Question

Use the method of isoclines to sketch the approximate integral curves of each of the differential equations. \(y^{\prime}=\frac{3 x-y}{x+y}\).

Short answer

To sketch the approximate integral curves using the method of isoclines for the given differential equation \(y' = \frac{3x - y}{x + y}\), we first find the isoclines by setting the slope function equal to a constant \(c\), and solve for \(y\). For various values of \(c\), we obtain the following isoclines: - For \(c = -1\), \(y = \frac{3x - x^2}{-2}\) - For \(c = 0\), \(y = -3x\) - For \(c \approx 1\), \(y = 3x - x^2\) - For \(c = 2\), \(y = \frac{3x + x^2}{1}\) - For \(c = 3\), \(y = \frac{9x + 3x^2}{2}\) - For \(c = 4\), \(y = \frac{12x + 4x^2}{3}\) Now, sketch the integral curves by drawing solution curves tangent to the isoclines, connecting them smoothly, and ensuring that the curves do not cross each other. This will provide an approximate visual representation of the integral curves of the given differential equation using the method of isoclines.

Step-by-step solution

Identify the given differential equation

The given differential equation is \(y' = \frac{3x - y}{x + y}\).

Define the slope function

We define the slope function \(f(x, y)\) as the right-hand side of the given differential equation: \(f(x, y) = \frac{3x - y}{x + y}\)

Calculate the isoclines

An isocline represents the curve having a constant slope for the given differential equation. For a constant slope \(c\), we set the slope function equal to \(c\), and solve for \(y\): \(c = \frac{3x - y}{x + y}\) To find \(y\) in terms of \(x\) and \(c\), multiply both sides by \(x + y\): \(c(x + y) = 3x - y\) Now, solve for \(y\): \(y(c - 1) = 3cx - cx^2\) \(y = \frac{3cx - cx^2}{c - 1}\)

Sketch the isoclines for different values of c

Starting with \(c = -1\), we can systematically increase the value of \(c\) and plot the isoclines: - For \(c = -1\), \(y = \frac{3x - x^2}{-2}\) - For \(c = 0\), \(y = -3x\) - For \(c = 1\), the denominator becomes zero, resulting in an undefined isocline. Instead, we can consider the limit as \(c\) approaches \(1\). Take the limit: \(\lim_{c \to 1} \frac{3cx - cx^2}{c - 1} = 3x - x^2\) So, for \(c \approx 1\), \(y = 3x - x^2\) Now, similarly, find the isoclines for \(c = 2\), \(3\), and \(4\): - For \(c = 2\), \(y = \frac{3x + x^2}{1}\) - For \(c = 3\), \(y = \frac{9x + 3x^2}{2}\) - For \(c = 4\), \(y = \frac{12x + 4x^2}{3}\)

Sketch the integral curves

Using the previous step's isoclines, we can now sketch the integral curves by drawing solution curves tangent to the isoclines. This can be done by connecting the isoclines smoothly, starting from one isocline to another, following the tangent properties of the isoclines. Make sure that the curves do not cross each other and are as smooth as possible. By performing this step, you will obtain an approximate visual representation of the integral curves of the given differential equation using the isoclines.

Key concepts

Isoclines

In differential equations, isoclines are curves along which the slope of a solution curve is constant. These are particularly handy when you want to visualize solutions to differential equations without solving them explicitly. To find an isocline for a given slope, you take the slope function, equate it to a constant value, and solve for one variable in terms of the other. In the exercise you have, the slope function was set equal to different constants like

• -1
• 0
• 1
• 2

This process allows you to sketch several lines or curves, each representing a constant slope. For instance, the line for slope 0 (\(c = 0\)) was directly \(y = -3x\). By drawing these isoclines on a graph, they guide you in sketching the integral curves of the differential equation, allowing for the visualization of how solutions behave across different regions in the plane.
This visualization helps in understanding the general direction and behavior of solutions over a field.

Integral Curves

Integral curves are the solution paths that satisfy a differential equation. Imagine you have a slope field where each point in the plane has a tiny arrow indicating the slope at that point. The integral curve is a path that travels through these arrows smoothly. As mentioned in your exercise, after identifying isoclines, drawing integral curves starts by making these curves tangent to the isoclines.
The key is to smoothly connect these isoclines without the curves intersecting each other. Keep in mind that the integral curve is unique through every point, provided the function satisfies certain conditions. Drawing these curves precisely can lead to a better understanding of the solutions' trajectories over time or space. This is especially useful for systems where explicit solutions are complex or unsolvable analytically. They provide a graphical representation, aiding comprehension of the behavior the differential equation models, in fields like physics or biology.

Slope Function

The slope function in a differential equation determines the steepness or direction of the curve at any given point. It is represented by the right-hand side of the differential equation, often written in terms of variables \(x\) and \(y\). For the given equation \(y' = \frac{3x - y}{x + y}\), the slope function is \(f(x, y) = \frac{3x - y}{x + y}\).
The slope function is crucial because it provides information about which way the solution curves should turn as they pass each point on the plane. If you understand this function and how it changes with \(x\) and \(y\), you can deduce the qualitative behavior of the solution without needing the exact formula for the solution curve. Knowing these slopes is foundational for sketching isoclines and integral curves, as they indicate what the solution looks like graphically. Thus, by working with the slope function, you can map how systems evolve in their respective environments.