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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

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  1. swissgirl Group Title
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    Well voila u just showed that its homogenous

    • one year ago
  2. swissgirl Group Title
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    Do you have to solve this

    • one year ago
  3. amistre64 Group Title
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    (x^2+xy+y^2)/x^2 1 + y/x + (y/x)^2 you might try a vx=y substition

    • one year ago
  4. Idealist Group Title
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    Then what's the answer for this problem?

    • one year ago
  5. Idealist Group Title
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    Do I plug that in for y?

    • one year ago
  6. amistre64 Group Title
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    no, you plug that in for y/x = v y' = v'x + v

    • one year ago
  7. Idealist Group Title
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    Wait a minute.

    • one year ago
  8. amistre64 Group Title
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    v'x + v = v^2 + v + 1 v'x = v^2 + 1 dv/(v^2 + 1) = dx/x

    • one year ago
  9. Idealist Group Title
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    How did you get v'x+v?

    • one year ago
  10. amistre64 Group Title
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    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
  11. Idealist Group Title
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    What's the next step and the answer?

    • one year ago
  12. amistre64 Group Title
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    i already presented you with the seperable form ... i would appreciate it if you took the time to work on it

    • one year ago
  13. amistre64 Group Title
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    this is implicit stuff right? as opposed to partials?

    • one year ago
  14. amistre64 Group Title
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    of course, the dy/dx suggests as such

    • one year ago
  15. Idealist Group Title
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    So the answer is just dv/(v^2+1)=dx/x?

    • one year ago
  16. amistre64 Group Title
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    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
  17. amistre64 Group Title
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    theny oull have all the parts needed to determine if this is homogenous (=0)

    • one year ago
  18. Idealist Group Title
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    So from x dv=(v^2+1)dx, I got x^2/2=v^3/3+v+c, right?

    • one year ago
  19. amistre64 Group Title
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    \[\int(\frac{1}{v^2+1})~dv=\frac1x ~dx\] \[tan^{-1}(v)=ln(x)+C\] \[v=tan(ln(x)+C)\] \[\frac yx=tan(ln(x)+C)\] \[y=x~tan(ln(x)+C)\]

    • one year ago
  20. Idealist Group Title
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    So the answer is y=x tan(ln abs(x)+c), right?

    • one year ago
  21. amistre64 Group Title
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    that is the solution for the diffy Q yes, but your question is to show its homogenous

    • one year ago
  22. Idealist Group Title
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    How do I do that?

    • one year ago
  23. amistre64 Group Title
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    define y' now that you know y

    • one year ago
  24. Idealist Group Title
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    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
  25. amistre64 Group Title
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    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
  26. amistre64 Group Title
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    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
  27. amistre64 Group Title
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    \[\tan^2(\ln(x)+c) + \tan(\ln(x)+c) + 1=\frac{x^2+xy+y^2}{x^2}\] \[y^2 + y + 1=\frac{x^2+xy+y^2}{x^2}\] \[x^2y^2 +x^2 y + x^2=x^2+xy+y^2\] \[x^2y^2 +x^2 y=+xy+y^2\] does it work out?

    • one year ago
  28. Idealist Group Title
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    But y=x tan(ln abs(x)+c), not tan(ln abs(x)+c).

    • one year ago
  29. amistre64 Group Title
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    good eye ... so that should fix it better :) replace tan(ln(x)+c) with y/x

    • one year ago
  30. Idealist Group Title
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    Wait a minute.

    • one year ago
  31. Idealist Group Title
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    Yes, so the final answer would be y^2+xy+x^2=x^2+xy+y^2, right?

    • one year ago
  32. amistre64 Group Title
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    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
  33. Idealist Group Title
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    Thank you so much for the big help. I appreciate it.

    • one year ago
  34. amistre64 Group Title
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    youre welcome :)

    • one year ago
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