# Homogeneous Differential Equations Inspiration

Homogeneous first order differential equations can be solved by substituting $y/x = v$. I was wondering what is the inspiration for this. I am trying to understand the thinking behind this substitution. It is a known property of homogeneous polynomials that they can be expressed as functions of $y/x$, but how could you be sure that once you do that, the variables are separable? Again, the homogeneous polynomials $M$ and $N$ have to be of the same degree, otherwise, they are not separable. What's the thinking behind that? HFODEs are one of simplest type, so I was wondering if there's an intuitive way one could think about this (or is there any other analytic algebraic property of homogeneous polynomials that I am missing?)

• If a function $f(x, y)$ of TWO variables is homogeneous of order $n$, then it always has a special form $f(x, y) = x^n f(1, y/x)$. That means if I define $g(z) = f(1, z)$, a function of ONE variable, I can write $f(x, y) = x^n g(y/x)$. Very informally speaking, you need less information to describe $f(x, y)$ than a general function with $2$ variables. You only need $g$ (a function with 1 variable) and $n$ (an integer) to describe $f$. So why the substitution $v = y/x$? That's kind of hinted by $f(x, y) = x^n g(y/x)$, isn't it? – Tunococ Jan 18 '13 at 6:41

Tunococ, no offense intended, but what you have said is given in any standard textbook on differential equations. As I said earlier, some ideas become clear after you go through with them, and this analysis is precisely that. That's not what I am after.

At this point I should clarify my comment on your answer. When solving a DE, our primary concern is separation of variables. Simplification is a secondary concern. A substitution which transforms our equation into something more complex, but separates the variables is more useful than one which simplifies our DE without separating the variables. As far as simplification is concerned, transforming, $f(x)$ into $f(y/x)$ and then substituting $y/x = v$ is a logical step (not a natural guess!). But with respect to separation of variables, it is not.

Johann Bernoulli is said to be the first to suggest this method.The following statement is attributed to him "I attempt only to separate the indeterminate $x$ and it's differential $dx$, from the indeterminates $y$ and $dy$, which deserves the prize in this investigation, for otherwise the construction of the solution to the differential equation won't be achieved." (Ref - A History of Analysis - Hans Niels Jahnke). So there's a strong probability that when JB suggested putting $y/x = v$, he was looking at separation of variables, not simplification of the DE.

Now I am not insisting that he discovered this substitution in a a logical manner. What you suggested could have been true. I am only saying that, if such a possibility exists, we should explore it.

You did raise an important point however, that I overlooked - homogeneous functions need not be polynomials.As for degree not being an integer, that can be changed via substitution.

• The 'comment' in the above post, refers to one which I posted earlier (see below - Jan 19 6:32). Yes, I could have posted my earlier 'answers' as comments(because they were really comments on other answers). But I was new to the site, so..... .One of my earlier comments - (Jan 19 2:25 is not appearing in full (why?). I did not want that to happen to this comment, which was intended to clarify what I was looking for.That's why I posted it in the answer field. – Nikhil Panikkar Jan 24 '13 at 2:46
• "Not appearing in full": that's because when an comment-posted-as-answer is converted to an actual comment, the character limit is for comments is applied. Your comment-posted-as-answer is just a bit too long. Comments are encouraged to be succinct. – Willie Wong Jan 24 '13 at 11:37

Ok, here's how I can "be sure that once you do that, the variables are separable".

Suppose we're given a first-order ODE of the form $$M(x, y)dx + N(x, y)dy = 0.$$

If $M$ and $N$ are homogeneous functions of the same order, then $$\frac{dy}{dx} = -\frac{M(1, y/x)}{N(1, y/x)} \quad\text{ or }\quad \frac{dy}{dx} = -\frac{M(x/y, 1)}{N(x/y, 1)}.$$ You can choose to substitute either $u = y/x$ or $u = x/y$. Say you pick $u = y/x$. Then you substitute $dy = x du + u dx$ back in the equation to get

$$\frac{x du + u dx}{dx} = f(u)$$

where $f(u) = -\frac{M(1,u)}{N(1,u)}$ is determined by the previous equation. Rearrange to get

$$xdu = (u + f(u))dx \quad \Rightarrow \quad \frac{du}{u + f(u)} = \frac{dx}x.$$

You can now integrate. The solution is $$\int^{y/x} \frac{du}{u - \frac{M(1,u)}{N(1,u)}} = \log x.$$

P.S. Homogeneous functions do not have to be polynomials. The order of a homogeneous function doesn't have to be an integer.

• Tunococ, I appreciate your answering my question, but you are missing the central point of my question. It is true that once you make the substitution y/x = v, you need less information to represent M and N. But in the big picture you are still reducing, a differential equation of two variables - x and y to another differential equation in two variables, x and v. My question was, how do you know that the variables x and v are separable? Now I understand, that in problem solving,(whether in science, engineering or math), some ideas become clear only after you go through with them.But when you s – Nikhil Panikkar Jan 19 '13 at 2:25
• @NikhilPanikkar: As Tunococ says, the form $f(x,y)=x^ng(y/x)$ would become $f(x,v)=x^ng(v)$, so you've clearly separated them. Actually getting to this form is perhaps a different question, but one that also comes up elsewhere separately from ODEs, so finding it isn't quite as magical as it might seem. – Robert Mastragostino Jan 19 '13 at 2:31
• Thanks, for replying Robert. But I have to point out - just because two functions can be expressed as factors that doesn't mean that their variables are separable. In the equation we considered, if M and N are of different degree, then the variables are not separable.(just to make it clear, the equation is of the form - $dy/dx = \frac {M(x,y)} {N(x,y)}$ ) – Nikhil Panikkar Jan 19 '13 at 2:56
• Even though Robert's answer is incorrect, I think it has pointed me in the right direction. Consider the equation dy/dx = M(x,y)/N(x,y). Your first attempt will be to factorize M and N such that the equation becomes dy/dx = F1(x)G1(y)/F2(x)G2(y). Now the variables are separable. Now the question boils down to - when can a function M(x,y), homogeneous or not, can be factorized in to a function of only x and a function of only y? I am stuck here.Somebody help... – Nikhil Panikkar Jan 19 '13 at 6:20
• First of all, you should understand that from what I said, the substitution $v = y/x$ is a natural guess. And everything Robert said was correct. Please re-read his last sentence. – Tunococ Jan 19 '13 at 6:26

I want to share a small analysis, that I came up with yesterday.

Consider the equation $dy/dx = M(x,y)/N(x,y) - M$, $N$ are homogeneous with same degree.

$dy N(x,y) = dx M(x,y)$ -eq 1

Now let's try separating the variables:

$dy y^n N(x/y) = dx x^n M(y/x)$ - using the property of homogeneous polynomials -eq 2

Now if you look at the above equation, the terms $N(x/y)$ and $M(y/x)$ are the only terms that are not separated. Now if you could reduce them to the same variable, then maybe you might be able to separate them. It must be at this stage that JB might have hit upon the idea of substituting $y = vx$, for then you would be reducing them both to functions of $v$ (Actually on one side you have a function of $1/v$, but for polynomials this can be reduced to a function of $v$. I hope that more knowledgable people will check if this is true for non - polynomial functions.)

So putting $y = vx$

$d(vx) (vx)^n G(v) = dx x^n M(v)$

$= (vdx + xdv) (vx)^n = x^n T(v) dx$

Now if you simplify, you end up with:

$dx/x = -v/ (v^n+1 - T(v))$ .dv

$= F(x) dx = S(v) dv$

where variables have been separated.

Now I am not claiming a great logical approach here (because my algebraic manipulations vary from the conventional approach in a very minute manner), but I would like to point out a major difference - I am trying to separate the variables from the outset, and it is during this process, that I stumble upon the substitution $y = vx$, rather than the conventional approach where, you substitute $y = vx$ to simplify the DE, and stumble upon the separation of variables.

I am not claiming this as a final answer to my question, so please go through my analysis and comment - especially regarding the non - polynomial function case. If anyone can come up with a better analysis, please share.