Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Idealist
Group Title
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?
 one year ago
 one year ago
Idealist Group Title
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?
 one year ago
 one year ago

This Question is Closed

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Well voila u just showed that its homogenous
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Do you have to solve this
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
(x^2+xy+y^2)/x^2 1 + y/x + (y/x)^2 you might try a vx=y substition
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Then what's the answer for this problem?
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Do I plug that in for y?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
no, you plug that in for y/x = v y' = v'x + v
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Wait a minute.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
v'x + v = v^2 + v + 1 v'x = v^2 + 1 dv/(v^2 + 1) = dx/x
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
How did you get v'x+v?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
What's the next step and the answer?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
i already presented you with the seperable form ... i would appreciate it if you took the time to work on it
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
this is implicit stuff right? as opposed to partials?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
of course, the dy/dx suggests as such
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
So the answer is just dv/(v^2+1)=dx/x?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
theny oull have all the parts needed to determine if this is homogenous (=0)
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
So from x dv=(v^2+1)dx, I got x^2/2=v^3/3+v+c, right?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\int(\frac{1}{v^2+1})~dv=\frac1x ~dx\] \[tan^{1}(v)=ln(x)+C\] \[v=tan(ln(x)+C)\] \[\frac yx=tan(ln(x)+C)\] \[y=x~tan(ln(x)+C)\]
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
So the answer is y=x tan(ln abs(x)+c), right?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
that is the solution for the diffy Q yes, but your question is to show its homogenous
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
How do I do that?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
define y' now that you know y
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
I did product rule and got sec^2(ln abs(x)+c)+tan(ln abs(x)+c), is that the final answer?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
or to clean it up, we already know that tan(ln(x)+c) = y so you could go that route as well
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\tan^2(\ln(x)+c) + \tan(\ln(x)+c) + 1=\frac{x^2+xy+y^2}{x^2}\] \[y^2 + y + 1=\frac{x^2+xy+y^2}{x^2}\] \[x^2y^2 +x^2 y + x^2=x^2+xy+y^2\] \[x^2y^2 +x^2 y=+xy+y^2\] does it work out?
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
But y=x tan(ln abs(x)+c), not tan(ln abs(x)+c).
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
good eye ... so that should fix it better :) replace tan(ln(x)+c) with y/x
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Wait a minute.
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Yes, so the final answer would be y^2+xy+x^2=x^2+xy+y^2, right?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

Idealist Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much for the big help. I appreciate it.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
youre welcome :)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.