Question

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?

Answer 1

Well voila u just showed that its homogenous

Answer 2

Do you have to solve this

Answer 3

(x^2+xy+y^2)/x^2

1 + y/x + (y/x)^2

you might try a vx=y substition

Answer 4

Then what's the answer for this problem?

Answer 5

Do I plug that in for y?

Answer 6

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

y' = v'x + v

Answer 7

Wait a minute.

Answer 8

v'x + v = v^2 + v + 1

v'x = v^2 + 1

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

Answer 9

How did you get v'x+v?

Answer 10

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

Answer 11

What's the next step and the answer?

Answer 12

i already presented you with the seperable form ... i would appreciate it if you took the time to work on it

Answer 13

this is implicit stuff right? as opposed to partials?

Answer 14

of course, the dy/dx suggests as such

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

\[\int(\frac{1}{v^2+1})~dv=\frac1x ~dx\]
\[tan^{-1}(v)=ln(x)+C\]
\[v=tan(ln(x)+C)\]
\[\frac yx=tan(ln(x)+C)\]
\[y=x~tan(ln(x)+C)\]

Answer 20

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

Answer 21

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

Answer 22

How do I do that?

Answer 23

define y' now that you know y

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

\[\tan^2(\ln(x)+c) + \tan(\ln(x)+c) + 1=\frac{x^2+xy+y^2}{x^2}\]
\[y^2 + y + 1=\frac{x^2+xy+y^2}{x^2}\]
\[x^2y^2 +x^2 y + x^2=x^2+xy+y^2\]
\[x^2y^2 +x^2 y=+xy+y^2\]
does it work out?

Answer 28

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

Answer 29

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

Answer 30

Wait a minute.

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

youre welcome :)