Chapter 5: Problem 6
Find the general solution and three particular solutions. \(y^{\prime}=\frac{3}{x}+x^{2}-x^{4}\)
Short Answer
Expert verified
The general solution is \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} + C \). Three particular solutions are for \( C = 0, 1, -2 \).
Step by step solution
01
Identify the Type of ODE
The given differential equation is \( y^{\prime} = \frac{3}{x} + x^2 - x^4 \). This is a first-order ordinary differential equation, which is non-homogeneous and separable.
02
Integrate to Find the General Solution
To find the general solution, integrate the right-hand side of the equation with respect to \( x \). The integral becomes: \( \int \left( \frac{3}{x} + x^2 - x^4 \right) \, dx = \int \frac{3}{x} \, dx + \int x^2 \, dx - \int x^4 \, dx \).
03
Evaluate Each Integral
Compute each integral separately:1. \( \int \frac{3}{x} \, dx = 3 \ln |x| \)2. \( \int x^2 \, dx = \frac{x^3}{3} \)3. \( \int x^4 \, dx = \frac{x^5}{5} \)
04
Construct the General Solution
Combine the results of the integrals from Step 3 to form the general solution:\( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} + C \), where \( C \) is an arbitrary constant.
05
Find Three Particular Solutions
To find particular solutions, choose specific values for \( C \). Here are three examples:1. If \( C = 0 \), then \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} \).2. If \( C = 1 \), then \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} + 1 \).3. If \( C = -2 \), then \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} - 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
General Solution
In the context of differential equations, the **general solution** represents a family of functions that satisfy a given ordinary differential equation (ODE). It encompasses every possible solution, distinguished by arbitrary constants. In our example, the integration process transforms the differential equation into a form where a constant, denoted as \( C \), is included. This constant represents the infinite number of solutions consistent with the general solution. The general solution for our given first-order ODE is:
- \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} + C \)
Particular Solutions
A **particular solution** is a specific instance of the general solution, found by assigning a definite value to the arbitrary constant \( C \) present in the general solution. This turns the family of solutions into a single, specific trajectory that can be plotted or used for further analysis. By selecting certain values for \( C \), we tailor the general solution to meet particular conditions or constraints. Here are examples of particular solutions derived from our general solution:
- For \( C = 0 \): \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} \)
- For \( C = 1 \): \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} + 1 \)
- For \( C = -2 \): \( y(x) = 3 \ln |x| + \frac{x^3}{3} - \frac{x^5}{5} - 2 \)
First-Order ODE
A **first-order ordinary differential equation** (ODE) is a relationship involving a function and its first derivative. It provides a powerful tool for modeling dynamic processes in which the rate of change of a quantity depends on the quantity itself and possibly other variables. The given differential equation is of the form \( y' = f(x, y) \), where only the first derivative (\( y' \)) is present, making it a first-order ODE. We can break it down further as non-homogeneous because it includes terms that solely depend on \( x \) (such as \( \frac{3}{x}, x^2, -x^4 \)). This means that besides the function \( y \), the equation also involves explicit terms of \( x \). First-order ODEs like the one in our exercise often require straightforward techniques like integration for their solutions. Understanding the structure and behavior of first-order ODEs equips us to analyze simple yet insightful models in fields like physics, biology, and economics.