Answer Exercise 7.4.7 when L=3 D^2-2 D-5.

Name: Problem 8, Chapter 7: Applied Linear Algebra (Undergraduate Texts in Mathematics)
Uploaded: 2020-05-20T18:02:57-08:00
Duration: 3 min 48 s
Channel: Ryan Mcalister
Description: Answer Exercise 7.4.7 when L=3 D^2-2 D-5.

Key Concepts

Characteristic Equations

The characteristic equation is obtained by substituting an exponential trial solution into a linear differential equation, turning the differential equation into an algebraic equation. The roots of this polynomial provide key insights into the behavior of the solutions, such as oscillatory or exponential decay/growth, and help in constructing the general solution of the homogeneous equation.

Method of Undetermined Coefficients

The method of undetermined coefficients is a systematic approach for finding a particular solution to non-homogeneous linear differential equations with constant coefficients. It involves guessing a form of the particular solution based on the non-homogeneous term and then determining the unknown coefficients by substitution into the original equation.

Linear Differential Equations

Linear differential equations are equations in which the unknown function and its derivatives appear linearly. This property leads to the superposition principle, meaning that the general solution can be written as a sum of independent solutions, often split into homogeneous and particular parts. They are central in modeling a wide range of phenomena in science and engineering.

Differential Operators

Differential operators are algebraic expressions involving derivatives, typically represented by symbols such as D for differentiation. They serve as a compact notation for writing differential equations and allow the use of algebraic techniques to analyze and solve these equations, especially when the operators have constant coefficients.

Transcript

00:01 This problem asks us to find the product of a monomial and a polynomial.

00:06 We're given the terms two -thirds, a to the fourth, that's the monomial, times the polynomial, 6a to the fifth minus 12a to the third minus 5 eighths.

00:27 To solve for this, we're going to use the distributive law of multiplication, which means we're going to take this monomial as on the outside and multiply it.

00:35 It by every term in the polynomial.

00:41 So that's going to look something like this.

00:45 2 thirds, a to the 4th times 6a to the 5th.

00:52 Now i'm going to do something a little bit different to make it easier or more clear.

00:57 Instead of minus 12a to the 3rd, i'm going to write this as plus negative 12a to the 3rd.

01:03 That way, we're sure that this is a plus sign.

01:06 So plus two -thirds a to the fourth times negative 12 a cubed.

01:17 The reason i do that is because if there's a negative, if there's a minus in front of the 12a cubed, sometimes it's easy to get confused with the negative goes here or a negative goes here.

01:29 With a positive sign, it's always clear.

01:32 And i'll do the same thing with the last term.

01:35 Instead of minus 5 eighth, it'll do plus negative 5 eighths.

01:38 So we have plus two -thirds a to the fourth times negative five -eighths.

01:51 So now each of these is something that we know.

01:55 We know how to do multiplication of coefficients and variables.

01:59 So we can solve for each term individually...

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