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Well voila u just showed that its homogenous

Do you have to solve this

(x^2+xy+y^2)/x^2
1 + y/x + (y/x)^2
you might try a vx=y substition

Then what's the answer for this problem?

Do I plug that in for y?

no, you plug that in for y/x = v
y' = v'x + v

Wait a minute.

v'x + v = v^2 + v + 1
v'x = v^2 + 1
dv/(v^2 + 1) = dx/x

How did you get v'x+v?

What's the next step and the answer?

this is implicit stuff right? as opposed to partials?

of course, the dy/dx suggests as such

So the answer is just dv/(v^2+1)=dx/x?

theny oull have all the parts needed to determine if this is homogenous (=0)

So from x dv=(v^2+1)dx, I got x^2/2=v^3/3+v+c, right?

So the answer is y=x tan(ln abs(x)+c), right?

that is the solution for the diffy Q yes, but your question is to show its homogenous

How do I do that?

define y' now that you know y

I did product rule and got sec^2(ln abs(x)+c)+tan(ln abs(x)+c), is that the final answer?

or to clean it up, we already know that tan(ln(x)+c) = y so you could go that route as well

But y=x tan(ln abs(x)+c), not tan(ln abs(x)+c).

good eye ... so that should fix it better :) replace tan(ln(x)+c) with y/x

Wait a minute.

Yes, so the final answer would be y^2+xy+x^2=x^2+xy+y^2, right?

Thank you so much for the big help. I appreciate it.

youre welcome :)