Differential equation modeling question

[sergey90]

Hi guys, just embarking on the journey of modeling with differential equations and solving them. I have a more conceptual quesiton. Linear differential equations in my book are defined to be ones where F(t, y, y', y'',...,yn) = 0 is a linear function of the dependent var y and all of its derivatives. Does this mean that if in the equation we have say "...ty" it is still linear? If yes, then how does this translate to linear algebra where in a system of linear equations the functions have to be linear for all the variables, including the "free" ones?

 

[DelCrossE]

A linear differential equation has the form:

a$_{n}$(t)$\frac{d^{n}y}{dt^{n}}$ + a$_{n-1}$(t)$\frac{d^{n-1}y}{dt^{n-1}}$ +...+ a$_{1}$(t)$\frac{dy}{dt}$ + a$_{0}$(t)y = F(t)

If a$_{0}$(t) = t, then "...ty" is linear if the entire equation can be written in the form above.

As for how this relates to linear algebra, it is important to note that y is not a "free variable", it is a function of t.

 

[diazona]

Hi guys, just embarking on the journey of modeling with differential equations and solving them. I have a more conceptual quesiton. Linear differential equations in my book are defined to be ones where F(t, y, y', y'',...,yn) = 0 is a linear function of the dependent var y and all of its derivatives. Does this mean that if in the equation we have say "...ty" it is still linear? If yes, then how does this translate to linear algebra where in a system of linear equations the functions have to be linear for all the variables, including the "free" ones?

I'm guessing you're familiar with linear algebra in the form of matrix algebra? Where the important objects are vectors and matrices, and you can multiply a matrix by a vector to get a different vector? Well, linear differential equations work the same way, except just replace vectors with functions and matrices with operators.

In this sense, an operator is something which acts on a function to give a different function, just like a matrix acts on a vector to produce a different vector. Specifically, that means it has to be linear: if $\mathbf{A}$ and $\mathbf{B}$ are operators and $f$ and $g$ are functions, you need
$$\begin{align}\mathbf{A}[f(x)+g(x)] &= \mathbf{A}f(x)+\mathbf{A}g(x) \\ (\mathbf{A}+\mathbf{B})f(x) &= \mathbf{A}f(x) + \mathbf{B}f(x)\end{align}$$
among other properties. Subject to these constraints, there are basically two fundamental things you can do to a function: either multiply it by some quantity which may depend on $x$, or take its derivative with respect to $x$. Any linear transformation of one function into another can be represented by a linear combination of multiplications and derivatives.

Now, as you know, when you solve a linear system of algebraic equations using matrix algebra, you're solving a system of the form
$$A\vec{x} = \vec{b}$$
A linear differential equation is the same thing, only $\vec{x}$ and $\vec{b}$ become functions and $A$ becomes an operator:
$$\mathbf{A}f(x) = g(x)$$
Remember that $\mathbf{A}$ is a linear combination of multiplicative factors and derivatives. So in general, you can have something like
$$\mathbf{A}f(x) = c(x) f(x) + a_1f'(x) + a_2f''(x) + \cdots$$
Hopefully you can see that with this definition, the differential equation above can be written
$$F(x, f', f'', \cdots) = -g(x) + c(x) f(x) + a_1f'(x) + a_2f''(x) + \cdots = 0$$
The inhomogeneous term $g(x)$ and the multiplicative coefficient $c(x)$ can be any arbitrary (fixed) functions of $x$, they just can't depend on the function $f$.