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Chapter 1: Problem 38

Use the concept that \(y=c,-\infty < x < \infty,\) is a constant function if and only if \(y^{\prime}=0\) to determine whether the given differential equation possesses constant solutions. $$y^{\prime}=y^{2}+2 y-3$$

Short Answer

Expert verified
The constant solutions are \( y = 1 \) and \( y = -3 \).

Step by step solution

01

Understand the Problem

We need to find if there are any constant solutions for the differential equation given by \( y' = y^2 + 2y - 3 \). A constant function, \( y = c \), will have a derivative \( y' = 0 \).
02

Set the Derivative to Zero

Since a constant solution means \( y' = 0 \), set \( y^2 + 2y - 3 = 0 \) and solve for \( y \). This will identify potential constant values \( c \) for which \( y = c \) is a constant solution.
03

Solve the Quadratic Equation

The equation \( y^2 + 2y - 3 = 0 \) is a quadratic in \( y \). Use the quadratic formula to solve it: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -3 \).
04

Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \) to check if there are real solutions. Here, it is \( 2^2 - 4(1)(-3) = 4 + 12 = 16 \), which is positive, indicating two real solutions.
05

Find the Roots

Substitute into the quadratic formula to find the roots: \[ y = \frac{-2 \pm \sqrt{16}}{2} \]. This gives \[ y = \frac{-2 \pm 4}{2} \], resulting in the solutions \( y = 1 \) and \( y = -3 \).
06

Identify Constant Solutions

The solutions \( y = 1 \) and \( y = -3 \) represent the potential constant solutions of the differential equation because for these values, the derivative \( y^\prime = 0 \).

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Key Concepts

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

Differential Equations
A differential equation involves an equation that relates a function with its derivatives. In this exercise, the differential equation is given by \( y' = y^2 + 2y - 3 \), where \( y' \) represents the derivative of the function \( y \) with respect to \( x \). Differential equations are powerful in modeling real-world phenomena, like population growth or physical systems dynamics.

Finding constant solutions to a differential equation can provide insights into the system's steady states or equilibria, where no change occurs over time.

Steps to Determine Constant Solutions:
  • If a function \( y = c \) is constant, then its derivative \( y' \) must be 0.
  • Set \( y' = 0 \) and solve for \( y \) to find potential constant solutions.
Quadratic Equation
A quadratic equation is a second-order polynomial equation of the form \( ax^2 + bx + c = 0 \). The quadratic equation emerges in the context of determining constant solutions for differential equations.

For this exercise, when we set \( y' = 0 \), we end up with a quadratic equation \( y^2 + 2y - 3 = 0 \). Solving a quadratic equation involves finding the values of \( y \) that satisfy it. This is often done using the quadratic formula or factoring.

General Structure:
  • \( a \) is the coefficient of \( y^2 \)
  • \( b \) is the coefficient of \( y \)
  • \( c \) is the constant term
Discriminant
The discriminant is a component of the quadratic formula that indicates the nature of the roots of a quadratic equation. It is given by \( b^2 - 4ac \). Understanding the discriminant is crucial in identifying whether a quadratic equation has real or complex solutions.

Discriminant Analysis:

  • If the discriminant is positive, the quadratic equation has two distinct real roots.
  • If it is zero, there is exactly one real root (a double root).
  • If negative, the roots are complex and not real.
In our exercise, the discriminant of \( y^2 + 2y - 3 = 0 \) is 16, which is positive, confirming two distinct real constant solutions.
Derivative
The derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It is the foundation of differential equations, like the one in this exercise \( y' = y^2 + 2y - 3 \). Calculating derivatives is essential for finding constant solutions.

In simpler terms:
  • The derivative \( y' \) represents the rate of change of the function \( y \).
  • For a constant solution, \( y' = 0 \), indicating no change in \( y \).
Derivatives provide insights into the behavior of functions and are pivotal in fields like physics, engineering, and economics.

To check for constant solutions in the exercise, we set the derivative \( y' = 0 \) and analyzed the resulting quadratic to find suitable constants.

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Most popular questions from this chapter

Verify that the indicated expression is an implicit solution of the given first-order differential equation. Find at least one explicit solution \(y=\phi(x)\) in each case. Use a graphing utility to obtain the graph of an explicit solution. Give an interval \(I\) of definition of each solution \(\phi\). $$\frac{d X}{d t}=(X-1)(1-2 X) ; \quad \ln \left(\frac{2 X-1}{X-1}\right)=t$$

Use a CAS to compute all derivatives and to carry out the simplifications needed to verify that the indicated function is a particular solution of the given differential equation. $$\begin{aligned} &x^{3} y^{\prime \prime \prime}+2 x^{2} y^{\prime \prime}+20 x y^{\prime}-78 y=0\\\ &y=20 \frac{\cos (5 \ln x)}{x}-3 \frac{\sin (5 \ln x)}{x} \end{aligned}$$

Consider the differential equation \(y^{\prime}=y^{2}+4\) (a) Explain why there exist no constant solutions of the DE. (b) Describe the graph of a solution \(y=\phi(x)\). For example, can a solution curve have any relative extrema?

Verify that the indicated function \(y=\phi(x)\) is an explicit solution of the given first-order differential equation. Proceed as in Example \(6,\) by considering \(\phi\) simply as a function and give its domain. Then by considering \(\phi\) as a solution of the differential equation, give at least one interval \(I\) of definition. $$y^{\prime}=25+y^{2} ; \quad y=5 \tan 5 x$$

Determine by inspection at least two solutions of the given first-order IVP. $$y^{\prime}=3 y^{2 / 3}, \quad y(0)=0$$
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