/*! 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 19 Solve. \(y^{\prime \prime \pri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve. \(y^{\prime \prime \prime}-6 y^{\prime \prime}+3 y^{\prime}-18 y=0\)

Short Answer

Expert verified
The solution is \(y = C_1 e^{3x} + C_2 e^{\sqrt{3}x} + C_3 e^{-\sqrt{3}x}\).

Step by step solution

01

Identify the characteristic equation

For the differential equation given, find the corresponding characteristic equation. The characteristic equation for the given third-order linear homogeneous differential equation is obtained by replacing each derivative term with a power of \(r\). Thus, \(y^{\text{'}\text{'}}\text{'}-6y^{\text{'}\text{'}}+3y^{\text{'}}-18y=0\) transforms to \(r^3 - 6r^2 + 3r - 18 = 0\).
02

Solve the characteristic equation

Next, solve the characteristic equation \(r^3 - 6r^2 + 3r - 18 = 0\). This can be done by finding the roots of the polynomial equation. Use methods like factoring, synthetic division, or the rational root theorem to identify the roots.
03

Factor or use the Rational Root Theorem

Using the Rational Root Theorem, we test possible rational roots \(\pm{1},\pm{2},\pm{3},\pm{6},\pm{9},\pm{18}\). Testing, we find that \(r=3\) is a root. Perform synthetic division or polynomial division to factor out \(r-3\).
04

Simplify the polynomial

After factoring out \(r-3\), the remaining quadratic equation is \(r^2 - 3=0\). Solve this quadratic equation using the quadratic formula or factorization. We get \(r = \pm{\sqrt{3}}\).
05

Write the general solution

The general solution of the differential equation is a linear combination of the solutions corresponding to each root. Since the roots are real and distinct, the complementary solution will be \(y = C_1 e^{3x} + C_2 e^{\sqrt{3}x} + C_3 e^{-\sqrt{3}x}\).

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.

Characteristic Equation
A characteristic equation is fundamental in solving linear differential equations with constant coefficients. For a given differential equation, you transform the original equation into a polynomial equation by substituting each derivative term with a corresponding power of a variable, often denoted as \(r\). This transforms the problem from one involving derivatives into one involving algebra, specifically polynomial roots. For the example equation \(y^{\text{'}\text{'}}\text{'}-6y^{\text{'}\text{'}}+3y^{\text{'}}-18y=0\), each derivative is replaced by powers of \(r\), resulting in the characteristic equation: \(r^3 - 6r^2 + 3r - 18 = 0\).
Rational Root Theorem
The Rational Root Theorem is a useful tool for finding possible rational roots of polynomial equations. It tells us that any potential rational root of a polynomial equation with integer coefficients is a factor of the constant term divided by a factor of the leading coefficient. For our characteristic equation \(r^3 - 6r^2 + 3r - 18 = 0\), we look at the constant term, -18, and the leading coefficient, 1. The possible rational roots are the factors of -18, which are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\). By testing these, we find that \(r = 3\) is a root. After finding one root, we can factor the polynomial by dividing it by \(r - 3\) and proceed to solve the resulting simpler polynomial.
General Solution
Once the characteristic equation is solved, you use the roots to construct the general solution of the original differential equation. Each root corresponds to a particular solution of the differential equation. For distinct roots, \(r_1, r_2,..., r_n\), the general solution is \(y = C_1 e^{r_1 x} + C_2 e^{r_2 x} + ... + C_n e^{r_n x}\). In our example, the roots of the equation \(r^3 - 6r^2 + 3r - 18 = 0\) are \(3, \sqrt{3}, \text{and} -\sqrt{3}\). Thus, the general solution is \(y = C_1 e^{3x} + C_2 e^{\sqrt{3}x} + C_3 e^{-\sqrt{3}x}\). This set of solutions forms the complete solution set for the differential equation because it incorporates all possible linearly independent solutions based on the polynomial's roots.

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

Find the general solution of the system of equations. \(x^{\prime}=-3 x+5 y, y^{\prime}=-x+y\)

Sketch trajectories of the solutions of \(z^{\prime}=A z\) given the eigenvalues and eigenvectors of A. Classify the origin and assess its stability. \(r_{1}=-1, r_{2}=-4, \mathbf{v}_{1}=\left[\begin{array}{r}2 \\\ -1\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\)

Find the equilibrium points and assess the stability of each. \(x^{\prime}=x-e^{y}, y^{\prime}=2 \ln x+y-6\)

The matrix method may also be used for systems of three or more functions. For Exercises \(44-49\), find the general solution. \(x^{\prime}=x-3 y, y^{\prime}=-3 x-3 y-4 z\) \(z^{\prime}=3 x+5 y+6 z\)

Chemicals enter a house's basement air at the rate of \(0.1 \mathrm{mg}\) per min. Let \(F(t)\) and \(B(t)\) denote the total amount of chemical present in the first-story air and the basement air after \(t\) minutes, respectively. Both the first floor and the basement have volumes of \(200 \mathrm{~m}^{3}\). Air flows from the basement into the first floor at the rate of \(2 \mathrm{~m}^{3}\) per min, while air flows through the first foor to the outside at the rate of \(4 \mathrm{~m}^{3}\) per min. Uncontaminated air from the outside replenishes the air that flows out of the house. \({ }^{8}\) a) Draw a two-compartment model for \(B\) and \(F\). b) Show that \(B(t)\) and \(F(t)\) satisfy the system of equations \(B^{\prime}=-0.01 B+0.1\) and \(F^{\prime}=0.01 B-0.02 F\) c) Suppose that no chemicals are initially present in either floor. Solve for \(B(t)\) and \(F(t)\). d) Show that the equilibrium values of \(B\) and \(F\) are \(10 \mathrm{mg}\) and \(5 \mathrm{mg}\), respectively.

See all solutions

Recommended explanations on Biology 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.