First of all, I am not asking about $v=y/x$ transformation kinda homogeneous.

Can we say this nonlinear differential equation is homogeneous $$y^{\prime}=ty^2.$$

Here there is no term without $y$, this is okay. But this is nonlinear equation.

I saw here some discussions and some people said it is just for linear equations. I want to give two links:

Dummies Series (assumes it can be nonlinear) (I also see some universities documents too).

Wolfram Math World (assumes it has to be linear)

Which one is the definition:

1) No term without $y$,

2) Linear and No term without $y$.

3) $y$ is a solution, then $\lambda y$ is solution for all $\lambda \in \mathbb R$.

Considering the references, I will use the 3rd one. This option is also meaningful.

up vote 2 down vote accepted
+100

The answer to the title-question is: Yes, it can. Here are two references.

The paper On the second order homogeneous quadratic differential equation by Roger Chalkley considers homogeneous quadratic differential equations of the form \begin{align*} Q(y)\equiv a\left(y^{\prime\prime}\right)^2+by^{\prime\prime}y^{\prime}+cy^{\prime\prime}y+d\left(y^{\prime}\right)^2 +ey^\prime y+fy^2=0\tag{1} \end{align*}

from which we conclude linearity and homogeneity are two concepts which do not exclude each other.

Note: The Math world page does not include the term linear in the definition of Homogeneous Ordinary Differential Equation. It rather uses a linear differential equation as an easy to follow example for a homogeneous ordinary differential equation.

Another example is the paper

Classification and Analysis of Two-Dimensional Real Homogeneous Quadratic Differential Equation Systems by Tsutomu Date.

  • Thanks for your answer. In order to make it clear: (I am understanding like that) Homogeneous differential equation means that after rewriting the (any) differential equation in the form $Q(y)=0$, then if $Q(\lambda y)=\lambda^r Q(y)$ for some real number $r$, we will call the equation as homogeneous. – student forever Mar 11 '17 at 19:17
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    @studentforever: Yes, homogeneous means if $y$ fits $Q(y)$ then so does $\lambda y$ for any constant $\lambda$. If the right side were a nonzero function then $\lambda y$ would no longer be a solution and is then called inhomogeneous. – Markus Scheuer Mar 11 '17 at 19:23
  • By the way, the link you give about homogeneity gives the equation in the form of linear ODE. – student forever Mar 11 '17 at 21:59
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    @studentforever: As indicated in the note, I don't think this is essential with respect to the definition of homogeneity. – Markus Scheuer Mar 11 '17 at 22:05
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    I am actually grateful. The knowledge is priceless. Many thanks. – student forever Mar 17 '17 at 22:41

It is to some extent a matter of convention and as with many things different conventions exist.

Consider the following related notion: a function $f : \mathbb R \to \mathbb R$ is called homogeneous of degree $\alpha$, if $f(kx) = k^\alpha f(x)$. We could apply this to ODEs and say that an ODE $$ y'(t) = f(t, y(t)) $$ is homogeneous, if the right hand side is homogeneous in the $y$-variable. With this definition your example would be homogeneous of degree 2.

I would not call the ODE $$ y'''' + ty'' + y^2 = 0$$ in the first example you linked homogeneous. Informally, the term homogeneous implies that we have some form of scaling behaviour.

I have come across homogeneous ODEs only in the context of linear equations, but it is possible that more general notions exist.

  • Thanks for your answer. I actually think similar. But I don't call first one homogenous of degree two. Left hand side is degree one and right hand side is degree two. I shall call $(y^{\prime\prime})^2=yy^\prime$ is homegenous of degree $2$ . Here if $y$ is solution, then $\lambda y$ is solution. Even solution space is not a vector space, it has this nice property. – student forever Mar 9 '17 at 0:33

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