Homogeneous Diff. Eqn Finding Solution

  • Thread starter Thread starter Arman777
  • Start date Start date
  • Tags Tags
    Homogeneous
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
Arman777
Insights Author
Gold Member
Messages
2,163
Reaction score
191

Homework Statement


##(2xy+3y^2)dx-(2xy+x^2)dy=0##

Homework Equations

The Attempt at a Solution


It's a homogeneous equation since
we can write,
##M(x,y)=(2xy+3y^2)## and ##M(tx,ty)=t^2M(x,y)## and ##N(x,y)=(2xy+x^2)## and ##N(tx,ty)=t^2N(x,y)##
since orders of t are same they are homogeneous.

Now I can say that ##\frac {dy} {dx}=g(\frac {y} {x})##

where ##v=\frac {y} {x}##.
Now I think there is kind of 2 ways to solve the equation. First way is just put the y=vx in the main equation and try to solve it like that and later by doing separation we can get the result. But I couldn't find a solution
Here it comes,

##(2xvx+3v^2x^2)dx-(2xvx+x^2)(dvx+vdx)=0##
let's gather the terms with dx and dv and separate them,
##(v^2x^2+2x^2v-x^2v)dx-(2x^3v+x^3)dv=0## which later on I don't know what to do. I know that I should separate them but it seems confusing.

Other way is to write in the from of ##\frac {dy} {dx}=\frac {(2xy+3y^2)} {(2xy+x^2)}##

so ##v+\frac {dv} {dx}x=\frac {(2xy+3y^2)} {(2xy+x^2)}##
but then I couldn't separate them so that we can get ##\frac {y} {x}## in the right side.
 
Last edited:
Physics news on Phys.org
Orodruin said:
I suggest replacing ##y## by ##vx## on the rhs of your last expression.
Yes I just now realized that, then I can cancel ##x^2##.
 
Arman777 said:
so ##v+\frac {dv} {dx}x=\frac {(2xy+3y^2)} {(2xy+x^2)}##
After your substitution v = y/x (or equivalently y = vx), you shouldn't end up with x, y, and v. The whole reason for this substitution is to get to a separable equation in x and v, and then later undo the substitution.
 
  • Like
Likes   Reactions: Arman777
Mark44 said:
After your substitution v = y/x (or equivalently y = vx), you shouldn't end up with x, y, and v. The whole reason for this substitution is to get to a separable equation in x and v, and then later undo the substitution.
I see your point, yes you are right. Our teacher used the first method for examples and in the book it was using different method (second one) so at first it made me confused.

It was stupid for me to change left side with y=vx but not the right side.