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Chapter 8: Problem 23

Find a solution to the initial-value problem. \(y^{\prime}-y^{2} \equiv 0, y(1)=2 \quad[\) Hint: Assume the solution has an inverse function \(x=x(y)\). Find, and solve, a differential equation that involves \(x^{t}(y)\).]

Short Answer

Expert verified
The solution is \( y = \frac{-1}{x - \frac{3}{2}} \).

Step by step solution

01

Recognize the structure of the differential equation

The given differential equation is \( y' - y^2 = 0 \), which can be rewritten as \( y' = y^2 \). This is a separable differential equation.
02

Separate variables

Separate the variables to allow integration. We do this by rewriting the equation as \( \frac{dy}{y^2} = dx \). Our goal is to move all terms involving \( y \) to one side and all terms involving \( x \) to the other.
03

Integrate both sides

Integrate both sides of the equation separately. The integral of \( \frac{dy}{y^2} \) is \( -\frac{1}{y} + C_1 \), and the integral of \( dx \) is \( x + C_2 \). Combine these to get \( -\frac{1}{y} = x + C \) where \( C = C_2 - C_1 \).
04

Solve for the inverse relation \( x = x(y) \)

From the rearranged equation \( -\frac{1}{y} = x + C \), solve for \( x \). This gives \( x = -\frac{1}{y} - C \).
05

Apply the initial condition to find \( C \)

Using the initial condition \( y(1) = 2 \), substitute \( y = 2 \) and \( x = 1 \) into the equation \( 1 = -\frac{1}{2} - C \). Solving for \( C \), we find \( C = -\frac{3}{2} \).
06

Rewrite the inverse function

Substitute the value of \( C \) into \( x = -\frac{1}{y} - C \), resulting in \( x = -\frac{1}{y} + \frac{3}{2} \).
07

Write the solution for \( y \) in terms of \( x \)

Re-arrange \( -\frac{1}{y} = x - \frac{3}{2} \) to solve for \( y \). Therefore, \( y = \frac{-1}{x - \frac{3}{2}} \).

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Key Concepts

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

Separable Differential Equations
To solve an initial-value problem involving a separable differential equation, it's important to understand what a separable differential equation is. These equations can be rewritten so that all terms involving one variable (typically y) appear on one side of the equation, and terms involving the other variable (typically x) are on the opposite side. During this process, the equation can be expressed in the form \( \frac{dy}{dx} = g(y) h(x) \).
  • We start by separating the variables, giving us two integrals to solve.
  • After separation, the goal is to integrate both sides to find a general solution.
In this exercise, the separation transforms the equation \( y' = y^2 \) into \( \frac{dy}{y^2} = dx \). After separating, each side can be integrated independently to move closer to the solution.
Integration
Integration is a fundamental concept when dealing with differential equations, especially separable ones. After separating the variables, our main task is to integrate both sides to find a relationship between the variables.
  • On the left, the integral \( \int \frac{dy}{y^2} \) is evaluated, resulting in \( -\frac{1}{y} + C_1 \).
  • On the right, the integral of \( dx \) is straightforward, resulting in \( x + C_2 \).
These results are combined as \( -\frac{1}{y} = x + C \), where \( C \) accounts for the integration constants. Integration helps bridge the gap from the differential equation to an expression that can be manipulated to find solutions satisfying any initial conditions.
Inverse Function
The concept of inverse functions comes into play when the solution to a differential equation is expressed in terms of a relation like \( x = x(y) \). In this problem, solving \( -\frac{1}{y} = x + C \) gives the inverse function \( x = -\frac{1}{y} - C \).
  • Inverse functions essentially "flip" the roles of x and y.
  • In our solution, we first find \( x \) in terms of \( y \), then reconstruct \( y \) in terms of \( x \).
This two-step process helps define the solution more clearly, allowing us to easily apply initial conditions to find any constants involved. Understanding this concept makes it easier to navigate through similar differential equations.
Initial Condition
The initial condition is crucial in determining the particular solution to an initial-value problem. Given an initial condition, such as \( y(1) = 2 \), we can solve for the constant \( C \) found in the expression derived from the integration process.
  • Substitute the initial values into the inverse function equation.
  • In our case, setting \( y = 2 \) and \( x = 1 \) helps us determine \( C \).
Solving \( 1 = -\frac{1}{2} - C \) yields \( C = -\frac{3}{2} \). This specific value of \( C \) allows us to write a solution that not only satisfies the differential equation but also fits the initial conditions given in the problem.

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Most popular questions from this chapter

In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. \(y^{\prime \prime}-4 y^{\prime}+4 y=0\) (a) \(e^{2 x}\) and \(x e^{2 x}\) (b) \(c_{1} e^{2 x}+c_{2} x e^{2 x}\left(c_{1}, c_{2}\right.\) constants \()\)

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) $$ \frac{d y}{d x}-\frac{y^{2}-y}{\sin x}=0 $$

A rocket, fired upward from rest at time \(t=0\), has an initial mass of \(m_{0}\) (including its fuel). Assuming that the fuel is consumed at a constant rate \(k\), the mass \(m\) of the rocket, while fuel is being burned, will be given by \(m=m_{0}-k t\). It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed \(c\) relative to the rocket, then the velocity \(v\) of the rocket will satisfy the equation $$ m \frac{d v}{d t}=c k-m g $$ where \(g\) is the acceleration due to gravity. (a) Find \(v(t)\) keeping in mind that the mass \(m\) is a function of \(t\). (b) Suppose that the fuel accounts for \(80 \%\) of the initial mass of the rocket and that all of the fuel is consumed in \(100 \mathrm{~s}\). Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) and \(\left.c=2500 \mathrm{~m} / \mathrm{s} .\right]\)

Determine whether the statement is true or false. Explain your answer. If a population is growing exponentially, then the time it takes the population to quadruple is independent of the size of the population.

Suppose that \(30 \%\) of a certain radioactive substance decays in 5 years. (a) What is the half-life of the substance in years? (b) Suppose that a certain quantity of this substance is stored in a cave. What percentage of it will remain after \(t\) years?
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