Example (1)Solve The Differential Equation:${ \frac{d Y}{d X} = \frac{3 X^2 - 2}{2 Y - 1} }$

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Introduction

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Differential equations are a fundamental concept in mathematics, and solving them is a crucial skill for students and professionals alike. In this article, we will focus on solving a specific differential equation, which is given by:

$$\frac{d y}{d x} = \frac{3 x^2 - 2}{2 y - 1}$$

This equation is a first-order differential equation, and it can be solved using various techniques. In this article, we will use the method of separation of variables to solve this equation.

Understanding the Equation

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Before we dive into the solution, let's understand the equation. The equation is a first-order differential equation, which means that it involves a derivative of a function with respect to a single variable. In this case, the derivative is with respect to x, and the function is y.

The equation is given by:

$$\frac{d y}{d x} = \frac{3 x^2 - 2}{2 y - 1}$$

This equation can be rewritten as:

$$\frac{d y}{d x} = \frac{3 x^2 - 2}{2 y - 1}$$

$$\frac{d y}{d x} (2 y - 1) = 3 x^2 - 2$$

$$2 y \frac{d y}{d x} - \frac{d y}{d x} = 3 x^2 - 2$$

Separation of Variables

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To solve this equation, we will use the method of separation of variables. This method involves separating the variables x and y, and then integrating both sides of the equation.

Let's start by separating the variables. We can do this by multiplying both sides of the equation by (2y - 1), which is the denominator of the right-hand side.

$$2 y \frac{d y}{d x} - \frac{d y}{d x} = 3 x^2 - 2$$

$$\frac{d y}{d x} (2 y - 1) = 3 x^2 - 2$$

Now, we can separate the variables by dividing both sides of the equation by (2y - 1).

$$\frac{d y}{2 y - 1} = \frac{3 x^2 - 2}{2 y - 1} d x$$

Integrating Both Sides

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Now that we have separated the variables, we can integrate both sides of the equation.

$$\int \frac{d y}{2 y - 1} = \int \frac{3 x^2 - 2}{2 y - 1} d x$$

To integrate the left-hand side, we can use the substitution method. Let's substitute u = 2y - 1.

$$\int \frac{d y}{2 y - 1} = \int \frac{d u}{u}$$

$$\int \frac{d y}{2 y - 1} = \ln |u| + C_1$$

$$\int \frac{d y}{2 y - 1} = \ln |2 y - 1| + C_1$$

To integrate the right-hand side, we can use the substitution method. Let's substitute u = 2y - 1.

$$\int \frac{3 x^2 - 2}{2 y - 1} d x = \int \frac{3 x^2 - 2}{u} d x$$

$$\int \frac{3 x^2 - 2}{2 y - 1} d x = \int \frac{3 x^2 - 2}{u} d x$$

$$\int \frac{3 x^2 - 2}{2 y - 1} d x = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\int \frac{3 x^2 - 2}{2 y - 1} d x = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

Evaluating the Integrals

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Now that we have integrated both sides of the equation, we can evaluate the integrals.

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

To evaluate the integral on the right-hand side, we can use the substitution method. Let's substitute u = 2y - 1.

$$\frac{1}{2} \int \frac{3 u^2 - 4}{u} d x = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\frac{1}{2} \int \frac{3 u^2 - 4}{u} d x = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\frac{1}{2} \int \frac{3 u^2 - 4}{u} d x = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

Simplifying the Equation

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Now that we have evaluated the integrals, we can simplify the equation.

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

Solving for y

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Now that we have simplified the equation, we can solve for y.

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

$$\ln |2 y - 1| + C_1 = \frac{1}{2} \int \frac{3 u^2 - 4}{u} d x$$

\ln |2 y - 1<br/> # **Frequently Asked Questions: Solving the Differential Equation** ===========================================================

Q: What is a differential equation?

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A: A differential equation is a mathematical equation that involves an unknown function and its derivatives. It is a fundamental concept in mathematics and is used to model a wide range of phenomena in physics, engineering, economics, and other fields.

Q: What is the method of separation of variables?

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A: The method of separation of variables is a technique used to solve differential equations by separating the variables and integrating both sides of the equation. It is a powerful method that can be used to solve a wide range of differential equations.

Q: How do I know if a differential equation can be solved using the method of separation of variables?

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A: To determine if a differential equation can be solved using the method of separation of variables, you need to check if the equation can be written in the form:

$$\frac{d y}{d x} = f(x) g(y) </span></p> <p>If the equation can be written in this form, then it can be solved using the method of separation of variables.</p> <h2><strong>Q: What is the substitution method?</strong></h2> <hr> <p>A: The substitution method is a technique used to solve differential equations by substituting a new variable for one of the variables in the equation. It is a powerful method that can be used to solve a wide range of differential equations.</p> <h2><strong>Q: How do I use the substitution method to solve a differential equation?</strong></h2> <hr> <p>A: To use the substitution method to solve a differential equation, you need to substitute a new variable for one of the variables in the equation. You then need to integrate both sides of the equation and solve for the new variable. Finally, you need to substitute back the original variable and solve for the original variable.</p> <h2><strong>Q: What is the general solution of a differential equation?</strong></h2> <hr> <p>A: The general solution of a differential equation is a solution that contains one or more arbitrary constants. It is a solution that satisfies the differential equation for all values of the independent variable.</p> <h2><strong>Q: What is the particular solution of a differential equation?</strong></h2> <hr> <p>A: The particular solution of a differential equation is a solution that satisfies the differential equation for a specific value of the independent variable. It is a solution that is obtained by substituting a specific value for the independent variable into the general solution.</p> <h2><strong>Q: How do I find the particular solution of a differential equation?</strong></h2> <hr> <p>A: To find the particular solution of a differential equation, you need to substitute a specific value for the independent variable into the general solution. You then need to solve for the arbitrary constants and substitute back the original value for the independent variable.</p> <h2><strong>Q: What is the initial condition of a differential equation?</strong></h2> <hr> <p>A: The initial condition of a differential equation is a condition that is specified at a specific value of the independent variable. It is a condition that is used to determine the particular solution of the differential equation.</p> <h2><strong>Q: How do I use the initial condition to find the particular solution of a differential equation?</strong></h2> <hr> <p>A: To use the initial condition to find the particular solution of a differential equation, you need to substitute the initial condition into the general solution. You then need to solve for the arbitrary constants and substitute back the original value for the independent variable.</p> <h2><strong>Q: What is the final answer of a differential equation?</strong></h2> <hr> <p>A: The final answer of a differential equation is the solution that satisfies the differential equation for all values of the independent variable. It is a solution that contains no arbitrary constants.</p> <h2><strong>Q: How do I find the final answer of a differential equation?</strong></h2> <hr> <p>A: To find the final answer of a differential equation, you need to substitute the initial condition into the general solution. You then need to solve for the arbitrary constants and substitute back the original value for the independent variable. Finally, you need to simplify the solution and obtain the final answer.</p> <h2><strong>Q: What are some common mistakes to avoid when solving differential equations?</strong></h2> <hr> <p>A: Some common mistakes to avoid when solving differential equations include:</p> <ul> <li>Not checking if the equation can be written in the form of a separable differential equation</li> <li>Not using the correct method to solve the differential equation</li> <li>Not checking if the solution satisfies the initial condition</li> <li>Not simplifying the solution to obtain the final answer</li> </ul> <h2><strong>Q: What are some tips for solving differential equations?</strong></h2> <hr> <p>A: Some tips for solving differential equations include:</p> <ul> <li>Always check if the equation can be written in the form of a separable differential equation</li> <li>Use the correct method to solve the differential equation</li> <li>Always check if the solution satisfies the initial condition</li> <li>Always simplify the solution to obtain the final answer</li> <li>Practice, practice, practice! The more you practice solving differential equations, the more comfortable you will become with the methods and techniques.</li> </ul>$$