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Sometimes it arrives to me that I try to solve a linear differential equation for a long time and in the end it turn out that it is not homogeneous in the first place.

Is there a way to see directly that a differential equation is not homogeneous?

Please, do tell me.

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    $\begingroup$ "Homogeneous" means that the only entities present are the unknown function and its derivatives (possibly with some coefficients). Thus $y''=xy$ is homogeneous; $y''=xy+x+1$ is not, since $x+1$ doesn't "involve" $y$ or its derivatives. $\endgroup$ – J. M. isn't a mathematician Dec 21 '11 at 5:21
  • $\begingroup$ homogeneous means you can prove the space of solutions is a vector space with your eyes closed in $2$ seconds. $\endgroup$ – mercio Nov 27 '15 at 19:30
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For a linear differential equation $$a_n(x)\frac{d^ny}{dx^n}+a_{n-1}(x)\frac{d^{n-1}y}{dx^{n-1}}+\cdots+a_1(x)\frac{dy}{dx}+a_0(x)y=g(x),$$ we say that it is homogenous if and only if $g(x)\equiv 0$. You can write down many examples of linear differential equations to check if they are homogenous or not. For example, $y''\sin x+y\cos x=y'$ is homogenous, but $y''\sin x+y\tan x+x=0$ is not and so on. As long as you can write the linear differential equation in the above form, you can tell what $g(x)$ is, and you will be able to tell whether it is homogenous or not.

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    $\begingroup$ I like this answer, thank you a lot. $\endgroup$ – VVV Dec 21 '11 at 5:23
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    $\begingroup$ This was a very good explanation!!! Cleared up a bit for sure! $\endgroup$ – user41920 Sep 20 '12 at 14:09
  • $\begingroup$ Finally clicked! Thank you. $\endgroup$ – iceburger Nov 4 at 3:42
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The simplest test of homogeneity, and definition at the same time, not only for differential equations, is the following:

An equation is homogeneous if whenever $\varphi$ is a solution and $\lambda$ scalar, then $\lambda\varphi$ is a solution as well.

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  • $\begingroup$ could you tell how the term 'homogeneous' came in mathematics? $\endgroup$ – justin Nov 13 '14 at 9:12
  • $\begingroup$ @justin I am not sure about terminology, but it seems because all scalar multiples of a solution are solutions and thus they "mix" as "one" not "separately". $\endgroup$ – Hasan Saad May 12 '15 at 10:13
  • $\begingroup$ @HasanSaad:Could you give an example.Is it only for linear equations? $\endgroup$ – justin May 12 '15 at 10:17
  • $\begingroup$ I am not sure it holds for non-linear ones, but for linear ones, you can easily just plug $\lambda\phi$ in the equation and it "gets out" of the derivatives, so it'll work. As for an example, $y'+y=0$, then $y=ce^{-x}$ holds for all scalars $c$. $\endgroup$ – Hasan Saad May 12 '15 at 10:20
  • $\begingroup$ @HasanSaad:Could you give an simple example with equations without log or derivatives or any calculus or exponential terms. $\endgroup$ – justin May 12 '15 at 12:05
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A homogeneous differential equation have same power of $X$ and $Y$ example :$- x+y dy/dx= 2y$

$X+y$ have power $1$ and $2y$ have power $1$ so it is an homogeneous equation.

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The best and the simplest test for checking the homogeneity of a differential equation is as follows :-->

Take for example we have to solve $$\frac{dy}{dx}=\frac{y+ (x^2+y^2)^\frac12}{x}$$

  1. Put $x=ax$ and $y=ay$ where a is any arbitrary constant.

  2. Now from the numerator and denominator take the constant as common with maximum power possible from both numerator and denominator each.

  3. If the constant gets cancelled throughout and we obtain the same equation again then that particular differential equation is homogeneous and the the power of constant which remains after cutting it to lowest degree is the degree of homogeneity of that equation. Hope That helped!

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  • $\begingroup$ What's the theory that supports this process you're suggesting here @aditya2222? $\endgroup$ – Neisy Sofía Vadori Jun 1 '18 at 16:38
  • $\begingroup$ What's happens if $\alpha<0$. I though a possible problem with the definition because $\sqrt{\alpha^2}=|\alpha|=-\alpha\not=\alpha$. $\endgroup$ – JimmyJP Jul 12 '19 at 14:25
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Any function like y and its derivatives are found in the DE then this equation is homgenous

ex. y"+5y´+6y=0 is a homgenous DE equation

But y"+xy+x´=0 is a non homogenous equation becouse of the X funtion is not a function in Y or in its derivatives

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    $\begingroup$ Well, it seems to me that you are a bit confused on either the definition of "homogeneous" or the explanation of your thoughts. Please, take your time to better elaborate your post. Also, having a look at other people's answers might help. $\endgroup$ – user228113 May 12 '15 at 10:16
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I think a differential equation is homogeneous if every term contains y or derivatives of y in the equation

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    $\begingroup$ Please, do not answer to four-year-old questions unless you are adding something substantial to the discussion. $\endgroup$ – user228113 Nov 27 '15 at 20:05
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if you are given an ODE say $f(x,y)=x^2-3xy+5y^2$ and they ask you to show if it is homogeneous or not here is how to do it

If a function $f$ has the property that $f(tx,ty)=t^nf(x,y)$ then the function is homogeneous of degree n...to prove if the above DE is homogeneous here is how to do it \begin{align*} f(x,y)&=x^2-3xy+5y^2\\ f(tx,ty)&=(tx)^2-3(tx)(ty)+5(ty)^2\\ &=t^2x^2-3t^2xy+5t^2y^2\text{ if we factor $t^2$ we get}\\ &=t^2[x^2-3xy+5y^2]\\ &=t^2f(x,y) \end{align*} hence the function is homogeneous of degree $n$

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The equations in the form $f(xy)$ can be said to be homogeneous also if they can be put in the form $dy/dx =f(y/x)$ or in other cases $f(x,y )=x^n g(y/x)$

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