/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Find the general solution to the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the general solution to the given differential equation. $$z^{\prime \prime}+2 z=0$$

Short Answer

Expert verified
The general solution is \( z(t) = C_1 \cos(\sqrt{2} t) + C_2 \sin(\sqrt{2} t) \).

Step by step solution

01

Identify the Form of the Differential Equation

The differential equation given is \( z'' + 2z = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients.
02

Write the Characteristic Equation

Rewrite the differential equation in terms of its characteristic equation by assuming a solution of the form \( z = e^{rt} \), where \( r \) is a constant. Plugging this into the differential equation, we obtain \( r^2 e^{rt} + 2e^{rt} = 0 \). Divide through by \( e^{rt} \) (non-zero), resulting in the characteristic equation: \( r^2 + 2 = 0 \).
03

Solve the Characteristic Equation

Solve the characteristic equation \( r^2 + 2 = 0 \) for \( r \). This can be rewritten as \( r^2 = -2 \), which yields complex roots \( r = \pm i\sqrt{2} \).
04

Write the General Solution

With complex roots \( r = \pm i\sqrt{2} \), the general solution to the differential equation is given by \( z(t) = C_1 \cos(\sqrt{2} t) + C_2 \sin(\sqrt{2} t) \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Second-Order Linear Differential Equation
A second-order linear differential equation is one that involves the second derivative of a function. In the given problem, we have the differential equation \( z'' + 2z = 0 \). This type of equation is termed "linear" because each term is either a constant or a product of a constant and a function or its derivatives. These equations are widely used because they can describe many natural phenomena, such as oscillations and the movement of mechanical systems.
To be classified as second-order, the highest derivative in the equation must be the second derivative. This is important because the order of a differential equation directly influences the complexity and type of solutions one can expect. Solving them involves methods like finding the characteristic equation, which we will discuss later.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is dependent on the function or its derivatives. In our equation, \( z'' + 2z = 0 \), each term involves the function \( z \) or its derivatives, and there is no isolated constant term.
Homogeneous equations are special because they imply that any non-zero multiple of a solution is also a solution. This characteristic leads to family-like sets of solutions. The concept of homogeneity also plays a crucial role in solving these equations because it affects how you set up and solve the characteristic equation. When dealing with second-order homogeneous equations, your solutions often involve exponential functions or trigonometric functions if complex roots are present.
Characteristic Equation
The characteristic equation is a way to transform a differential equation into an algebraic equation, simplifying the problem. For the equation \( z'' + 2z = 0 \), you assume a solution of the form \( z = e^{rt} \). Substituting this into the differential equation and simplifying gives the characteristic equation \( r^2 + 2 = 0 \).
Solving the characteristic equation is a key step in finding solutions to the differential equation. It reduces the differentiation problem to finding the roots of a simple polynomial, usually quadratic in nature. The nature of these roots, whether real or complex, guides the form of the general solution.
Complex Roots
Complex roots arise when solving the characteristic equation in cases where the discriminant (the part under the square root in the quadratic formula) is negative. This situation appears in the example \( r^2 + 2 = 0 \), giving us roots \( r = \pm i\sqrt{2} \).
When roots are complex, the general solution to the differential equation includes trigonometric functions, rather than simple exponentials. For the given problem, the solution is \( z(t) = C_1 \cos(\sqrt{2} t) + C_2 \sin(\sqrt{2} t) \). These solutions are often used to describe oscillatory systems, as with waves or harmonic oscillators. The constants \( C_1 \) and \( C_2 \) are typically determined by initial or boundary conditions specific to the context of the problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Give an example of: A system of differential equations for the profits of two companies if each would thrive on its own but the two companies compete for business. Let \(x\) and \(y\) represent the profits of the two companies.

The motion of a mass on the end of a spring satisfies the differential equation $$ \frac{d^{2} s}{d t^{2}}+2 \frac{d s}{d t}+3 s=0 $$ (a) Give the general solution to the differential equation. (b) Solve the differential equation if the initial height is +2 and the initial velocity is +5 (c) How low does the mass at the end of the spring go? How high does it go? (d) How long does it take until the spring stays within 0.1 unit of equilibrium?

Explain what is wrong with the statement. For a system of differential equations for \(x\) and \(y,\) at the point \((2,3),\) we have \(d x / d t<0\) and \(d y / d t>0\) and \(d y / d x>0\)

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. $$\begin{aligned}&\frac{d x}{d t}=x\left(1-y-\frac{x}{3}\right)\\\&\frac{d y}{d t}=y\left(1-\frac{y}{2}-x\right)\end{aligned}$$

Morphine is often used as a pain-relieving drug. The half-life of morphine in the body is 2 hours. Suppose morphine is administered to a patient intravenously at a rate of \(2.5 \mathrm{mg}\) per hour, and the rate at which the morphine is eliminated is proportional to the amount present. (a) Use the half-life to show that, to three decimal places, the constant of proportionality for the rate at which morphine leaves the body (in mg/hour) is \(k=-0.347\) (b) Write a differential equation for the quantity, \(Q,\) of morphine in the blood after \(t\) hours. (c) Use the differential equation to find the equilibrium solution. (This is the long-term amount of morphine in the body, once the system has stabilized.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.