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Chapter 5: Problem 5

$$\text { Find the general solution and three particular solutions.}$$ $$y^{\prime}=\frac{8}{x}-x^{2}+x^{5}$$

Short Answer

Expert verified
The general solution is \( y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + C \). Three particular solutions are for C = 0, 1, and -1.

Step by step solution

01

- Recognize the Differential Equation

Identify the given differential equation: \[ y' = \frac{8}{x} - x^2 + x^5 \]
02

- Integrate Both Sides

Integrate both sides with respect to x to find the general solution. \[ \begin{aligned} y &= \int \left( \frac{8}{x} - x^2 + x^5 \right) \, dx \ &= \int \frac{8}{x} \, dx - \int x^2 \, dx + \int x^5 \, dx \ &= 8 \int \frac{1}{x} \, dx - \int x^2 \, dx + \int x^5 \, dx \ &= 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + C \ &= 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + C \end{aligned} \]
03

- General Solution

Summarize the general solution: \[ y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + C \]
04

- Particular Solutions

Find three particular solutions by assigning specific values to the constant C. 1. For C = 0: \[ y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} \] 2. For C = 1: \[ y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + 1 \] 3. For C = -1: \[ y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} - 1 \]

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Key Concepts

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

integration
Integration is a fundamental concept in calculus. It is the process of finding the integral of a function, which can be thought of as the reverse process of differentiation. When we perform integration, we are essentially finding the original function from its derivative.

In our problem, we begin by identifying the differential equation: \[ y' = \frac{8}{x} - x^2 + x^5 \]
Here, we need to integrate both sides with respect to x. This involves breaking down the equation into simpler parts that can be integrated separately.

  • The integral of \( \frac{8}{x} \) is \( 8 \ln |x| \)
  • The integral of \( - x^2 \) is \( - \frac{x^3}{3} \)
  • The integral of \( x^5 \) is \( \frac{x^6}{6} \)

Combining these results, we obtain the general integral of the given differential equation.
particular solutions
A particular solution to a differential equation is a specific solution obtained by assigning particular values to the constants in the general solution. The general solution we previously found is:

\[ y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + C \]
Here, C is a constant. By assigning different values to C, we get particular solutions. Let's explore three particular solutions:

  • For C = 0: The equation simplifies to \( y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} \)
  • For C = 1: The equation becomes \( y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} + 1 \)
  • For C = -1: We have \( y = 8 \ln |x| - \frac{x^3}{3} + \frac{x^6}{6} - 1 \)

These particular solutions demonstrate how the integration constant affects the form of the solution.
constant of integration
The constant of integration, often denoted as C, appears when we integrate a function. This constant reflects the fact that there are infinitely many antiderivatives for any given function, all differing by a constant.

In the context of our problem, the constant C is added to the general solution after integrating each term. Let's summarize:

  • After integrating \( \frac{8}{x} \), \( - x^2 \), and \( x^5 \), we include a constant C in the general solution.
  • This constant can take any real number value, giving us different particular solutions.
  • The value of C can be determined if an initial condition is provided. In this exercise, we found different particular solutions by assigning values to C.

Recognizing the role of the constant of integration helps understand how differential equations can have multiple valid solutions, depending on initial or boundary conditions.

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Most popular questions from this chapter

Demand for aluminum ore (bauxite). In \(2005(t=0)\) bauxite production was approximately 153 million metric tons, and the demand was growing exponentially at a rate of \(2.5 \%\) per year. (Source: U.S. Energy Information Administration.) If the demand continues to grow at this rate, how many tons of bauxite will the world use from 2005 to \(2030 ?\)

Radioactive implant treatments. In the treatment of prostate cancer, radioactive implants are often used. The implants are left in the patient and never removed. The amount of energy that is transmitted to the body from the implant is measured in rem units and is given by $$ E=\int_{0}^{a} P_{0} e^{-k t} d t$$ where \(k\) is the decay constant for the radioactive material, a is the number of years since the implant, and \(P_{0}\) is the initial rate at which energy is transmitted. Use this information for Exercises 41 and 42 Suppose that the treatment uses palladium-103, which has a half-life of 16.99 days. a) Find the decay rate, \(k,\) of palladium- 103 b) How much energy (measured in rems) is transmitted in the first month if the initial rate of transmission is 10 rems per year? c) What is the total amount of energy that the implant will transmit to the body?

Determine whether each improper integral is convergent or divergent, and calculate its value if it is convergent. $$\int_{1}^{\infty} \frac{d x}{\sqrt{x}}$$

Find the volume generated by rotating about the \(x\) -axis the regions bounded by the graphs of each set of equations. $$y=2 \sqrt{x}, x=1, x=2$$

Solve. $$e^{-1 / x} \cdot \frac{d y}{d x}=x^{-2} \cdot y^{2}$$
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