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

Short Answer

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In summary, to sketch the approximate integral curves of the differential equation \(y' = \frac{-x}{y}\), using the method of isoclines, follow these steps: 1. Find the isocline expression by setting the derivative equal to a constant value \(k\). In this case, the isocline expression is \(y = \frac{-x}{k}\). 2. Sketch the isoclines on the graph for different values of \(k\), creating a set of parallel lines. 3. Sketch the approximate integral curves by following the direction of the isoclines, ensuring that the integral curves are tangent to the isoclines and do not cross them.

Step by step solution

01

Understand Isoclines and Integral Curves

Isoclines are curves where the function representing the differential equation has the same slope. To find isoclines, we will set the derivative equal to a constant value and then solve the resulting equation for \(y\). Integral curves represent the set of functions satisfying the given differential equation. The shape of these curves can be determined by the isoclines' direction.
02

Find the Isocline Expression

The given differential equation is \(y' = \frac{-x}{y}\). To find the isoclines, set the derivative equal to a constant value \(k\). \(\frac{-x}{y} = k\) Now solve for \(y\): \(y = \frac{-x}{k}\) This is the expression for the isoclines. Each isocline will have a different slope value \(k\), which defines a family of parallel lines with varying intercepts.
03

Sketch the Isoclines

Sketch the isoclines on the graph for different values of \(k\). For example, when \(k = 1\), the isocline will be \(y = -x\); when \(k = -1\), the isocline will be \(y = x\). These lines will give you an idea of the direction at which the integral curves will travel.
04

Sketch the Approximate Integral Curves

Using the direction of the isoclines, start sketching the approximate integral curves following the isoclines' direction. Make sure that the integral curves are tangent to the isoclines and do not cross them. As you draw more integral curves, you will begin to see the pattern and shape of the integral curves for this differential equation. In conclusion, using the method of isoclines, you can sketch the approximate integral curves of the given differential equation \(y' = \frac{-x}{y}\). Remember that the integral curves represent the set of functions satisfying the given differential equation, while the isoclines are lines with the same constant slope.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Curves
Integral curves are the foundational concept in understanding how solutions behave for a differential equation. An integral curve is a graph of a function that satisfies a given differential equation. In simple terms, it's like a path traced out by the solutions of the differential equation. These paths smoothly connect points so that at any point on the curve, the slope of the curve matches the value given by the differential equation.

For the differential equation in our problem, such as \(y' = \frac{-x}{y}\), the integral curves are crucial. They help illustrate how the solutions behave over different points in space. The sketching of integral curves involves discerning the direction and trajectory of these curves based on information like isoclines and slope fields. As you sketch integral curves, ensure they align with the slope provided by the differential equation at every point, integrating over all possible slopes (isoclines) in the plane.
Differential Equations
Differential equations like \(y' = \frac{-x}{y}\) are equations involving derivatives of a function. They are powerful tools used to model many real-world phenomena, such as population dynamics, motion, and heat transfer. A differential equation provides a relationship between a function and its derivatives, typically indicating how a quantity changes over time or space.

Solutions to differential equations are not just single functions but may involve whole families of functions that satisfy the equation under different conditions. This is why concepts like isoclines and integral curves are important. They offer graphical ways to visualize the possible behavior of solutions. In this problem, the set of integral curves gives us a picture of all possible solutions that match the differential equation \(y' = \frac{-x}{y}\). Understanding how to solve differential equations is essential for predicting systems' behavior and solving scientific problems.
Slope Field
A slope field is a visual tool to help us comprehend the behavior of differential equations. It is a grid of small line segments or arrows drawn on a coordinate plane, where each segment's slope is determined by the differential equation at that point. This can make it much easier to see the general direction in which the integral curves are heading.

In the problem, using the slope field might involve drawing small lines at evenly spaced points which have the slope given by \(y' = \frac{-x}{y}\). This shows at each point the direction in which a solution to the differential equation moves. Consequently, by examining the slope field, one can sketch integral curves more accurately, ensuring they match the slopes at every point.

Slope fields can sometimes cover an entire plane, offering a comprehensive picture of how solutions evolve. They act as a guide, helping to intuitively predict and sketch the paths that potential solutions might take without solving the differential equation analytically. As a study aid, they make the abstract nature of differential equations much more accessible and easier to grasp visually.

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Most popular questions from this chapter

For each initial-value problem below, use the Euler method and a calculator to approximate the values of the exact solution at each given \(x .\) Obtain the exact solution \(\phi\) and evaluate it at each \(x\). Compare the approximations to the exact values by calculating the errors and percentage relative errors. \(y^{\prime}=y \sin x, \quad y(0)=0.5 . \quad\) Approximate \(\phi\) at \(x=0.2,0.4, \ldots, 1.0\). \((h=0.2)\)

Obtain a power series solution in powers of \(x\) of each of the initial-value problems by (a) the Taylor series method and (b) the method of undetermined coefficients. \(y^{\prime}=x+y, \quad y(0)=1\).

For each initial-value problem below, use the Euler method and a calculator to approximate the values of the exact solution at each given \(x .\) Obtain the exact solution \(\phi\) and evaluate it at each \(x\). Compare the approximations to the exact values by calculating the errors and percentage relative errors. \(y^{\prime}=\frac{x}{y}, \quad y(1)=0.2\) Approximate \(\phi\) at \(x=1.2,1.4, \ldots, 2.0 .\) \((h=0.2)\)

For each initial-value problem below, use the Runge-Kutta method and a calculator to approximate the values of the exact solution at each given \(x .\) Obtain the exact solution \(\phi\) and evaluate it at each \(x .\) Compare the approximations to the exact values by calculating the errors and percentage relative errors. \(y^{\prime}=\frac{x}{y}, \quad y(1)=0.2 . \quad(h=0.2)\) Approximate \(\phi\) at \(x=1.2,1.4, \ldots, 2.0\)

For each initial-value problem below, use the ABAM method and a calculator to approximate the values of the exact solution at each given \(x .\) Obtain the exact solution \(\phi\) and evaluate it at each \(x .\) Compare the approximations to the exact values by calculating the errors and percentage relative errors. Values for the approximations at \(x_{1}, x_{2}\), and \(x_{3}\) have been found by the Runge-Kutta method and are listed following each problem. \(y^{\prime}=x-2 y, \quad y(0)=0\). Approximate \(\phi\) at \(x=0.25,0.5, \ldots, 1.5\). \((h=0.25)\) \(\left(y_{1}=0.0266927083, y_{2}=0.0920427110, y_{3}=0.1808488325 .\right)\)

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