Homogeneous Equations

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 4.2 – Homogeneous Linear Equations: The General Solution

Expected Educational Results

• Objective 10–1: I can identify if two or more functions are linearly-independent.
• Objective 10–2: I can identify the characteristic equation for $n^{\text{th}}$-degree homogeneous linear ODEs.
• Objective 10–3: I can find the most general solution to $n^{\text{th}}$-degree homogeneous linear ODEs.

Algebra

Fundamental Theorem of Algebra

Theorem: Fundamental Theorem of Algebra

A polynomial of degree $n$, $P(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0$, $a_n\ne 0$, has exactly $n$ factors, counting multiplicities.

Corollary: Fundamental Theorem of Algebra

$P(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0=0$, where $a_n\ne 0$, has exactly $n$ solutions, counting multiplicities.

Solving Polynomials

Distributive Property and Zero Product Property of Real Numbers

Example 01

$x^3-x=0 \Rightarrow x(x^2-1)=0\\ \Rightarrow x=0 \text{ or } x^2-1=0\\ \Rightarrow x=0 \text{ or } x=-1 \text{ or } x=1$

Example 02

$x(x-3)=4 \Rightarrow x^2-3x=4\\ \Rightarrow x^2-3x-4=0\\ \Rightarrow (x-4)(x+1)=0\\ \Rightarrow x-4=0 \text{ or } x+1=0 \\ \Rightarrow x=4 \text{ or } x=-1$

Quadratic Equations

Example 03

Distributive Property and Zero Product Property of Real Numbers

$ax^2+bx+c=0 \Rightarrow (x+s)(x+t)=0, \text{ where } (s)(t)=c \text{ and } s+t=b\\ \Rightarrow x+s=0 \text{ or } x+t=0\\ \Rightarrow x=s \text{ or } x=t$

Example 04

Distributive Property and Zero Product Property of Real Numbers

$x^2-1=0 \Rightarrow (x+1)(x-1)=0 \\ \Rightarrow x=-1 \text{ or } x=1$

Example 05

Quadratic Formula

$ax^2+bx+c=0 \Rightarrow x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Quadratic-Like Equations

Definition: Quadratic-Like Equation

Quadratic-like equations have the form: $\displaystyle ax^{2n}+bx^{n}+c=0, n\in\mathbb{R}$.

Method to Solve Quadratic-Like Equations

1. Let $\displaystyle u=x^{n}$ and substitute into the quadratic-like equation;
2. Factor: $\displaystyle au^2+bu+c=0 \Rightarrow (u+s)(u+t)=0$, where $(s)(t)=c$ \text{ and } $s+t=b$;
3. Solve in terms of $u$: $\displaystyle (u+s)(u+t)=0 \Rightarrow (u+s)(u+t)=0 \Rightarrow u=-s \text{ or } u=-t$;
4. Solve in terms of $x$: $\displaystyle u=-s \Rightarrow x^{n}=-s \Rightarrow x=(-s)^{\frac{1}{n}}$ $\displaystyle u=-t \Rightarrow x^{n}=-t \Rightarrow x=(-t)^{\frac{1}{n}}$

Example 06

$x^4-5x^2+4=0 \text{, Let } u=x^2 \\ \Rightarrow u^2-5u+4=0 \\ \Rightarrow (u-4)(u-1)=0 \\ \Rightarrow u=4 \text{ or } u=1 \\ \Rightarrow x^2=4 \text{ or } x^2=1 \\ \Rightarrow x=\pm 2 \text{ or } x=\pm 1$

Finding Zeros Using Rational Root Theorem and Polynomial or Synthetic Division

Given the polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots + a_2x^2+a_1x+a_0=0,a_n\ne 0$

Let $p_1,p_2,p_3,\ldots$ be the factors of $a_0$. Let $q_1,q_2,q_3,\ldots$ be the factors of $a_n$. Then the set of potential rational roots of the polynomial are

$\left\{\dfrac{p_1}{q_1},-\dfrac{p_1}{q_1},\dfrac{p_1}{q_2},-\dfrac{p_1}{q_2},\dfrac{p_2}{q_1},-\dfrac{p_2}{q_1},\ldots\right\}$

Example 07

Find all three real roots of: $12x^3-25x^2+x+2=0$.

The factors of $a_0=2$ are $p=-1,1,-2,2$.

The factors of $a_3=12$ are $q=-1,1,-2,2,-3,3,-4,4,-6,6,-12,12$.

Then the set of distinct potential rational roots of the polynomial are:

$\left\{\dfrac{-1}{-1},\dfrac{-1}{1},\dfrac{-2}{-1},\dfrac{-2}{1},\dfrac{-1}{-2},\dfrac{-1}{2},\dfrac{-1}{-3},\dfrac{-1}{3},\dfrac{-2}{-3},\dfrac{-2}{3},\dfrac{-1}{-4},\dfrac{-1}{4},\dfrac{-1}{-6},\dfrac{-1}{6},\dfrac{-1}{-12},\dfrac{-1}{12}\right\}$

Now use polynomial or synthetic division to find the three factors.

Using Polynomial Division

a. Divide the polynomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-(-1)$:

$\dfrac{12x^3-25x^2+x+2}{x+1}$

Since the above polynomial division has a non-zero remainder, then $x+1$ is not a factor of $12x^3-25x^2+x+2=0$.

b. Divide the polynomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-2$:

Since the above polynomial division has a zero remainder, then $x-2$ is a factor of $12x^3-25x^2+x+2=0$.

c. Divide the polynomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-\frac{1}{3}$:

Since the above polynomial division has a zero remainder, then $x-\frac{1}{3}$ is a factor of $12x^3-25x^2+x+2=0$.

d. Divide the polynomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-\left(-\frac{1}{4}\right)$:

Since the above polynomial division has a zero remainder, then $x+\frac{1}{4}$ is a factor of $12x^3-25x^2+x+2=0$.

Thus, $\displaystyle 12x^3-25x^2+x+2 = 12(x-2)\left(x-\frac{1}{3}\right)\left(x+\frac{1}{4}\right)$.

Finding Zeros Using Synthetic Division

a. Divide the polynomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-(-1)$:

The steps above are provided to show the method of synthetic division. Your work should only look like the following:

Since the above polynomial division has a non-zero remainder (the right-most number on the last row), then $x+1$ is not a factor of $12x^3-25x^2+x+2=0$.

b. Divide the polynomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-2$:

Since the above polynomial division has a zero remainder (the right-most number on the last row), then $x-2$ is a factor of $12x^3-25x^2+x+2=0$.

c. Divide the polyomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-\frac{1}{3}$:

Since the above polynomial division has a zero remainder (the right-most number on the last row), then $x-\frac{1}{3}$ is a factor of $12x^3-25x^2+x+2=0$.

d. Divide the polyomial by $x-\left(\text{ rational root}\right)$, i.e., divide by $x-\left(-\frac{1}{4}\right)$:

Since the above polynomial division has a zero remainder (the right-most number on the last row), then $x-\left(-\frac{1}{4}\right)$ is a factor of $12x^3-25x^2+x+2=0$.

Thus, $\displaystyle 12x^3-25x^2+x+2 = 12(x-2)\left(x-\frac{1}{3}\right)\left(x+\frac{1}{4}\right)$.

Practice 01

Solve, i.e., find all solutions, to the following equations:

1. $\displaystyle x-3=0$
2. $\displaystyle x^2-3=0$
3. $\displaystyle x^2-4x+4=0$
4. $\displaystyle x^2-5x+6=0$
5. $\displaystyle x^2+6x=-4$
6. $\displaystyle x^3+3x^2+2x=0$
7. $\displaystyle x^4-5x^3-6x^2=0$
8. $\displaystyle x^2+2x-2=0$
9. $\displaystyle x^3-x^2-5x=3$
10. $\displaystyle x^3-7x-6=0$
11. $\displaystyle x^4+4x^3+3x^2-4x-4=0$
12. $\displaystyle x^4+5x^3+9x^2+7x+2=0$

Determinants

Let $A$ be a $n\times n$ square matrix. The determinant of $A$ is represented by det($A$) or $|A|$.

Properties of Determinants

If $\text{det}(A) = 0$, then the following are all equivalent:

• the columns of $A$ are linearly dependent vectors in $\mathbb{R}^n$
• the rows of $A$ are linearly dependent vectors in $\mathbb{R}^n$
• the matrix $A$ is not invertible
• the system of linear equations either has no solution or infinitely-many solutions

If $\text{det}(A) \ne 0$, then the following are all equivalent:

• the columns of $A$ are linearly independent vectors in $\mathbb{R}^n$
• the rows of $A$ are linearly independent vectors in $\mathbb{R}^n$
• the matrix $A$ is invertible
• the system of linear equations has a unique solution

Computation of Determinants

$2 \times 2$ Matrix

Let $A= \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$

Then, $\text{det}\left(A\right) = |A| = ad - bc$

$3 \times 3$ Matrix

Let $B= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & k \end{array} \right)$

Then, using cofactors, $\text{det}\left(B\right) = |B| = a \left| \begin{array}{cc} e & f \\ h & k \end{array} \right| -b \left| \begin{array}{cc} d & f \\ g & k \end{array} \right| + c \left| \begin{array}{cc} d & e \\ g & h \end{array} \right|$

Practice 02

Find the determinants of the following matrices:

1. $\displaystyle A= \left( \begin{array}{cc} x^2 & e^x \\ 2x & e^x \end{array} \right)$
2. $\displaystyle B= \left( \begin{array}{cc} e^x & e^{2x} \\ e^x & 2e^{2x} \end{array} \right)$
3. $\displaystyle C= \left( \begin{array}{cc} e^x & xe^{x} \\ e^x & xe^{x}+e^x \end{array} \right)$
4. $\displaystyle D= \left( \begin{array}{cc} \sin{(x)} & e^x \\ \cos{(x)} & e^x \end{array} \right)$
5. $\displaystyle E= \left( \begin{array}{cc} \sin{(2x)} & \cos{(2x)} \\ 2\cos{(2x)} & -2\sin{(2x)} \end{array} \right)$
6. $\displaystyle F= \left( \begin{array}{ccc} 1 & e^x & e^{2x} \\ 0 & e^x & 2e^{2x} \\ 0 & e^x & 4e^{2x} \end{array} \right)$
7. $\displaystyle F= \left( \begin{array}{ccc} x^2 & e^x & \sin{(2x)} \\ 2x & e^x & 2\cos{(2x)} \\ 2 & e^x & -4\sin{(2x)} \end{array} \right)$
8. $\displaystyle F= \left( \begin{array}{ccc} x^2+3x+4 & x^2+5 & e^x \\ 2x+3 & 2x & e^x \\ 2 & 2 & e^x \end{array} \right)$

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Last Modified: Monday, 6 September 2020 13:33 EDT