Section 2.2 Linear & Nonlinear Terms
Subsection Linear Terms
Let \(y\) be the dependent variable. A term is linear if it appears as either:
\begin{equation}
a(t), \quad a(t)y, \quad \text{or} \quad a(t)y^{(n)}\text{,}\tag{2.1}
\end{equation}
where the coefficient, \(a(t)\text{,}\) can be a constant or a function of the independent variable \(t\) only, and \(y^{(n)}\) is the \(n\)-th derivative of \(y\text{.}\)
For example, the following terms are linear:
\begin{equation*}
\us{a(t)}{\us{\uparrow}{\ul{23}}}, \quad
\us{a(t)}{\us{\uparrow}{\ul{2t^3}}}y'', \quad
\us{a(t)}{\us{\uparrow}{\ul{\frac{1}{t}}}}y^{(5)}, \quad
\us{a(t)}{\us{\uparrow}{\ul{e^{2t}}}}y, \quad
\us{a(t)}{\us{\uparrow}{\ul{7\sin(t)}}}y'.
\end{equation*}
The coefficient of the term actually has no effect on whether it’s linear, regardless of how complicated \(a(t)\) appears.
Checkpoint 14.
Subsection Nonlinear Terms
A term is nonlinear if it is not in one of the forms in (2.1). Assuming \(y\) is the dependent variable, some telltale signs that a term is nonlinear include:
-
\(y\) or \(y^{(n)}\) are raised to a power other than 1 \(\left(\text{e.g.,}\ y^2,\ (y')^{-4}\right)\text{.}\)
-
\(y\) or \(y^{(n)}\) appear inside a nonlinear function \(\left(\text{e.g.,}\ \ln(y),\ \sin(y),\ e^{y'}\ \right)\text{.}\)
-
\(y\) or \(y^{(n)}\) are multiplied or divided by each other \(\left(\text{e.g.,}\ y y''',\ \sfrac{y'}{y}\right)\text{.}\)
Let’s apply these strategies by breaking down the linearity of each term in the following differential equation:
\begin{equation*}
y\frac{d^5y}{dt^5} + 2t^3\ \frac{d^2y}{dt^2} + \ln\left(\frac{dy}{dt}\right) + y^2 = t^3
\end{equation*}
The terms are:
- \(\ds y\frac{d^5y}{dt^5}\)
- Nonlinear. This term involves a product of the dependent variable and its derivative, violating the rule that only one of them can appear in a term.
- \(\ds 2t^3\frac{d^2y}{dt^2}\)
- Linear. The derivative \(y''\) appears by itself and to the first power. The coefficient \(2t^3\) does not affect linearity.
- \(\ds \ln\left(\frac{dy}{dt}\right)\)
- Nonlinear. The derivative \(y'\) appears inside a nonlinear function, the natural logarithm.
- \(\ds y^2\)
- Nonlinear. The dependent variable appears raised to a power other than one.
- \(\ds t^3\)
- Linear. This term doesn’t involve the dependent variable or any of its derivatives.
Checkpoint 15.
See if you can identify the linearity of the terms in the following examples before looking at the solutions.
🌌 Example 16. Identify the Linearity of the Terms.
Determine whether each term in the equation is linear or nonlinear:
\(\dfrac{1}{t}y'' + y^2 + \ln(y') = e^t\)
Solution.
Focus on the dependent variable, \({\color{BurntOrange} y}\text{,}\) and its derivatives.
\begin{gather*}
\underset{\overline{\text{linear}}}{\frac{1}{t}{\color{BurntOrange} y''}} +
\underset{\overline{\text{nonlinear}}}{{\color{BurntOrange} y}^2} +
\underset{\overline{\text{nonlinear}}}{\ln({\color{BurntOrange} y'})} =\
\underset{\overline{\text{linear}}}{e^t}
\end{gather*}
\(P^{(6)} + \dfrac{m P'}{P} = (m - 1)^2\)
Solution.
Focus on the dependent variable, \({\color{BurntOrange} P}\text{,}\) and its derivatives.
\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}}{\underline{(m - 1)^2}}
\end{gather*}
Checkpoint 17.
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