Consider The Differential Equation D Y D X = Y − 1 X 2 \frac{dy}{dx} = \frac{y-1}{x^2} D X D Y = X 2 Y − 1 , Where X ≠ 0 X \neq 0 X  = 0 .A) On The Axes Provided, Sketch A Slope Field For The Given Differential Equation At The Nine Points Indicated.B) Find The Particular Solution

Mar 10, 2025 by ADMIN 286 views

**Solving Differential Equations: A Step-by-Step Guide**

Introduction

Differential equations are a fundamental concept in mathematics, used to describe the behavior of physical systems over time. In this article, we will explore the solution to the differential equation $\frac{dy}{dx} = \frac{y-1}{x^2}$ , where $x \neq 0$ . We will first sketch a slope field for the given differential equation at the nine points indicated, and then find the particular solution.

Sketching the Slope Field

To sketch the slope field, we need to evaluate the differential equation at the nine points indicated. The slope field is a graphical representation of the solution to the differential equation, showing the direction of the solution at each point.

Evaluating the Differential Equation at Each Point

Point	x	y
1	1	0
2	1	1
3	1	2
4	2	0
5	2	1
6	2	2
7	3	0
8	3	1
9	3	2

We will evaluate the differential equation at each point using the formula $\frac{dy}{dx} = \frac{y-1}{x^2}$ .

Evaluating the Differential Equation at Point 1

At point 1, $x = 1$ and $y = 0$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{0-1}{1^2} = -1

The slope at point 1 is -1.

Evaluating the Differential Equation at Point 2

At point 2, $x = 1$ and $y = 1$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{1-1}{1^2} = 0

The slope at point 2 is 0.

Evaluating the Differential Equation at Point 3

At point 3, $x = 1$ and $y = 2$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{2-1}{1^2} = 1

The slope at point 3 is 1.

Evaluating the Differential Equation at Point 4

At point 4, $x = 2$ and $y = 0$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{0-1}{2^2} = -\frac{1}{4}

The slope at point 4 is $-\frac{1}{4}$ .

Evaluating the Differential Equation at Point 5

At point 5, $x = 2$ and $y = 1$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{1-1}{2^2} = 0

The slope at point 5 is 0.

Evaluating the Differential Equation at Point 6

At point 6, $x = 2$ and $y = 2$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{2-1}{2^2} = \frac{1}{4}

The slope at point 6 is $\frac{1}{4}$ .

Evaluating the Differential Equation at Point 7

At point 7, $x = 3$ and $y = 0$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{0-1}{3^2} = -\frac{1}{9}

The slope at point 7 is $-\frac{1}{9}$ .

Evaluating the Differential Equation at Point 8

At point 8, $x = 3$ and $y = 1$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{1-1}{3^2} = 0

The slope at point 8 is 0.

Evaluating the Differential Equation at Point 9

At point 9, $x = 3$ and $y = 2$ . Plugging these values into the differential equation, we get:

\frac{dy}{dx} = \frac{2-1}{3^2} = \frac{1}{9}

The slope at point 9 is $\frac{1}{9}$ .

Plotting the Slope Field

Using the slopes calculated above, we can plot the slope field for the given differential equation.

Finding the Particular Solution

To find the particular solution, we need to integrate the differential equation with respect to x.

\frac{dy}{dx} = \frac{y-1}{x^2}

Integrating both sides with respect to x, we get:

\int \frac{dy}{dx} dx = \int \frac{y-1}{x^2} dx

y = \int \frac{y-1}{x^2} dx

Using the substitution method, we can rewrite the integral as:

y = \int \frac{y-1}{x^2} dx = \int \frac{y}{x^2} dx - \int \frac{1}{x^2} dx

y = \frac{y}{x} - \frac{1}{x} + C

where C is the constant of integration.

Solving for y

To solve for y, we can rearrange the equation as:

y - \frac{y}{x} = -\frac{1}{x} + C

y\left(1 - \frac{1}{x}\right) = -\frac{1}{x} + C

y = \frac{-\frac{1}{x} + C}{1 - \frac{1}{x}}

Simplifying the expression, we get:

y = \frac{-1 + Cx}{x - 1}

Conclusion

Introduction

Differential equations are a fundamental concept in mathematics, used to describe the behavior of physical systems over time. In our previous article, we explored the solution to the differential equation $\frac{dy}{dx} = \frac{y-1}{x^2}$ , where $x \neq 0$ . In this article, we will answer some frequently asked questions about differential equations and provide additional insights into solving these types of equations.

Q&A

Q: What is a differential equation?

A: A differential equation is an equation that involves an unknown function and its derivatives. It is used to describe the behavior of physical systems over time.

Q: What is the difference between a differential equation and a regular equation?

A: A regular equation is an equation that involves only known quantities, whereas a differential equation involves an unknown function and its derivatives.

Q: How do I solve a differential equation?

A: To solve a differential equation, you need to find the unknown function that satisfies the equation. This can be done using various methods, such as separation of variables, integration, or numerical methods.

Q: What is the slope field?

A: The slope field is a graphical representation of the solution to a differential equation. It shows the direction of the solution at each point.

Q: How do I plot the slope field?

A: To plot the slope field, you need to evaluate the differential equation at various points and plot the resulting slopes.

Q: What is the particular solution?

A: The particular solution is the solution to a differential equation that satisfies the initial conditions.

Q: How do I find the particular solution?

A: To find the particular solution, you need to integrate the differential equation with respect to x and apply the initial conditions.

Q: What is the general solution?

A: The general solution is the solution to a differential equation that does not satisfy the initial conditions.

Q: How do I find the general solution?

A: To find the general solution, you need to integrate the differential equation with respect to x and leave the constant of integration arbitrary.

Q: What is the difference between the particular solution and the general solution?

A: The particular solution is the solution to a differential equation that satisfies the initial conditions, whereas the general solution is the solution to a differential equation that does not satisfy the initial conditions.

Q: Can I use numerical methods to solve differential equations?

A: Yes, you can use numerical methods to solve differential equations. These methods involve approximating the solution using numerical techniques.

Q: What are some common numerical methods for solving differential equations?

A: Some common numerical methods for solving differential equations include Euler's method, Runge-Kutta method, and finite difference method.

Q: How do I choose the right numerical method for my problem?

A: To choose the right numerical method, you need to consider the complexity of the problem, the desired accuracy, and the computational resources available.

Conclusion

In this article, we have answered some frequently asked questions about differential equations and provided additional insights into solving these types of equations. We hope that this article has been helpful in clarifying some of the concepts and methods involved in solving differential equations.

Additional Resources

For more information on differential equations, we recommend the following resources:

Textbooks: "Differential Equations and Dynamical Systems" by Lawrence Perko, "Ordinary Differential Equations" by Morris Tenenbaum and Harry Pollard
Online Courses: "Differential Equations" by MIT OpenCourseWare, "Ordinary Differential Equations" by Stanford University
Software: MATLAB, Mathematica, Python libraries such as SciPy and NumPy

We hope that this article has been helpful in your journey to learn about differential equations. Happy solving!