# Can't understand solution by substitution, any help?

1. Feb 2, 2015

### MarcL

So I just started my DE class and I'm kinda stuck on solutions by substitutions. My book explains it as just having a homogeneous function of degree α, we can also write M(x,y) = xαM(1,u) and N(x,y) = xα (1,u) where u = y/x

I don't understand how the substitution simplifies our life ( there's no proof in my book , it just says that a function can have y = g(x,u) and replace g(x,u) in f(x,y)). To me it seems more complicated than anything.

Secondly, how do I find my xα. I tried doing many examples, but without success I never get the correct coefficient, so there is something I'm not understand. I just have to factor out the highest degree of x, no?

Any help would be greatly appreciated

2. Feb 3, 2015

### epenguin

Your question gives so little context that's it will have to be answered by someone who knows the background so well he knows what you are not saying.

But I can say that if you have a formula in two variables, if it is homogeneous you can extract xα (or yα) and be able.e to work with a formula in one variable. You might find one easier to deal with than two, no?

(In algebra quite often the opposite is done. A one-variable polynomial is changed into a two-variable homogeneous one because you get a. more convenient-to-work-with symmetry.)

Last edited: Feb 3, 2015
3. Feb 4, 2015

### Simon Bridge

Try it and see.

4. Feb 25, 2015

### LCKurtz

You don't have to find $\alpha$. The point of the substitution is that it always makes the given DE separable. There are lots of resources and examples on the net. Here's one, for example:
http://www.cliffsnotes.com/math/dif...r-equations/first-order-homogeneous-equations

5. Mar 1, 2015

### MarcL

Late reply but yeah, there are a lot of ressources' however, sometimes it is nice to hear a "explain like I'm 5 " version. but thank you!

6. Mar 1, 2015

### lurflurf

This allows us to write the functions of two variables as products of two one variable functions which is much easier.
You do not show the equation, so I assume a usual one.

M(x,y) dx+N(x,y) dy=0
using
M(x,y) = xαM(1,y/x)
N(x,y) = xαN(1,y/x)
dy=d(x y/x)=(y/x)dx+x d(y/x)
we have
M(x,y) dx+N(x,y) dy=0
xαM(1,y/x) dx+xαN(1,y/x) [(y/x)dx+x d(y/x)]=0
dx/x+N(1,y/x)/[M(1,y/x)+N(1,y/x)] d(y/x)=0
which is separated