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 one year ago
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
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?

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

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

amistre64
 one year ago
Best 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

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

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

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

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1its 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

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

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

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

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

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

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

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

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

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

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

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

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

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

amistre64
 one year ago
Best 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?

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

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

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1looks 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

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