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Substitute \(y=e^{r x}\) into the given differential equation to determine all values of the constant \(r\) for which \(y=e^{r x}\) is a solution of the equation. $$ 4 y^{\prime \prime}=y $$

Short Answer

Expert verified
The values of \(r\) are \(\pm \frac{1}{2}\).

Step by step solution

01

Differentiate the Expression for y

Given that \(y = e^{rx}\), start by differentiating this expression to find \(y'\) and \(y''\). The first derivative is \(y' = r e^{rx}\) and the second derivative is \(y'' = r^2 e^{rx}\).
02

Substitute into the Differential Equation

Substitute \(y = e^{rx}\) and \(y'' = r^2 e^{rx}\) into the differential equation \(4y'' = y\). This yields the equation \(4(r^2 e^{rx}) = e^{rx}\).
03

Simplify the Equation

Divide both sides of the equation \(4r^2 e^{rx} = e^{rx}\) by \(e^{rx}\), assuming \(e^{rx} eq 0\). This simplifies to \(4r^2 = 1\).
04

Solve for r

Solve the equation \(4r^2 = 1\) by dividing both sides by 4, resulting in \(r^2 = \frac{1}{4}\). Taking the square root of both sides gives \(r = \pm \frac{1}{2}\).

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Key Concepts

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

Exponential Functions
Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In the form \( y = e^{r x} \), \( e \) is the base of the natural logarithm, approximately equal to 2.718. Exponential functions are prevalent in many areas of mathematics, especially in solving differential equations because of their unique property of differentiating into multiples of themselves.
When solving differential equations, exponential functions often serve as potential solutions due to this property. For example, when you differentiate \( y = e^{rx} \), the derivative is \( y' = r e^{rx} \) and the second derivative is \( y'' = r^2 e^{rx} \). These expressions preserve the form of the original function, allowing equations to be simplified easily during substitution steps.
Characteristic Equation
The characteristic equation is a crucial concept in solving linear differential equations with constant coefficients. It emerges from substituting a trial solution, usually an exponential function like \( y = e^{rx} \), into the differential equation.
  • In the given exercise, after substituting \( y \) and its derivatives into the differential equation, we reach the expression \( 4r^2 e^{rx} = e^{rx} \).
  • Simplifying by dividing both sides by \( e^{rx} \) (not equal to zero), leads to \( 4r^2 = 1 \).
The equation \( 4r^2 = 1 \) is what we call the characteristic equation. Solving this equation for \( r \) gives the values for which the exponential function \( y = e^{rx} \) is a solution to the differential equation. In this case, the possible values of \( r \) are \( \pm \frac{1}{2} \).
Characteristic equations are powerful as they transform differential equations into polynomial equations, which are generally easier to solve.
Second Order Differential Equations
Second order differential equations are differential equations that involve the second derivative of a function. They are a step up in complexity from first order differential equations, as they provide a richer set of solutions and behaviors.
  • The hallmark of second order differential equations is the presence of the second derivative \( y'' \). In our example, the equation is \( 4y'' = y \).
  • These equations often appear in physical systems, such as in models of spring motion or electrical circuits, where the relation between position, velocity, and force (or similar quantities) involves second derivatives.
The goal in solving a second order differential equation is to find a function \( y(x) \) that satisfies the equation for all values in a given domain. With an equation like \( 4y'' = y \), applying techniques such as substituting trial solutions like exponential functions, is common to find general solutions. The solutions for \( r \) resulting from the characteristic equation guide us to the full set of possible solutions for the differential equation. Here, \( y = e^{\frac{1}{2} x} \) and \( y = e^{-\frac{1}{2} x} \) are the solutions.

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Most popular questions from this chapter

A certain moon rock was found to contain equal numbers of potassium and argon atoms. Assume that all the argon is the result of radioactive decay of potassium (its half-life is about \(1.28 \times 10^{9}\) years) and that one of every nine potassium atom disintegrations yields an argon atom. What is the age of the rock, measured from the time it contained only potassium?

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Determine by inspection at least one solution of the given differential equation. That is, use your knowledge of derivatives to make an intelligent guess. Then test your hypothesis. $$ y^{\prime \prime}+y=0 $$

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