I came across an interesting set of problems as follows:
\begin{gather}
\dfrac{dy}{dx} =\dfrac{1-xy^2}{2x^2y} \tag{1}\label{1} \\
\dfrac{dy}{dx}=\dfrac{2+3xy^2}{4x^2y} \tag{2}\label{2} \\
\dfrac{dy}{dx}=\dfrac{y-xy^2}{x+x^2y}. \tag{3}\label{3}
\end{gather}
We know these can be solved via Exact Equations and Integrating factors; however, the point of this exercise is to use the following substitution: $z=\dfrac{y}{x^n}$ where $n$ is any real number.
The difficulty I am facing is "how to determine the value of $n$ other than just guessing".
For instance, for \eqref{1} and \eqref{2}: $n=-\frac{1}{2}$ and for \eqref{3}: $n=-1$.
(Note for \eqref{2}, the textbook uses $n=\frac{3}{4}$, but when I tried $n=-\frac{1}{2}$, it worked just fine.)
Just to be clear, the chapter mainly discusses "Homogeneous Equations", but the last exercises invite us to try to extend this by using techniques that convert the differential equation into a separable equation (sometimes "homogeneous equation"), and this substitution indeed does that, but the value of $n$ is the "weird" part, as it seems determining the value of $n$ is by "guessing or trial and improvement".
Searching the internet, I deduced this substitution is related to Quasi-homogeneous equations.
For instance, substitution to homogeneous equation asks the same questions as I stated above; however, I am having the same difficulty in determining the degree of $x$ and degree of $y$ in the theorem in the reply.
Here is the theorem for convenience:
Theorem: A quasi-homogeneous equation $y'=F(x,y)$
with weight $\deg x=\alpha$, $\deg y=\beta$
can be reduced to an equation with separable variables by passing to the coordinates $(x,y^\alpha/x^\beta)$
in the domain $x>0$.
In the theorem the equation is quasi-homogeneous if and only if $F(e^{\alpha\xi}x,e^{\beta\xi}y)=e^{(\beta-\alpha)}\xi F(x,y)$. In polynomials you should only count degrees; see Wikipedia.
Ok, I am aware that the theorem is not meant for "introduction to DE", as it is found in Arnold's DE book. But, for polynomials, the only Issue I face now is how to determine $\alpha$ and $\beta$?
Searching further on the internet, I found a more concrete way to understanding the "quasi-homogeneous equations" : Solving a differential equation related to quasihomogeneous polynomials.
projectilemotion's answer discusses two methods, the issue in the first method is what I initially have, $\alpha$ and $\beta$ are called "weights" and the answer does not justify how to find them.
And for the second method discussed, it does not work for Equation \eqref{1}, listed above for example.
I.e., $\deg_x f = \operatorname{Max}(1,2) = 2 $ and $\deg_y f = \operatorname{Max}(2,1) = 2$.
Hence, the substitution should be $y=zx$; however, we know the substitution is $y=\sqrt{x}z$ and trying $y=zx$ does not work.
Maybe because issue in the method used to determine the degree, he said, by "convention."
However, I wonder if this convention works for DEs?
Now, how I solve these is as follows:
$$f(\lambda^\alpha x, \lambda^\beta y)=\lambda^{\beta – \alpha}f(x,y)$$
Which is the definition of quasi-homogeneous.
What I do is substitute these into the differential equation and by equating both sides I get for instance, $\alpha+2\beta=0$ , implying $\frac{\beta}{\alpha}=-\frac{1}{2}$, which is the correct substitution for \eqref{1}. But, this also assumes that the DE is Quasi-homogeneous, before knowing it.
Sorry for the very long post, but I was just trying to explain exactly where the "issue" lies, and that the previously asked questions regarding these topics did not clarify the degrees.
Note: these problems come from the second chapter (First Order Equations) in G. Simmons book: Differential Equations with Applications and Historical Notes; thus, for solving them, it does not require any knowledge of (very) "advanced" mathematics.
Thank you in advance.
Also asked by me on MO.
Best Answer
$\DeclareMathOperator\wt{wt}$To make \eqref{1} quasi-homogeneous, you need $1/(2x^2 y)$ and $(x y^2)/(2x^2 y)$ to have the same weight, so you need $\wt(x) + 2\wt(y) = 0$. If you want the substitution $z = y^\alpha/x^\beta$ to be of the form $z = y/x^n$, then you should take $\alpha = \wt(x) = 1$, so that $n = \beta = \wt(y)$ is $-1/2$. The same analysis applies for \eqref{2}.
To make \eqref{3} quasi-homogeneous, you need $x$ and $x^2 y$ to have the same weight, so you need $\wt(x) = 2\wt(x) + \wt(y)$; and then you need $y$ and $x y^2$ to have the same weight, so that you need $\wt(y) = \wt(x) + 2\wt(y)$. The solutions of these two equations lie on the line $\wt(y) = -\wt(x)$, so that taking $\alpha = \wt(x) = 1$ gives $n = \beta = \wt(y) = -1$.
On further reading, I see that you already discussed this analysis for \eqref{1}, and rejected it. Your ground made a lot of sense: this analysis seems to assume that the differential equation is quasi-homogeneous. Note, however, that, in the way I have phrased it, I never say that the differential equation is quasi-homogeneous; I just speculate on what it would take to show that it is so. If my goal were to prove that it was quasi-homogeneous, then I would have to go back and check that my idea works; but, if my goal is just to solve the differential equation, then no-one can stop me from making whatever substitution I like and seeing if it works.
That is, in this case we don't have to check the hypotheses of the theorem, since we aren't directly using its conclusion, only using it as inspiration for a method. But it's a good idea to get in the habit of checking hypotheses anyway, so let's do that. One easy way would have been for me to remember to say "the equation is quasi-homogeneous if and only if …" rather than "the equation is quasi-homogeneous only if …" up front.
But I didn't do that, so let's charge ahead. In \eqref{1} and \eqref{2}, we have that $\wt(1/(2x^2 y))$ is $-(2\wt(x) + \wt(y)) = -(2(1) + (-1/2)) = -3/2$ and $\wt((x y^2/(2x^2 y))$ is $\wt(x y^2/(2x^2 y))$ is $(\wt(x) + 2\wt(y)) - (2\wt(x) + \wt(y)) = ((1) + 2(-1/2)) - (2(1) + (-1/2)) = -3/2$. So we win, even if our idea to check this was motivated. The checks for \eqref{2} and \eqref{3} are similar.
If your question was the more basic "OK, but how did I know to check for quasi-homogeneity in the first place?", then an entirely unsatisfying answer is "because you're doing practice problems involving quasi-homogeneity." A slightly less unsatisfying answer is that the whole point of having a bag of differential-equations tricks is to be able to try various ones on various problems, without automatically knowing in advance which ones fit. (A more satisfying answer would be: here's a way automatically to test a differential equation for quasi-homogeneity. This amounts to the consistency of a certain system of linear equations, as above, but since the bare consistency is unlikely to be useful by itself, why not just try to solve the systems anyway, and either wind up with a solution or with the knowledge that you don't have a quasi-homogeneous equation after all?)