Homogeneous Differential Equations Inspiration

Question

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?)

Answer 1

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.

Answer 2

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.

Answer 3

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.