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Chapter 17: Problem 5

Find the general solution of the given equation. $$y^{\prime \prime}-4 y=0$$

Short Answer

Expert verified
The general solution is \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\).

Step by step solution

01

Recognize the Type of Differential Equation

The given differential equation is a second-order linear homogeneous differential equation with constant coefficients. It can be identified as \(y'' - 4y = 0\).
02

Write the Characteristic Equation

For a second-order differential equation of the form \(y'' - ay = 0\), the characteristic equation is given by \(r^2 - a = 0\). Here, \(a = 4\), so the characteristic equation is \(r^2 - 4 = 0\).
03

Solve the Characteristic Equation

Solve the characteristic equation \(r^2 - 4 = 0\). Rewriting it gives \((r - 2)(r + 2) = 0\), which results in two roots: \(r_1 = 2\) and \(r_2 = -2\).
04

Write the General Solution

The general solution for the differential equation with real and distinct roots \(r_1\) and \(r_2\) is given by \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). Substituting the roots found, we have the general solution \(y(t) = C_1 e^{2t} + C_2 e^{-2t}\), where \(C_1\) and \(C_2\) are arbitrary constants.

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Key Concepts

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

Second-Order Differential Equations
Second-order differential equations are equations that involve the second derivative of a function. This means they describe how an unknown function's rate of change itself changes over time or space. They are called "second-order" because the highest derivative is the second derivative, typically denoted as \(y''\).
In the context of physical systems, second-order differential equations often model motion, such as vibrations or waves. Their general form is \(a y'' + b y' + c y = 0\), where \(a\), \(b\), and \(c\) are constants.
  • These equations can model a variety of real-world phenomena, like electrical circuits and mechanical systems.
  • Second-order equations can have a wide range of solutions, depending on additional conditions and terms in the equation.
Understanding the nature and structure of these equations helps in solving and applying them to various fields.
Homogeneous Differential Equations
A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives. For example, any term with no dependent variables or derivatives is absent. When examining an equation like \(y'' - 4y = 0\), it qualifies as homogeneous due to its structure.
The term 'homogeneous' indicates that it can be represented as \(f(y', y, ") = 0\) with no extra constant terms disrupting this equality.
  • Being homogeneous means the solutions are functions dependent on arbitrary constants, representing a family of solutions.
  • The simplicity of homogeneous equations often makes them preferable to work with as they are easier to solve analytically than non-homogeneous ones.
These characteristics allow special methods to solve them, which leads to systematic solution techniques like finding the characteristic equation.
Characteristic Equation
The characteristic equation is crucial when solving linear differential equations, especially for those with constant coefficients like \(y'' - 4y = 0\). It changes the differential equation into an algebraic form.
For a second-order homogeneous differential equation of the form \(y'' - ay = 0\), the characteristic equation is derived as \(r^2 - a = 0\). By solving for \(r\), we determine the type of solution the differential equation will have.
  • The roots of the characteristic equation, like \(r^2 - 4 = 0\), provide essential insights into the behavior of the solution.
  • Depending on whether the roots are real, complex, or repeated, different formulas for the general solution are used.
This process of solving the characteristic equation transforms a complex problem into simpler algebra, making differential equations more manageable.
Constant Coefficients
When a differential equation has constant coefficients, each term involves the differential function multiplied by a constant. This property significantly simplifies the solution process.
In an equation like \(y'' - 4y = 0\), the coefficients are constant (with values 1 for \(y''\) and -4 for \(y\)), making it straightforward to apply standard solution techniques.
  • Constant coefficients ensure that the characteristic equation is polynomial with constant terms, simplifying algebraic operations.
  • These equations yield solutions that are commonly expressed in terms of exponential functions, which are easier to analyze and work with.
Thus, constant coefficients provide a structured path to solving differential equations, linking them naturally with exponential-based solutions.

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Most popular questions from this chapter

Use power series to find the general solution of the differential equation. $$y^{\prime \prime}+4 y=0$$

A \(10-\mathrm{kg}\) mass is attached to a spring having a spring constant of 140 N/m. The mass is started in motion from the equilibrium position with an initial velocity of \(1 \mathrm{m} / \mathrm{s}\) in the upward direction and with an applied external force given by \(f(t)=5 \sin t\) (in newtons). The mass is in a viscous medium with a coefficient of resistance equal to \(90 \mathrm{N}-\mathrm{s} / \mathrm{m}\). Formulate an initial value problem that models the given system; solve the model and interpret the results.

An inductor of 2 henrys is connected in series with a resistor of 12 ohms, a capacitor of \(1 / 16\) farad, and a 300 volt battery. Initially, the charge on the capacitor is zero and the current is zero. Formulate an initial value problem modeling this electrical circuit.

A series circuit consisting of an inductor, a resistor, and a capacitor is open. There is an initial charge of 2 coulombs on the capacitor, and 3 amperes of current is present in the circuit at the instant the circuit is closed. A voltage given by \(E(t)=20 \cos t\) is applied. In this circuit the voltage drops are numerically equal to the following: across the resistor to 4 times the instantaneous change in the charge, across the capacitor to 10 times the charge, and across the inductor to 2 times the instantaneous change in the current. Find the charge on the capacitor as a function of time. Determine the charge on the capacitor and the current at time \(t=10.\)

Solve the equations by the method of undetermined coefficients. $$y^{\prime \prime}+3 y^{\prime}+2 y=e^{-x}+e^{-2 x}-x$$
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