Definition of homogeneous ODE

Question

In my lecture notes, it gives this following definition of a homogeneous ODE:

A differential equation is called homogeneous if it can be written in the form $x′=f(\frac{x}{t})$

Then in one of the questions, it says that $\frac{dy}{dt} + t^2y = y$ is a homogenous ODE.

Rearranging, we get $\frac{dy}{dt} = y(1-t^2) = y(1-t)(1+t)$

How can this be written in the form $\frac{dy}{dt}=f(\frac{y}{t})$?

Answer 1

Unfortunately, there are two different uses of the word homogeneous in the context of differential equations.

Most generally, let use suppose we have an ordinary differential equation of first order of the form $$ x' = F(x,t). $$ Sometimes there is a natural symmetry to the equations such that there exists $p$, $q$ real numbers such that:

whenever $x(t)$ is a solution, so is the function $y(t) = \lambda^p x(\lambda^q t)$ for any $\lambda$.

(Note that without loss of generality we can assume that at least one of $p,q$ is $1$.)

Necessarily for this to be true, by plugging into the equation, we have that

$$ y'(t) = \lambda^{p+q} x'(\lambda^q t) $$

which implies

$$ \lambda^{p+q} F( x,\lambda^q t) = F(\lambda^p x, t) $$

which gives a functional relation for $F$ and restricts the form $F$ can take.

---

The two distinct meanings of the word homogeneous refers to

1. The case where $p = 1$ and $q = 0$. That is to say: whenever $x(t)$ is a solution, so is $\lambda x(t)$. This seems to be the sense in which the "question" is using.

2. The case where $p = 1$ and $q = -1$. Here we have $F(\lambda x,t) = F(x,\lambda^{-1}t)$. This implies that there exists some function $f$ such that $F(x,t) = f(x/t)$.

Answer 2

A linear differential equation is called homogenous if any linear combination of a given set of solutions is also a solution.

$y'(t)=y(t)(t^{2}-1)$ is an example of such an equation.

Answer 3

If $y$ is a solution of the ODE $L(y) = 0$ and we try for $c y$ with some constant $c$, we get

$$ L (c y) = (c y)' + t^2(c y) - (cy) = c (y' + t^2 y - y) = c L(y) = c \cdot 0 = 0 $$

so the multiple is a solution too. It is homogenous, which correspponds to the more often seen definition that there is no term $f(t)$ (the inhomogenity, which is usually written on the right hand side).

I see no way to write $$ y' = (1 -t^2) y = (t - t^3) \frac{y}{t} $$ just as $y' = F(y/t)$ with $F$ having no $y$ and $t$ used.