/*! 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 24 Solve. \(y^{\prime \prime}+4 y... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 24

Solve. \(y^{\prime \prime}+4 y^{\prime}+13 y=0\)

Short Answer

Expert verified
The general solution is \(y(t) = e^{-2t} ( C_1 \cos{3t} + C_2 \sin{3t} )\).

Step by step solution

01

Identify the characteristic equation

For the given differential equation \(y^{\backprime \backprime}+4 y^{\backprime}+13 y=0\), the characteristic equation can be formed as \(r^2 + 4r + 13 = 0\).
02

Solve the characteristic equation

The characteristic equation is a quadratic equation, so solve for \(r\) using the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = 4\), and \(c = 13\). Substituting these values, we get: \[r = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i\].
03

Form the general solution

Since the roots are complex conjugates \(-2 + 3i\) and \(-2 - 3i\), the general solution to the differential equation is: \[y(t) = e^{-2t} ( C_1 \cos{3t} + C_2 \sin{3t} )\].

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 91影视!

Key Concepts

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

Characteristic Equation
When solving a second-order linear differential equation like, \(y'' + 4y' + 13y = 0\),you must first form its characteristic equation. The characteristic equation is a quadratic equation derived from the given differential equation by assuming a solution of the form \(y = e^{rt}\).In this case, substituting \(y = e^{rt}\) into the differential equation gives \(r^2 + 4r + 13 = 0\).This characteristic equation helps transition from the differential form to a simpler algebraic form. Its roots determine the nature of the solutions to the original differential equation.
Complex Roots
Step 2 in solving the characteristic equation involves finding its roots. For \(r^2 + 4r + 13 = 0\), we use the quadratic formula: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].Here, \(a = 1\),\(b = 4\), and \(c = 13\). Plugging in these values, we get:\[r = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i\].These complex roots, \(-2 + 3i\) and \(-2 - 3i\),are important because they indicate that the solutions are oscillatory due to the imaginary parts.
General Solution
The complex roots \(-2 + 3i\)and \(-2 - 3i\)allow us to form the general solution to the differential equation. When the roots are complex conjugates, the general solution has the form: \[y(t) = e^{\alpha t} ( C_1 \cos{\beta t} + C_2 \sin{\beta t} )\].For our equation, \(\alpha\) is the real part of the roots, \(-2\),and \(\beta\) is the imaginary part, \(3\).Thus, the general solution is:\[y(t) = e^{-2t} ( C_1 \cos{3t} + C_2 \sin{3t} )\].In this solution, \(C_1\) and\(C_2\) are arbitrary constants determined by initial conditions or boundary values. This form combines exponential decay due to \(e^{-2t}\)with oscillations given by the sine and cosine terms.

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

Solve. \(y^{\prime \prime}-y^{\prime}-2 y=x^{3}-1\)

Let \(x\) and \(y\) represent the populations (in thousands) of two species that share a habitat. For each system of equations: a) Find the equilibrium points and assess their stability. Solve only for equilibrium points representing nonnegative populations. b) Give the biological interpretation of the asymptotically stable equilibrium point(s). \(x^{\prime}=x(0.01-0.0001 x-0.0007 y)\) \(y^{\prime}=y(0.02-0.0006 x-0.0002 y)\)

Determine the eigenvalues for the system of differential equations. If the eigenvalues are real and distinct, find the general solution by determining the associated eigenvectors. If the eigenvalues are complex or repeated, solve using the reduction method. \(x^{\prime}=-2 x-2 y, y^{\prime}=x\)

Determine the eigenvalues for the system of differential equations. If the eigenvalues are real and distinct, find the general solution by determining the associated eigenvectors. If the eigenvalues are complex or repeated, solve using the reduction method. \(x^{\prime}=-2 x+4 y, y^{\prime}=-x-7 y\)

Let \(x\) and \(y\) represent the populations (in thousands) of prey and predators that share a habitat. For the given system of differential equations, find and classify the equilibrium points. \(x^{\prime}(t)=0.5 x-0.2 x y, y^{\prime}(t)=-0.4 y+0.1 x y\)
See all solutions

Recommended explanations on Biology Textbooks

Control of Gene Expression

Read Explanation

Ecology

Read Explanation

Reproduction

Read Explanation

Cell Cycle

Read Explanation

Bioenergetics

Read Explanation

Heredity

Read Explanation
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.