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Section 2.3 Linearity

Now that we’ve defined what it means for a term to be linear, we can describe what it means for an entire differential equation to be linear.
Essentially, a differential equation is linear if every term in the equation is linear. If just one term is nonlinear, the entire equation is nonlinear. We formally define linear differential equations as follows:

Checkpoint 18.

Select the true statement.
  • An equation is nonlinear if at least one of its terms is nonlinear.
  • Correct! A differential equation is nonlinear if any of its terms are nonlinear.
  • An equation is linear if at least one of its terms is linear.
  • Incorrect. A differential equation needs all of its terms to be linear.
  • As long as there are more linear terms than nonlinear terms, the equation is linear.
  • Incorrect.

πŸ“™ Definition 19. Linear Differential Equations.

A differential equation is linear if it can be written as a sum of linear terms involving the dependent variable and its derivatives:
 1 
This structure is known as a linear combination of the dependent variable and its derivatives.
\begin{equation} \underset{\text{linear term}}{\underbrace{a_n(t)\ y^{(n)}}} + \cdots + \underset{\text{linear term}}{\underbrace{a_2(t)\ y''}} + \underset{\text{linear term}}{\underbrace{a_1(t)\ y'}} + \underset{\text{linear term}}{\underbrace{a_0(t)\ y}} = f(t)\tag{2.2} \end{equation}
where each \(a_k(t)\) is a function of the independent variable only, and \(f(t)\) is the input or forcing term.
For example, the following differential equations are linear:
\begin{equation} 3\frac{dy}{dx} + 2y = 0, \qquad y''' - 2y' - \frac{7}{x}y = e^x, \qquad w'' - \tan(t) = 3t.\tag{2.3} \end{equation}
Each term involving the dependent variable or its derivatives is linear; none appear raised to a power, inside a function, or multiplied together.
In contrast, the following equations are nonlinear because they include at least one nonlinear term (✷):
\begin{equation} y' + \us{\large ✷}{\ul{ t y^2 }} = t, \qquad t \frac{dP}{dx} + \us{\large ✷}{\ul{ \sin(P) }} = t^2, \qquad \us{\large ✷}{\ul{\omega\omega^{(5)}}} + s = \us{\large ✷}{\ul{ e^\omega }}\tag{2.4} \end{equation}
In each case, it only took one term to violate the definition of a linear differential equation.
Let’s practice identifying the linearity of entire equations with a few examples.

🌌 Example 20. Identify the Linearity of the Equation.

Determine whether each of the following differential equations is linear:
\begin{equation*} y^{(5)} + \frac{y'}{x^2} + 5y = \ln x, \qquad P^{(6)} + \frac{m P'}{P} = (m - 1)^2 \end{equation*}
Solution.
In the first equation, all terms involving \(y\) or its derivatives appear linearly:
\begin{gather*} \underset{\overline{\text{linear}}}{\vphantom{\frac11}\color{BurntOrange} y^{(5)}} + \underset{\overline{\text{linear}}}{\frac{1}{x^2} \color{BurntOrange} y'} + \underset{\overline{\text{linear}}}{5 \color{BurntOrange} y} =\ \underset{\overline{\text{linear}}}{\ln x} \end{gather*}
So the equation is linear.
In the second equation, the term \(\frac{m P'}{P}\) is nonlinear, since \(P\) appears in both the numerator and denominator:
\begin{gather*} \underset{\text{linear}}{\underline{\color{BurntOrange} P^{(6)}}} + \underset{\overline{\text{nonlinear}}}{\frac{m \color{BurntOrange} P'}{\color{BurntOrange} P}} =\ \underset{\text{linear}}{(m - 1)^2} \end{gather*}
This makes the entire equation nonlinear.

Checkpoint 21.

(a) Identify the Linear Equations.

Hint.
Remember that a linear differential equation contains only linear terms. Four of these equations are linear.

(b) Identify the Nonlinear Equations.

Hint.
First, identify the dependent variable, then carefully examine each term to determine whether it is nonlinear. There are six nonlinear equations here.
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