# Definition of a Linear ODE

1. Sep 2, 2014

### JeweliaHeart

I am learning how to solve 1st order linear ODEs using the integrating factor.However, I run into confusion at the definition of a linear ODE.

According to a reliable source, a linear ODE must have the form:

(dy/dt) + p(t)y= g(t)

I don't understand what it means for an ODE to be linear or why the formula above makes an ODE linear.

2. Sep 2, 2014

### Pythagorean

That is only 1st order, a more general linear ODE would be

$$\frac{d^n y}{dt^n} + A_1 \frac{d^{n-1} y}{dt^{n-1}} + ... + A_{n-1} \frac{dy}{dt} + A_n y$$

The linear operator here is the series of integer derivatives on y. The A's are the coefficient functions (like your p(t)).

3. Sep 2, 2014

### JeweliaHeart

Okay. I guess what I am really asking is why 'linear' is used to describe this form. Does it mean that every time you graph the equation you get a curve in the shape of a line? If so, why?

4. Sep 3, 2014

### ShayanJ

A linear ODE, is an ODE that has the following properties:
1- If $y(x)$ is one of its solutions, so is $ay(x)$ for constant a.
2- If $y_1(x)$ and $y_2(x)$ are two of its solutions, $y_1(x)+y_2(x)$ is also a solution.

5. Sep 3, 2014

### HallsofIvy

In general "linear vector spaces" (which is, in essence the basic study of "linearity") a "linear transformation" is a function v= f(u) that maps one vector space to another with the properties that
a) f(u+ v)= f(u)+ f(v)
b) f(au)= af(u) where a is a number.

The set of all functions itself, with the usual addition and multiplication by a number, forms a vector space and the linear differential operators are the "linear transformations" for that vector space.

That means that dealing with linear differential equations, we have the entire theory of "linear vector spaces" to work with. (And is why "Linear Algebra" should be a pre-requisite to "Differential Equations".)