Ok this is when I'm learning about solving first order differential equations by substitution.
I searched for answers but found mainly these points:
It's homogeneous if it can be expressed as $dy/dx = f(x/y)$;
It's homogeneous if $f(tx,ty)=t^{\alpha}\cdot f(x,y)$;
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 !
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$\begingroup$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
$\endgroup$ 9 $\begingroup$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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