Question

In each of Problems 15 through 18 determine the values of \(r\) for which the given differential equation has solutions of the form \(y=e^{t} .\) $$ y^{\prime}+2 y=0 $$

Short answer

Question: For the first-order linear differential equation \(y^{\prime} + 2y = 0\), find the value of r for which the solution has the form \(y=e^{rt}\). Answer: The value of r for which the given differential equation has a solution of the form \(y=e^{rt}\) is \(r=-2\).

Step-by-step solution

Write the differential equation and given solution form

We are given the differential equation \(y^{\prime} + 2y = 0\) and the solution form \(y=e^{rt}\).

Find the derivative of the solution form

Differentiate the solution form with respect to t: $$\frac{dy}{dt} = re^{rt}$$

Substitute the solution form and its derivative into the differential equation

Substitute the function and its derivative into the given equation: $$re^{rt} + 2e^{rt} = 0$$

Factor out the common term

Factor out the term \(e^{rt}\) to simplify the equation: $$e^{rt}(r+2) = 0$$

Determine values of r

Since \(e^{rt}\) is never equal to 0, we can divide both sides of the equation by it. This results in the expression $$r+2=0$$ Solving for r, we get: $$r = -2$$

State the final answer

The value of r for which the given differential equation has a solution of the form \(y=e^{rt}\) is \(r=-2\).

Key concepts

Exponential Solutions

Understanding exponential solutions in the context of differential equations is crucial when finding specific types of solutions. These are solutions where the answer can be expressed in the form of an exponential function, like \( y = e^{rt} \). Here, \( e \) is an important mathematical constant known as Euler's number, and \( r \) and \( t \) are variables that can affect the solution's growth or decay.
Expanding on the step-by-step solution above:

• Step 1: The differential equation we're working with is \( y' + 2y = 0 \) along with the solution form \( y = e^{rt} \). This means that we hypothesize the solution is an exponential one and need to find the specific rate of growth or decay, \( r \).
• Step 2: Differentiating \( y = e^{rt} \) gives us \( \frac{dy}{dt} = re^{rt} \), showing how the rate \( r \) affects the slope of the function.

Knowing how to handle exponential solutions helps to simplify and solve many differential equations efficiently.

First Order Linear Differential Equation

A first-order linear differential equation is an equation that involves the first derivative of the unknown function, and no higher derivatives. In our example, we have \( y' + 2y = 0 \), which is a classic first-order linear differential equation.
These equations have the following properties:

• They are linear; meaning they can be written in the form \( y' + p(t)y = q(t) \).
• They involve only the first derivative of the function, \( y' \).
• They can be solved using different methods, one being looking for exponential solutions \( y = e^{rt} \), as demonstrated above.

Understanding how to identify and work with first-order linear differential equations is key to solving various types of mathematical and real-world problems.

Homogeneous Equation

A differential equation is considered homogeneous if it can be rewritten such that all the terms involve the function or its derivatives. The differential equation in question is \( y' + 2y = 0 \), a typical example of a homogeneous equation.
Key characteristics include:

• A homogeneous equation can be expressed in the form \( a(t) y' + b(t) y = 0 \), where \( a(t) \) and \( b(t) \) are coefficients that could depend on \( t \).
• The inability to have a constant term or terms independent of the function \( y \) or its derivatives.
• Homogeneous solutions often lead to exponential-type solutions, as they allow for factoring out terms to simplify the equation, leading to expressions like \( (r+2) = 0 \).

The process of identifying a differential equation as homogeneous can significantly streamline the method of finding its solutions by leveraging the properties of exponential and linear functions.