# How to prove an equation is homogenous

1. Dec 19, 2005

### ShaunP1989

Could someone please explain to me how to proove an equation is homogenous. We've done it in our AS class, but it still makes very little sense to me.

2. Dec 20, 2005

### HallsofIvy

Staff Emeritus
Since you haven't told us what equation you are talking about, all I can say is "use the definition".

The phrase "homogeneous equation" is actually used in two different ways. One way, which I think is not the one you are looking for is simply a linear equation in which every term involves the dependent variable, y, or one of its derivatives- that's pretty easy to spot.

The other refers to first order equations: a first order equation is homomogeneous if and only if replacing both x and y by $\lambda x$ and $\lambda y$ results in exactly the same equation: for example, if we replace both x and y by $\lambda x$ and $\lambda y$ respectively in
$$\frac{dy}{dx}= \frac{x+ 2y}{4x- y}$$
results in
$$\frac{dy}{dx}= \frac{\lambda x+2\lambda y}{4\lambda x- \lambda y}= \frac{\lambda(x+2y)}{\lambda(4x-y)}= \frac{x+2y}{4x-y}$$.

It occurred to me after I answered that this was posted under "precalculus mathematics" which makes no sense for a differential equations question- and I have no idea what an "AS" course is! If this is not the kind of "homogeneous" you are asking about, please tell us what you do mean!

Last edited: Dec 20, 2005
3. Dec 20, 2005

### ShaunP1989

well i posted it under general physics i think.
What we have been doing is taking an equation and then putting it in base units, and the from there proving it is homogeneous.
In my notes i have written as a definition for Homogenity the following:
I dont know if this will help you help me, its all very confusing for me at the moment, and kind of annoying because its the only part i really struggle with. I cant remember of the top of my head, but i think we are doing th Edexcel course.
Thanks for all the help :)

Shaun.

4. Dec 21, 2005

### HallsofIvy

Staff Emeritus
Now I am really confused! Apparently this is just a matter of checking to see if an equation is "dimensionally correct"- if each side has the right dimensional units. For example, while we know that "speed equals distance divided by time", v= d/t, a beginner might accidently think it should be d= v/t. But checking units, d is in meters, v in meters per second, m/s, and t is in seconds, s. That means that m/t is in m/s2 so d= v/t gives m= m/s2 which cannot be correct.

More complicated is Force= mass times acceleration: F= ma. In the mks system, m is in kg, a is in m/s2, and the mks unit for force is "Newton's" which in "base units" is kg m/s2 so that F= ma does indeed give kg m/s2= kg m/s2.

Caution: the fact that an equation is "dimensionally correct" does not mean the equation is true! But, of course, if an equation is not dimensionally correct, it can't be true.

5. Dec 21, 2005

### ShaunP1989

Ok its starting to make some sense, thanks alot for your help. I wish physics was easier to understand :(

Shaun