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Chapter 4: Problem 108

Use a CAS to solve the initial value problems in Exercises \(107-110\) . Plot the solution curves. $$y^{\prime}=\frac{1}{x}+x, \quad y(1)=-1$$

Short Answer

Expert verified
The solution is \( y(x) = \ln|x| + \frac{x^2}{2} - \frac{3}{2} \).

Step by step solution

01

Identify the Differential Equation and Initial Condition

The differential equation given is \( y' = \frac{1}{x} + x \) with the initial condition \( y(1) = -1 \). Our goal is to find the function \( y(x) \) that satisfies this differential equation and the given initial condition.
02

Use CAS to Solve the Differential Equation

Utilize a computer algebra system (CAS) like Mathematica, Maple, or Wolfram Alpha to solve the differential equation \( y' = \frac{1}{x} + x \). Solving it, we get the general solution \( y(x) = \ln|x| + \frac{x^2}{2} + C \), where \( C \) is a constant to be determined.
03

Apply the Initial Condition to Find C

Substitute the initial condition \( y(1) = -1 \) into the general solution to find the constant \( C \). This gives us \( -1 = \ln|1| + \frac{1^2}{2} + C \). Since \( \ln|1| = 0 \) and \( \frac{1^2}{2} = \frac{1}{2} \), we have \( -1 = \frac{1}{2} + C \). Solving for \( C \), we find \( C = -\frac{3}{2} \).
04

Write the Particular Solution

Using the value of \( C \) found in the previous step, write the particular solution. The function that satisfies both the differential equation and the initial condition is \( y(x) = \ln|x| + \frac{x^2}{2} - \frac{3}{2} \).
05

Plot the Solution Curve

Plot the solution curve \( y(x) = \ln|x| + \frac{x^2}{2} - \frac{3}{2} \) using a graphing tool or CAS. Ensure that the plot includes the initial condition point \((1, -1)\) to verify the correctness of the solution.

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Key Concepts

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

Initial Value Problems
Initial value problems are fundamental in differential equations. They involve finding a function that satisfies a given differential equation and meets specific initial conditions. Let's break it down:
  • Differential Equation: It can be of any form, involving derivatives of a function.
  • Initial Condition: This is the given value of the unknown function and possibly some of its derivatives at a specific point.For example, in the problem given, we have the differential equation \( y' = \frac{1}{x} + x \) with the initial condition \( y(1) = -1 \).
  • Solution: The goal is to find a function that satisfies both the differential equation and the initial condition.
Initial value problems are crucial in modeling real-world processes where initial conditions are known, like physics or engineering situations.
To solve these problems, you may use various techniques, either analytical or computational.
Particular Solution
A particular solution of a differential equation is a specific solution that satisfies the given initial condition. This is distinct from the general solution, which includes a family of possible solutions. Let's dive deeper:
  • General Solution: Initially, you solve the differential equation to get a general form, such as \( y(x) = \ln|x| + \frac{x^2}{2} + C \).

  • Determining the Particular Solution: You substitute the initial condition into the general solution to find the constant \( C \). In our case, \( y(1) = -1 \) helps find \( C = -\frac{3}{2} \), leading to the particular solution \( y(x) = \ln|x| + \frac{x^2}{2} - \frac{3}{2} \).
The particular solution allows for pinpointing one specific curve from the infinite solutions provided by the general solution and precisely solving the problem within its given constraints.
Computer Algebra System
A Computer Algebra System (CAS) is vital in tackling complex calculations and solving equations, especially in mathematics and engineering. Tools like Mathematica, Maple, or Wolfram Alpha can handle symbolic computation swiftly:
  • Capabilities: A CAS can solve differential equations, perform algebraic manipulation, and handle calculus operations.
  • Efficiency: Using these systems, complex differential equations like our example can be solved promptly, providing both general and particular solutions with ease.
  • Graphing: Most CAS platforms also allow for plotting functions, which aids in visualizing solution curves.
Leveraging a CAS enriches the process of solving differential equations, enabling both quick calculations and the verification of solutions by visual confirmation.
Solution Curves
Solution curves represent the graphical representation of solutions to a differential equation, helping visualize the behavior and characteristics of these solutions over a domain.
  • By plotting the particular solution \( y(x) = \ln|x| + \frac{x^2}{2} - \frac{3}{2} \), you can observe how the solution behaves over various ranges of \( x \).
  • These curves also illustrate how different initial conditions can lead to different solution paths.
  • For insight into real-world phenomena, solution curves can reveal stability, convergence, or other dynamics crucial in analysis.
Utilizing a CAS for plotting simplifies this task, providing immediate visual context to the equations and confirming initial conditions have been satisfied, as demonstrated by passing through the point \((1, -1)\).
General Solution
The general solution of a differential equation is the form encompassing all possible solutions, before applying any initial conditions.
  • It usually contains one or more constants, each representing degrees of freedom in the solution space. In our example, \( y(x) = \ln|x| + \frac{x^2}{2} + C \) is the general form.
  • This form shows the family of curves that can potentially satisfy the differential equation without autocompletely specifying them.
  • Initial conditions are then applied to this general solution to extract a particular one.
Understanding the general solution provides essential groundwork before isolating specific solutions, allowing for the breadth and diversity of the differential equation's effects to be appreciated and analyzed.

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Most popular questions from this chapter

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Use a CAS to solve the initial value problems in Exercises \(107-110\) . Plot the solution curves. $$y^{\prime}=\frac{1}{\sqrt{4-x^{2}}}, \quad y(0)=2$$$
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