultaneously, giving us a complete picture of the object's movement in space.