Question

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. \(A\) detailed direction field is not needed. $$y^{\prime}(t)=y-3, y(0)=1$$

Short answer

In this problem, we are asked to sketch the direction field for the given differential equation \(y^{\prime}(t)=y-3\) on the window \([-2,2] \times [-2,2]\) and then sketch the solution curve corresponding to the initial condition \(y(0)=1\). Following the steps outlined in the solution, we first create a grid of points in the given window, calculate the slope at each grid point using the given differential equation, and draw short line segments with the calculated slopes to create the direction field. Next, we use separation of variables to find the particular solution of the differential equation corresponding to the initial condition. We obtain the particular solution \(y(t)=3-2e^{t}\). Finally, we sketch the particular solution on the direction field by plotting the curve \(y(t)=3-2e^{t}\) and evaluating the function at different values of \(t\). This allows us to visualize the behavior of the solution as it satisfies both the differential equation and the given initial condition.

Step-by-step solution

Create a grid of points

Form a set of equally spaced points in the window \([-2,2] \times [-2,2]\) for the variables \(t\) (horizontal axis) and \(y\) (vertical axis). For simplicity, we can choose a step of 1, resulting in a 5 by 5 grid.

Calculate the slope at each grid point

Use the given differential equation \(y^{\prime}(t)=y-3\) to find the slope at each grid point. The slope is simply the expression \(y-3\).

Draw short line segments

At each grid point, draw a short line segment with the slope calculated in Step 2. Connect these line segments to create a visual representation of the direction field.

Find the particular solution

Use the method of separation of variables to find the particular solution corresponding to the initial condition \(y(0)=1\). Divide both sides of the equation by \(y-3\) and integrate with respect to \(t\): $$\int \frac{dy}{y-3} = \int dt$$ This results in: $$\ln|y-3|=t+C$$ where \(C\) is the integration constant. Solve for \(y\): $$y(t)=3+Ce^{t}$$ Now, apply the initial condition \(y(0)=1\) to find the specific value of \(C\): $$1=3+Ce^{0} \Rightarrow C=-2$$ Thus, the particular solution is: $$y(t)=3-2e^{t}$$

Sketch the particular solution on the direction field

Plot the curve \(y(t)=3-2e^{t}\) on the direction field by evaluating the function at different values of \(t\). You can start at \(t=0\) (since \(y(0)=1\)) and move along the curve using the calculated slopes from the direction field.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve an unknown function and its derivatives. They describe the rate of change of quantities and are widely used in various scientific fields such as physics, engineering, and economics. In this exercise, we are dealing with the first-order differential equation \(y'(t) = y - 3\), which relates the rate of change of \(y\) to its current value.

The nature of differential equations can vary from simple, where separation of variables can be used, to complex, which might require numerical methods to solve. This makes them a crucial tool in modeling behaviors and predicting future states of dynamic systems.

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with a given initial condition. The initial condition provides a specific point through which the solution curve must pass. In our example, the condition \(y(0) = 1\) specifies the value of \(y\) at time \(t = 0\).

• Initial conditions are crucial because they determine the particular solution out of an infinite family of solutions to the differential equation.
• These conditions ensure that the model reflects the actual state or position at a starting point.

Such information is essential for accurately modeling real-world scenarios. Without it, the model may fail to predict the correct future behavior.

Separation of Variables

Separation of variables is a method used to solve first-order differential equations. It involves rearranging the equation to separate variables on opposite sides, which can then be integrated independently.

For the equation \(y'(t) = y - 3\), we separate variables as follows: \(\frac{dy}{y-3} = dt\). By integrating both sides, we find \(\ln|y-3| = t + C\), where \(C\) is the constant of integration.

This method is particularly useful when the variables can be cleanly separated and allows us to find explicit solutions for non-linear equations. Remember: every time you integrate, consider the constant \(C\) that captures the general solution family. Then use the given initial condition to determine the specific solution.

Solution Curves

Solution curves are graphical representations of the solutions of differential equations. They show how the solution behaves over time and can be drawn on direction fields for visualization. In this example, the solution curve for \(y(t) = 3 - 2e^{t}\) is drawn on the direction field.

• The curve allows us to easily see how the solution behaves as \(t\) increases or decreases.
• By plotting this curve, one can identify the stability and other properties of the solution.

Sketching solution curves on a direction field adds depth to our understanding of differential equations, allowing us to appreciate the dynamics of the system beyond mathematics alone.