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Chapter 6: Problem 7

Finding a General Solution Using Separation of Variables In Exercises 5-18, find the general solution of the differential equation. $$\frac{d y}{d x}=\frac{x-1}{y^{3}}$$

Short Answer

Expert verified
The general solution of the given differential equation is \(y = \pm \sqrt[4]{4(\frac{x^{2}}{2} - x + C)}\).

Step by step solution

01

Separate Variables

Rewrite the equation in a format that separates the variables and places them on different sides of the equation. You should get \(y^{3} dy = (x-1) dx\).
02

Integrate Both Sides

Now integrate both sides of the equation. On the left side, integrate \(y^{3} dy\) with respect to \(y\), and on the right side, integrate \((x-1) dx\) with respect to \(x\). This gives you: \(\int y^{3} dy = \int (x-1) dx\).
03

Find the Antiderivatives

The antiderivative of \(y^{3} dy\) with respect to \(y\) is \(\frac{y^{4}}{4}\), and the antiderivative of \((x-1) dx\) with respect to \(x\) is \(\frac{x^{2}}{2} - x\). Therefore, now you have the equation, \(\frac{y^{4}}{4} = \frac{x^{2}}{2} - x + C\), where \(C\) is the constant of integration.
04

Solving for \(y\)

Solve the equation for \(y\). Since we want to solve for \(y\) (denoted usually as \(y = f(x)\) in function notation), it can be rewritten as \(y = \pm \sqrt[4]{4(\frac{x^{2}}{2} - x + C)}\). And this is the general solution of the given differential equation. Note that the 卤 sign comes into play because the square root of a number can be either positive or negative.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that involve derivatives of a function. They provide a way to model real-world situations where something is changing with respect to another variable. For example, how quickly a population grows or how the temperature changes over time.

In our problem, we are given a differential equation: \(\frac{d y}{d x}=\frac{x-1}{y^{3}}\). This notation means that the rate at which \(y\) changes with respect to \(x\) depends on both \(x\) and \(y\). The primary goal is to find a function of \(y\) in terms of \(x\).
  • The process begins with separating the variables so each variable and its derivative are on opposite sides of the equation.
  • This allows us to integrate each side, leading us to a solution that gives the relationship between \(y\) and \(x\).
General Solution
The general solution of a differential equation includes all possible solutions that satisfy the equation. It involves integrating the differential equation and includes a constant of integration, \(C\).

In the problem provided, after separating the variables and integrating, we reach the general solution: \(\frac{y^{4}}{4} = \frac{x^{2}}{2} - x + C\).
  • This form represents a family of curves. Each curve corresponds to different initial conditions or particular solutions.
  • The constant \(C\) can take any value, leading to an infinite number of solutions that are applicable.
To finalize the process, solving for \(y\) gives us a clearer picture of \(y\) as a function of \(x\).
Integration
Integration is the process of calculating the integral of a function. It is essentially the reverse operation of differentiation. When we integrate, we are finding the original function from its derivative.

In this exercise, once we have separated the variables, we integrate both sides: \( \int y^{3} dy = \int (x-1) dx \).
  • The left side represents the integral of \(y^3\) with respect to \(y\), which means we are determining how \(y\) changes relative to itself.
  • The right side, \((x-1)\), is integrated with respect to \(x\), describing changes in \(x\).
Each integral involves finding a function whose derivative would match the original function before integration.
Antiderivatives
Antiderivatives are functions that reverse the process of differentiation. They are also known as indefinite integrals. Finding an antiderivative involves identifying a function that will give the original function when differentiated.

In our problem, the antiderivative of \(y^{3}\) is \(\frac{y^{4}}{4}\), and the antiderivative of \((x-1)\) is \(\frac{x^{2}}{2} - x\).
  • These antiderivatives allow us to form the general solution by expressing \(y\) and \(x\) in terms of their integrated forms with a constant \(C\).
  • When we integrate, we are essentially reversing the differentiation process to gain the original expressions for \(y(x)\).
This process highlights the beauty of calculus in connecting changes through derivatives to fixed values via integration.

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Most popular questions from this chapter

Population In Exercises \(51-54,\) the population (in millions) of a country in 2015 and the expected continuous annual rate of change \(k\) of the population are given. (Source: U.S. Census Bureau, International Data Base) (a) Find the exponential growth model \(P=C e^{k t}\) for the population by letting \(t=5\) correspond to \(2015 .\) (b) Use the model to predict the population of the country in \(2030 .\) (c) Discuss the relationship between the sign of \(k\) and the change in population for the country. \begin{array}{l}{\text { Country }}&{\text { 2015 Population }}&&{\text { \(k\) }} \\ {Ukraine}& {\text { 44.4}}&&{{-0.006}}\end{array}

In Exercises \(49-56\) , find the general solution of the first-order differential equation for \(x>0\) by any appropriate method. $$y \cos x-\cos x+\frac{d y}{d x}=0$$

Determining if a Function Is Homogeneous In Exercises \(69-76,\) determine whether the function is homogeneous, and if it is, determine its degree. A function \(f(x, y)\) is homogeneous of degree \(n\) if \(f(x, t y)=t^{n} f(x, y) .\) $$f(x, y)=\tan \frac{y}{x}$$

Using a Logistic Equation In Exercises 53 and 54 , the logistic equation models the growth of a population. Use the equation to (a) find the value of \(k,(\) b) find the carrying capacity, (c) find the initial population, (d) determine when the population will reach 50% of its carrying capacity, and (e) write a logistic differential equation that has the solution \(P(t).\) $$P(t)=\frac{5000}{1+39 e^{-0.2 t}}$$

Bacteria Growth At time \(t=0,\) a bacterial culture weighs 1 gram. Two hours later, the culture weighs 4 grams. The maximum weight of the culture is 20 grams. $$\begin{array}{l}{\text { (a) Write a logistic equation that models the weight of the }} \\ {\text { bacterial culture. }} \\ {\text { (b) Find the culture's weight after } 5 \text { hours. }} \\ {\text { (c) When will the culture's weight reach } 18 \text { grams? }}\end{array}$$ $$\begin{array}{l}{\text { (d) Write a logistic differential equation that models the }} \\ {\text { growth rate of the culture's weight. Then repeat part (b) }} \\ {\text { using Euler's Method with a step size of } h=1 . \text { Compare }} \\ {\text { the approximation with the exact answer. }} \\\ {\text { (e) At what time is the culture's weight increasing most }} \\\ {\text { rapidly? Explain. }}\end{array}$$
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