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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?

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  1. swissgirl
    • one year ago
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    Well voila u just showed that its homogenous

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

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

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

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

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

  7. Idealist
    • one year ago
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    Wait a minute.

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

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

  10. amistre64
    • one year ago
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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

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

  12. amistre64
    • one year ago
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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

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

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

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

  16. amistre64
    • one year ago
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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

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

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

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

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

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

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

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

  24. Idealist
    • one year ago
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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?

  25. amistre64
    • one year ago
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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

  26. amistre64
    • one year ago
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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

  27. amistre64
    • one year ago
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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?

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

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

  30. Idealist
    • one year ago
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    Wait a minute.

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

  32. amistre64
    • one year ago
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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

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

  34. amistre64
    • one year ago
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    youre welcome :)

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