Can't understand solution by substitution, any help?

[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) = x^αM(1,u) and N(x,y) = x^α (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 x^α. 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

 

[epenguin]

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) = x^αM(1,u) and N(x,y) = x^α (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 x^α. 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

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 x^α (or y^α) 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.)

 

[Simon Bridge]

I don't understand how the substitution simplifies [the calculation]

Try it and see.

 

[LCKurtz]

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) = x^αM(1,u) and N(x,y) = x^α (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 x^α. 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

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/differential-equations/first-order-equations/first-order-homogeneous-equations

 

[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!

 

[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) = x^αM(1,y/x)
N(x,y) = x^αN(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
x^αM(1,y/x) dx+x^αN(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