The textbook I am reading (Zill's "A First Course in Differential Equations with Modeling Applications) describes classifying ODEs as linear vs. nonlinear with the following statement:

An nth-order ordinary differential equation $$F(x,y,y',...,y^n) = 0$$ is said to be linear if F is linear in $$y,y',...,y^n$$

This means that an nth-order ODE is linear when it is $$a_n(x)y^{(n)}+a_{n-1}(x)y^{n-1}+...+a_1(x)y'+a_0(x)y - g(x) = 0$$

I understand that the intuitive definition of linearity is that an ODE is linear if:

1) the highest degree of the dependent variable (y) and all of its derivatives is 1
2) the coefficient of the dependent variable and its derivatives is either a constant, or some term with at most the independent variable (x)
3) the dependent variable and its derivatives aren't inside of other functions like sin(y)

However, the mathematical general definition is almost incomprehensible to me. I don't understand what the word "in" means when the author says that the function F is "linear in" followed by y and its derivatives separated by commas. And it does not say what g(x) means, what the collection of a(x) functions mean, etc.

  • $\begingroup$ The $x$ really isn't doing anything there. An ODE is linear if it is of the form $F(y,y',\ldots,y^{(n)})=0$ for some linear function $F$.. $\endgroup$ – Git Gud Jan 22 '16 at 22:00

Is just as in linear algebra.

We say $F$ is a linear differential operator (don't you worry about the fancy word) if its linear in every derivative. \begin{multline} F(x,y^{(0)},\ldots, a_j y^{(j)} + b_j z^{(j)},\ldots,y^{(n)}) = \\ a_j F(x,y^{(0)},\ldots, y^{(j)},\ldots, y^{(n)}) + b_j F(x,y^{(0)},\ldots, z^{(j)},\ldots, y^{(n)}) \end{multline} for all $j = 1,\ldots,n$.

If $F$ involves, say, $\sin(y)$, then it is't linear, because $\sin(a y + b z) \neq a \sin(y) + b \sin(z)$.

Of course, all $y$'s must be functions of the same argument; if not, then is not a differential equation.

The function $g(x)$ is a function that depends exclusively on $x$, and not on $y$ or any of its derivatives.

In the given example, $$ F(x,z_0,z_1,\ldots,z_n) = -g(x) + a_0(x) z_0 + a_1(x) z_1+ \ldots a_n(x) z_n. $$

  • $\begingroup$ This probably explains my confusion - I haven't taken linear algebra yet. It wasn't considered to be a pre-requisite of this course. I do think I understand what you're saying though - thank you! $\endgroup$ – dt5591 Jan 23 '16 at 0:25

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