Ok this is when I'm learning about solving first order differential equations by substitution.

I searched for answers but found mainly these points:

  1. It's homogeneous if it can be expressed as $dy/dx = f(x/y)$;

  2. It's homogeneous if $f(tx,ty)=t^{\alpha}\cdot f(x,y)$;

  3. homogeneity is about a "scaling" property of the function;

But both my textbook and online videos are quite shallow with regard to the explanation of the meaning of these points. What exactly is $f(x/y)$ and how exactly does it relate to $t^{\alpha}$?

Also, I have such an equation to solve and it's supposed to be homogeneous: $2yy′+5=y^2+5x$

I just don't see why it is homogeneous and how after transforming it into the form: $\displaystyle y'= \frac{y^2-5+5x}{2y}$ the right hand side can be seen as $f(x/y)$?

Someone plz shed me some light thanks !

  • $\begingroup$ The function you give $f(x,y) = \frac{y^2 + 5x - 5}{2y}$ is not homogeneous as you suspected $\endgroup$ – DWade64 Sep 30 '18 at 17:37

Homogeneous function: functions which have the property for every $t$

$$ f(tx, ty) = t^n f(x,y) \tag{1}$$

This is a scaling feature. Remember working with single variable functions? Remember function transformations $f \to 2f$ would be a vertical stretch, $f \to f(2x)$ would be a horizontal compression, etc. You can do the same thing with multi-variable functions

$f(x,y) \to f(tx, ty)$ would be a uniform (same in all directions) "horizontal" compression of the original graph if $t > 1$. If $f$ is a homogeneous function, then such a compression is equivalent to the vertical stretch $t^n f(x,y)$ of degree $n$. Likewise, a "horizontal stretch" $f(tx,ty)$ for $-1 < t < 1$, would be equivalent to a vertical compression (with possible reflection as well depending on the degree $n$).

You've already had experience with one simply homogeneous function: $f(x) =x^2$. Because $f(3x)$, a horizontal compression of the graph, is equivalent to, $(3x)^2 = 9(x^2)$, a vertical stretch. Either or, both transformations end up doing the same thing. This is a homogeneous function.

Equivalent definition: $(1)$ is equivalent to, since $t \in \mathbb{R}$, we can make the substitution $t = 1/x$ since $1/x \in \mathbb{R}$ as well (Not quite. $t$ and $1/x$ are almost equivalent, but $1/x$ doesn't include $0$. You might think this is a problem but for what I'm trying to show, let's say it is not. If $t = 0$, this destroys the original graph anyways. $f(0x,0y) \to 0$ function. I will exclude such destruction). You get

$$ f(1, y/x) = \frac{1}{x^n}f(x,y) \\ F(y/x) = \frac{1}{x^n}f(x,y) $$

$$f(x,y) = x^n F(y/x) \tag{2} $$

$(1)$ implies $(2)$ and $(2)$ implies $(1)$ as you can check. So the definitions are equivalent. Likewise, with the substitution $t = 1/y$, you can show that an equivalent definition of a homogeneous function is

$$f(x,y) = y^n G(x/y) \tag{3} $$

I point all this out to show how the first definition relates to all this "division stuff" like $y/x$ that you are seeing. Therefore, if you have 2 homogenous functions $P$ and $Q$ of the same degree $n$, then

$$ F(x,y) = \frac{P(x,y)}{Q(x,y)} = \frac{P(tx,ty)/t^n}{Q(tx,ty)/t^n} = \frac{P(tx,ty)}{Q(tx,ty)} = F(tx,ty)$$

In other words, you have a homoegenous function of degree $0$. Continuing with this train of thought, using $t = 1/x$, you get that the division of $P/Q$ is a function of the form

$$ \frac{P(1, y/x)}{Q(1, y/x)} = F(y/x)$$

We could have gotten here more straightforwardly using the other definitions

$$ \frac{P(x,y)}{Q(x,y)} = \frac{x^n p(y/x)}{x^n q(y/x)} = F(y/x) $$

For your question, a homogeneous differential equation is of the form

$y' = F$ where $F$ is a homogeneous function of order $0$. Or if you like differentials,

$$P(x,y) dx = Q(x,y) dy$$ where $P$ and $Q$ are homogeneous functions of the same degree. I hope this helps

  • $\begingroup$ Thank you so much for the detailed answer! $\endgroup$ – user598895 Sep 30 '18 at 19:33
  • $\begingroup$ @Shuo, You're welcome! I give 3 different definitions of a homogeneous function. Their equivalency is not very rigorous. $(2)$ and $(3)$ definitely imply $(1)$ is true. However, going from $(1)$ to $(2)$ or $(3)$ is a little more shaky. Why? Besides the issue with $0$, because $x$ stands for a number in the domain of the function, and the domain of the function could be restricted (the domain is whatever it is), then $1/x$ might be limited in the possible values it can take while $t$ stands for any and all real numbers. However, our purpose is to solve the differential equation. I don't think $\endgroup$ – DWade64 Sep 30 '18 at 19:40
  • $\begingroup$ this issue is too serious or causes much problems. My textbook is also very shallow on homogeneous functions. However, my textbook (published 1963 and again 1985 -just pointing out that it's old and therefore "standard or accepted definitions" might change over time) does present these 3 definitions and hand-wavy claims they are all equivalent without proof $\endgroup$ – DWade64 Sep 30 '18 at 19:46
  • $\begingroup$ Yea I'm not concerned about the 0 as it doesn't change the general idea. Also for the P(x,y)/Q(x,y) part, I understood that it's gonna end up being F(y/x) if they have the same degree of n for t^n. But I don't see how that is related to the dy/dx = F(y/x) that I see in some videos. You made it clear that F(y/x) will imply that it is a division of 2 homogeneous equations of the same degree, but how does that imply that it is also the first differential dy/dx of a certain equation? On a side note, are most geometric shapes we know such as spheres and cubes homogeneous? $\endgroup$ – user598895 Sep 30 '18 at 19:56
  • 1
    $\begingroup$ yea that's what I mean :D Well your answer really shed some light into me, but I really hope those people writing books can actually take the effort to explain an important yet hard to grasp concept, not by just presenting it as it is and hope the students will get it eventually... That's wishful thinking lol. Anyway thank you! $\endgroup$ – user598895 Sep 30 '18 at 20:26

Similiar to algebraic equations differential equations are called homogeneous when you can rewrite them as $F(y^{(n)},\dots,y',y)=0$ or in other words when they are missing a "constant term" (which is in the context of DEs a term only consisting of $x$-terms and numbers).

For example the DE $y'=y$ is homegeneous where on the other hand the DE $y'=y+x$ is inhomogeneous. Your first point states the same as I have written.

The other two points are not the definition of the same homogeneity and to be honest I am not sure about them either.

Your given DE $2yy'+5=y^2+5x \Leftrightarrow 2yy'-y^2=5x-5 $ is not homogenous. It has to be written in the form $f\left(\frac{x}{y}\right)$ and not in the form $f(y,x)$. Furthermore this DE can in fact be solved by a subsititution; to be exact by setting $u=y^2$ and therefore $u'=2yy'$.


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