## Idealist 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?

1. swissgirl

Well voila u just showed that its homogenous

2. swissgirl

Do you have to solve this

3. amistre64

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

4. Idealist

Then what's the answer for this problem?

5. Idealist

Do I plug that in for y?

6. amistre64

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

7. Idealist

Wait a minute.

8. amistre64

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

9. Idealist

How did you get v'x+v?

10. 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

11. Idealist

What's the next step and the answer?

12. amistre64

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

13. amistre64

this is implicit stuff right? as opposed to partials?

14. amistre64

of course, the dy/dx suggests as such

15. Idealist

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

16. 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

17. amistre64

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

18. Idealist

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

19. amistre64

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

20. Idealist

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

21. amistre64

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

22. Idealist

How do I do that?

23. amistre64

define y' now that you know y

24. Idealist

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

25. 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

26. amistre64

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

27. 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?

28. Idealist

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

29. amistre64

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

30. Idealist

Wait a minute.

31. Idealist

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

32. 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

33. Idealist

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

34. amistre64

youre welcome :)