##
Section 6.1 Classification

Now that we’ve tackled some foundational techniques for solving differential equations, it’s time to introduce a new type: linear homogeneous differential equations with constant coefficients, or LHCC equations for short. But how do we recognize them? Let’s break down each part of the term *linear*, *homogeneous*, and *constant coefficients* to understand these equations fully.

First, recall that a linear differential equation looks like this:

\begin{equation}
{\color{blue} \os{\text{coefficient}}{\os{\downarrow}{a_n(x)}}} y^{(n)} + \cdots +
{\color{blue} \os{\text{coefficient}}{\os{\downarrow}{a_2(x)}}} y'' +
{\color{blue} \os{\text{coefficient}}{\os{\downarrow}{a_1(x)}}} y' +
{\color{blue} \os{\text{coefficient}}{\os{\downarrow}{a_0(x)}}} y =
\os{\text{constant term}}{\os{\downarrow}{f(x).}}\tag{24}
\end{equation}

The key idea here is that each term involves either a derivative of \(y\) or \(y\) itself, all multiplied by coefficients that depend on \(x\text{.}\)

Next, the equation is homogeneous if the constant term is zero, meaning \(f(x) = 0\text{.}\) This ensures the left-hand side of the equation equals zero.

Finally, if all the coefficients are constants (i.e., numbers, not functions of \(x\)), the equation is said to have constant coefficients.

Combining these ideas, we arrive at the formal definition of an LHCC equation.

##
Definition 1. LHCC Equation.

A linear homogeneous differential equation with constant coefficients (LHCC) is of the form:

\begin{equation}
a_n\ y^{(n)} + \cdots + a_2\ y'' + a_1\ y' + a_0\ y = 0\tag{25}
\end{equation}

where \(a_n,\ a_{n-1},\ \ldots,\ a_2,\ a_1,\ a_0\) are constants.

Now that we have the definition, let’s practice identifying these equations with a few examples.

##
Example 2.

Identify which of the following are linear homogeneous differential equations.

\begin{equation*}
3y'' - 2ty' + y = 0, \quad
\frac{dg}{dx} + 3x^2 = 0, \quad
\frac{d^2s}{dt^2} + \frac{ds}{dt} = 4s
\end{equation*}

## Solution. Solution

The trick is to check if the constant term is zero and the equation has only \(y\) and its derivatives on the left-hand side. Let’s rewrite each equation with the constant term isolated:

\begin{equation*}
3y'' - 2ty' + y = {\color{blue} 0}, \quad
\frac{dg}{dx} = {\color{blue} -3x^2}, \quad
\frac{d^2s}{dt^2} + \frac{ds}{dt} - 4s = {\color{blue} 0} \text{.}
\end{equation*}

The first and last equations are homogeneous, but the second is not because the right-hand side is non-zero.

##
Example 3.

Determine which of the following equations have constant coefficients.

\begin{equation*}
3y'' - 2\tau y' + y = \tau, \quad
\sqrt{t} - \frac{dP}{dt} - \frac{P^2}{2} = 1, \quad
s''' = \frac{7s}{w}
\end{equation*}

## Solution. Solution

Check if each coefficient is constant:

\begin{equation*}
\os{\text{yes}}{\os{\downarrow}{\ul{3}}} y'' -
\os{\text{no}}{\os{\downarrow}{\ul{2\tau}}}\ y' +
\os{\text{yes}}{\os{\downarrow}{\ul{1}}} y = \tau, \quad
\sqrt{t} -
\os{\text{yes}}{\os{\downarrow}{\ul{1}}} \frac{dP}{dt} -
\os{\text{yes}}{\os{\downarrow}{\ul{\frac{1}{2}}}} P^2 = 1, \quad
\os{\text{yes}}{\os{\downarrow}{\ul{1}}} s''' =
\os{\text{no}}{\os{\downarrow}{\ul{\frac{7}{w}}}} s\text{.}
\end{equation*}

The second equation has constant coefficients, while the others do not.

###
Reading Questions Check your Understanding

###
1. *Select all the true statements below*.

Select all the true statements below.An LHCC equation must have constant coefficients.

Correct! Constant coefficients are one of the defining features of LHCC equations.

If an equation contains non-linear terms like \( y^2 \), it can still be classified as LHCC.

Incorrect, LHCC equations are linear, meaning they cannot contain non-linear terms like \( y^2 \).

The equation \( y’ + 3y = 0 \) is both linear and homogeneous.

Correct! This is a first-order linear homogeneous differential equation.

A non-homogeneous equation has a constant term that is not zero.

Correct! If the constant term is not zero, the equation is non-homogeneous.

###
2. *Fill in the Blank: Classifying a DE*.

The differential equation

\begin{equation*}
y'' + 5y' + 6y = 0
\end{equation*}

is classified as a _______ homogeneous differential equation with _______ coefficients.

linear, constant

Correct! The equation is linear (because there are no powers or products of the dependent variable and its derivatives) and has constant coefficients.

non-linear, constant

Incorrect. Try again.

linear, variable

Incorrect. Try again.

non-linear, variable

Incorrect. Try again.

###
3. *Identifying a Homogeneous Equation*.

Which of the following equations is homogeneous?- \( y’’ + 4y^2 + 3 = 0 \)
Incorrect. This equation has a non-linear term (\(y^2\)), so it is not linear.

- \( y’’ + 4y’ + 3y = 7 \)
Incorrect. The equation has a non-zero constant term, making it non-homogeneous.

- \( y’ + 2y = x^2 \)
Incorrect. The right-hand side has a non-zero function of \(x\), so this is non-homogeneous.

- \( y’’ + 4y’ + 3y = 0 \)
Correct! This equation has no constant term and is homogeneous.

###
4. *Match the Label to the DE*.

Match each label on the left to an appropriate DE on the right.

Key: L = linear, H = homogeneous, CC = constant coefficient

*Note: Multiple matches can be correct, but there is only one perfect matching where all are correct.*

- CC
- \(\ds r'' + 3r' + 2r^2 = 0 \)
- LH, order 1
- \(\ds 3t r' = 0 \)
- LHCC
- \(\ds r'' + 3r' + 2r = 0 \)
- LH, order 2
- \(\ds r'' + 3t r' = 2r \)
- LCC
- \(\ds y''' - 7y = e^x \)
- L
- \(\ds t^5\frac{dw}{dt} = \sin(3t) \)

###
5. *Select the LHCC Equations*.

Select all of the LHCC Differential Equations.
\(\ds y'' + \sin(y) = 17t \) |
\(\ds y'' + \frac{y'}{t^2} + y = 7t \) |
\(\ds y'' + 3y' + 2y = 0 \) |

\(\ds y'' + y^2 = 17t \) |
\(\ds y'' + y'y = 0 \) |
\(\ds y = y' \) |

\(\ds \frac15 y^{(8)} - \sqrt{15}y' = y \) |
\(\ds s'''+\pi s = \frac{7}{w} \) |
\(\ds t\frac{dP}{dt} + P^2 = \sqrt{t} \) |

\(\ds \frac{dg}{dx} + 3x^2 = 0 \) |
\(\ds \frac15 y^{(8)} - \sqrt{15}x y' = y \) |
\(\ds \frac{d^2s}{dt^2} + \frac{ds}{dt} = 4s \) |

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