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Give an example of: A second-order differential equation which has \(y=e^{2 x}\) as a solution.

Short Answer

Expert verified
The differential equation is \(\frac{d^2y}{dx^2} - 4y = 0\).

Step by step solution

01

Start with the general form of a second-order differential equation

A second-order differential equation typically has the form \(a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0\), where \(a\), \(b\), and \(c\) are constants.
02

Differentiate the provided solution

Given the solution \(y = e^{2x}\), find the first and second derivatives. The first derivative is \(\frac{dy}{dx} = 2e^{2x}\), and the second derivative is \(\frac{d^2y}{dx^2} = 4e^{2x}\).
03

Substitute derivatives into the differential equation

Substitute \(y = e^{2x}\), \(\frac{dy}{dx} = 2e^{2x}\), and \(\frac{d^2y}{dx^2} = 4e^{2x}\) into the general equation. We aim for the left side to equal zero.
04

Find appropriate coefficients

Choose coefficients \(a\), \(b\), and \(c\) such that \(a\cdot4e^{2x} + b\cdot2e^{2x} + c\cdot e^{2x} = 0\). For instance, let \(a=1\), \(b=0\), \(c=-4\), resulting in \(\frac{d^2y}{dx^2} - 4y = 0\).
05

Verify the solution

Check that substituting \(y = e^{2x}\) makes the equation zero: \(4e^{2x} - 4e^{2x} = 0\), which confirms \(y = e^{2x}\) is a solution.

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Key Concepts

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

Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. Their main goal is to determine the unknown function given certain conditions or rules.

A **second-order differential equation** involves the second derivative and is commonly represented as \( a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = 0 \). These types of equations describe a wide variety of physical systems, from mechanics to electrical circuits.

Understanding the order and degree of a differential equation is crucial. In this context, 'order' refers to the highest derivative term present, while 'degree', if defined, is the power of the highest derivative after it has been made rational and integral.
Calculus
Calculus is the mathematical study of continuous change and is fundamental to differential equations. It provides the methods for differentiating and integrating functions, which are steps often necessary to solve differential equations.

With calculus, we explore the concepts of differentiation, which involves finding the rate at which a function is changing, and integration, which is about finding the area under a curve. In the context of differential equations, especially second-order ones like in the exercise, calculus aids in determining how various rates of change are connected by the equation.
Solution Verification
Verifying a solution to a differential equation means checking if the proposed solution satisfies the original equation. To do this, you substitute the solution and its derivatives back into the equation.

For example, let's say our solution is \( y = e^{2x} \). By substituting \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \) into the differential equation \( a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0 \), we aim to make the equation equal zero. If successful, it is verified that the function satisfies the differential equation.

Solution verification is critical for determining whether a proposed solution is correct and applicable to the differential equation given.
Differentiation
Differentiation is a core calculus concept used to find the rate of change of a function. It tells us how a function changes as its input changes.

In the given problem, differentiation is utilized by deriving the first and second derivatives of the solution \( y = e^{2x} \).
  • The first derivative is found using the rule for the derivative of an exponential function, resulting in \( \frac{dy}{dx} = 2e^{2x} \).
  • The second derivative, which involves finding the derivative of \( \frac{dy}{dx} \), gives \( \frac{d^2y}{dx^2} = 4e^{2x} \).
These derivatives are essential in substituting back into the differential equation for verification purposes.
Exponential Function
Exponential functions are functions of the form \( f(x) = e^{kx} \), where \( e \) is Euler's number, approximately equal to 2.71828, and \( k \) is a constant.

These functions are powerful in mathematics due to their ubiquitous properties and appearance in growth, decay, and many natural processes.

In second-order differential equations, exponential functions like \( y = e^{2x} \) often serve as solutions due to their properties when differentiated. Differentiating an exponential function returns a result that is proportional to the original function, which can satisfy the structure of differential equations.

Understanding exponential functions is essential in appreciating their role in various mathematical and scientific applications.

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Most popular questions from this chapter

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Analyze the phase plane of the differential equations for \(x, y \geq 0 .\) Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. $$\begin{aligned}&\frac{d x}{d t}=x(2-x-2 y)\\\&\frac{d y}{d t}=y(1-2 x-y)\end{aligned}$$

Use the idea of nullclines dividing the plane into sectors to analyze the equations describing the interactions of robins and worms: $$\begin{array}{l}\frac{d w}{d t}=w-w r \\\\\frac{d r}{d t}=-r+r w \end{array}$$

Consider the system of differential equations $$ \frac{d x}{d t}=-y \quad \frac{d y}{d t}=-x $$ (a) Convert this system to a second order differential equation in \(y\) by differentiating the second equation with respect to \(t\) and substituting for \(x\) from the first equation. (b) Solve the equation you obtained for \(y\) as a function of \(t ;\) hence find \(x\) as a function of \(t\).

We analyze world oil production. \(^{25}\) When annual world oil production peaks and starts to decline. major economic restructuring will be needed. We investigate when this slowdown is projected to occur. We define \(P\) to be the total oil production worldwide since 1859 in billions of barrels. In \(1993,\) annual world oil production was 22.0 billion barrels and the total production was \(P=724\) billion barrels. In 2013 , annual production was 27.8 billion barrels and the total production was \(P=1235\) billion barrels. Let \(t\) be time in years since 1993 (a) Estimate the rate of production, \(d P / d t,\) for 1993 and 2013 (b) Estimate the relative growth rate, \((1 / P)(d P / d \imath)\) for 1993 and 2013 (c) Find an equation for the relative growth rate. \((1 / P)(d P / d t),\) as a function of \(P,\) assuming that the function is linear. (d) Assuming that \(P\) increases logistically and that all oil in the ground will ultimately be extracted, estimate the world oil reserves in 1859 to the nearest billion barrels. (e) Write and solve the logistic differential equation modeling \(P\)

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