anonymous
  • anonymous
Show that dy/dx=(x^2+xy+y^2)/x^2 is homogeneous. I did x^2 dy-(x^2+xy+y^2)dx=0 but what's the next step?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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swissgirl
  • swissgirl
Well voila u just showed that its homogenous
swissgirl
  • swissgirl
Do you have to solve this
amistre64
  • amistre64
(x^2+xy+y^2)/x^2 1 + y/x + (y/x)^2 you might try a vx=y substition

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anonymous
  • anonymous
Then what's the answer for this problem?
anonymous
  • anonymous
Do I plug that in for y?
amistre64
  • amistre64
no, you plug that in for y/x = v y' = v'x + v
anonymous
  • anonymous
Wait a minute.
amistre64
  • amistre64
v'x + v = v^2 + v + 1 v'x = v^2 + 1 dv/(v^2 + 1) = dx/x
anonymous
  • anonymous
How did you get v'x+v?
amistre64
  • amistre64
its a substition: v = y/x ,,, therefore vx = y take the derivative of both sides (think of the product rule) v'x + v = y' sub in all the parts as needed
anonymous
  • anonymous
What's the next step and the answer?
amistre64
  • amistre64
i already presented you with the seperable form ... i would appreciate it if you took the time to work on it
amistre64
  • amistre64
this is implicit stuff right? as opposed to partials?
amistre64
  • amistre64
of course, the dy/dx suggests as such
anonymous
  • anonymous
So the answer is just dv/(v^2+1)=dx/x?
amistre64
  • amistre64
of course not, thats the seperable form that is very doable to determine what v equates to, then known what v equates to, equate that back to y/x
amistre64
  • amistre64
theny oull have all the parts needed to determine if this is homogenous (=0)
anonymous
  • anonymous
So from x dv=(v^2+1)dx, I got x^2/2=v^3/3+v+c, right?
amistre64
  • amistre64
\[\int(\frac{1}{v^2+1})~dv=\frac1x ~dx\] \[tan^{-1}(v)=ln(x)+C\] \[v=tan(ln(x)+C)\] \[\frac yx=tan(ln(x)+C)\] \[y=x~tan(ln(x)+C)\]
anonymous
  • anonymous
So the answer is y=x tan(ln abs(x)+c), right?
amistre64
  • amistre64
that is the solution for the diffy Q yes, but your question is to show its homogenous
anonymous
  • anonymous
How do I do that?
amistre64
  • amistre64
define y' now that you know y
anonymous
  • anonymous
I did product rule and got sec^2(ln abs(x)+c)+tan(ln abs(x)+c), is that the final answer?
amistre64
  • amistre64
you are building the parts you need to determine the proper answer :) lets rewrite that as: \[\tan^2(\ln(x)+c) + \tan(\ln(x)+c) + 1=\frac{x^2+xy+y^2}{x^2}\]plug in y and simplify to your hearts content
amistre64
  • amistre64
or to clean it up, we already know that tan(ln(x)+c) = y so you could go that route as well
amistre64
  • amistre64
\[\tan^2(\ln(x)+c) + \tan(\ln(x)+c) + 1=\frac{x^2+xy+y^2}{x^2}\] \[y^2 + y + 1=\frac{x^2+xy+y^2}{x^2}\] \[x^2y^2 +x^2 y + x^2=x^2+xy+y^2\] \[x^2y^2 +x^2 y=+xy+y^2\] does it work out?
anonymous
  • anonymous
But y=x tan(ln abs(x)+c), not tan(ln abs(x)+c).
amistre64
  • amistre64
good eye ... so that should fix it better :) replace tan(ln(x)+c) with y/x
anonymous
  • anonymous
Wait a minute.
anonymous
  • anonymous
Yes, so the final answer would be y^2+xy+x^2=x^2+xy+y^2, right?
amistre64
  • amistre64
looks that way yes, if a adjust for my mistake :) \[x^2\frac{y^2}{x^2} +x^2 \frac{y}{x}+x^2=x^2+xy+y^2\] \[y^2 +xy+x^2=x^2+xy+y^2\]which zeros out
anonymous
  • anonymous
Thank you so much for the big help. I appreciate it.
amistre64
  • amistre64
youre welcome :)

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