Problem 31

Question

Use the concept that \(y=c,-\infty

Step-by-Step Solution

Verified

Answer

Yes, the equation has constant solutions.

1Step 1: Understanding the Problem

We need to determine if the differential equation $ x y'' + 2 y' = 0 $ can have constant solutions. A constant function is one where $ y = c $, which implies that the derivative $ y' = 0 $. Given this, we will substitute $ y' = 0 $ into the differential equation to see if it holds.

2Step 2: Analyzing the Equation with Constant Solutions

Substitute $ y' = 0 $ into the equation: $ x y'' + 2(0) = 0 $. This simplifies to just $ x y'' = 0 $. Now, we'll analyze this result further.

3Step 3: Analyzing Simplified Equation

From $ x y'' = 0 $, we can see that for the equation to be true: either $ x = 0 $ or $ y'' = 0 $ must hold. If $ y'' = 0 $, that means $ y' = c_1 $ (a constant) and $ y = c_1 x + c_2 $. Since constant solutions must have $ y' = 0 $, $ c_1 $ must be 0. Therefore, $ y = c_2 $, a constant solution, is possible.

4Step 4: Conclusion

The given differential equation $ x y'' + 2 y' = 0 $ does possess constant solutions since $ y = c_2 $ satisfies the equation with $ y' = 0 $. Constant functions have $y' = 0$, which contributes to a correct simplification of 0 in the differential equation.

Key Concepts

Constant FunctionsSecond-Order Differential EquationsSolution AnalysisDerivatives and Continuity

Constant Functions

In mathematics, a constant function is one where the output or value does not change, no matter what input you provide. In simpler terms, it stays the same for any value of x. For a function to be constant, it must satisfy the condition that its first derivative $ y' $ is equal to 0 everywhere in its domain. This is because the derivative measures the rate of change; for a constant function, this rate needs to be zero since there is no change at all.

Some key points about constant functions include:

It is represented by $ y = c $, where $ c $ is a constant.
Graphically, constant functions are always horizontal lines.
They often appear in mathematics to help simplify equations and models.

In the context of differential equations, checking if a function is constant is a useful step in evaluating the nature of solutions to the equation.

Second-Order Differential Equations

Second-order differential equations include derivatives of second degree, such as $ y'' $, the second derivative of a function y with respect to x. These types of equations often describe systems with more complex behaviors than first-order differential equations.

They can model various physical phenomena:

Oscillations, like those of a pendulum.
Mechanical systems, like springs.
Electrical circuits.

The differential equation $ x y'' + 2 y' = 0 $ discussed here is second-order. The process to solve it involved substituting $ y' = 0 $, which simplifies the equation significantly. Second order is crucial here because it determines the need to look at both $ y'' $ and potential non-constant parts of the function when seeking solutions.

Solution Analysis

Analyzing the solution of a differential equation involves several steps, often tested by substituting potential solutions back into the equation. In the case of $ x y'' + 2 y' = 0 $, the process showed whether constant solutions were possible by setting $ y' = 0 $.

The steps are as follows:

Substitute any proposed solution into the original equation.
Simplify the equation with the proposed substitution.
Analyze the resulting terms – in this instance, we set $ x y'' = 0 $.
Determine the viability of solution components, e.g., constant derivatives.

The conclusion from the analysis revealed that a constant solution $ y = c_2 $ indeed satisfies the differential equation, meeting the condition when $ y' = 0 $.

Derivatives and Continuity

Understanding derivatives is essential when working with differential equations, as they represent rates of change and are foundational for solving and analyzing different types of equations. The first derivative $ y' $ corresponds to the slope or rate of change of the function y.

In context:

If $ y' = 0 $, the function y does not change, indicating continuity in terms of a constant function.
The second derivative $ y'' $ provides insight into the curvature or acceleration of the function.
Setting higher derivatives to zero, as shown in the exercise, helps identify constant behaviors.

Continuity itself is a property where a function remains smooth without breaks, jumps, or gaps. When checking for constant solutions in the differential equation scenario, continuity and derivatives go hand in hand. They help verify that the solution is valid and meets the mathematical requirements laid out by the problem.

Problem 31

Other exercises in this chapter

Problem 31

$$ x y^{\prime \prime}+2 y^{\prime}=0 $$

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Problem 31

(a) Verify that $y=-1 /(x+c)$ is a one-parameter family of solutions of the differential equation $y^{\prime}=y^{2}$. (b) Since $f(x, y)=y^{2}$ and \(\par

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Problem 31

In Problem 31 and 32, verify that the function defined by the definite integral is a particular solution of the given differential equation. In both problems, u

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Problem 32

Use the concept that \(y=c,-\infty

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