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melmel

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  • 8 months ago
  • 8 months ago

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  1. melmel
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    bernoulli's Equation

    • 8 months ago
  2. UnkleRhaukus
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    what do you get if you re arrange to dy/dx=

    • 8 months ago
  3. melmel
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    sorry but i dont know .................it equal to 1 dont know if I am correct >.<

    • 8 months ago
  4. UnkleRhaukus
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    after your get the left hand side to be dy/dx sub v=y/x

    • 8 months ago
  5. melmel
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    \[\frac{ xdy }{ y-xdx }=1\]

    • 8 months ago
  6. melmel
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    its right?

    • 8 months ago
  7. UnkleRhaukus
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    yeah, now move the other terms to the other side

    • 8 months ago
  8. melmel
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    \[\frac{ dy }{ dx }= \frac{ y-x }{ x }\]

    • 8 months ago
  9. melmel
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    Im correct?

    • 8 months ago
  10. UnkleRhaukus
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    good

    • 8 months ago
  11. melmel
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    then?

    • 8 months ago
  12. melmel
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    let v

    • 8 months ago
  13. melmel
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    how?

    • 8 months ago
  14. UnkleRhaukus
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    \[(y-x)/x = y/x - x/x\]

    • 8 months ago
  15. melmel
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    yes

    • 8 months ago
  16. melmel
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    after that

    • 8 months ago
  17. UnkleRhaukus
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    x/x simplifies to ...

    • 8 months ago
  18. melmel
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    1

    • 8 months ago
  19. UnkleRhaukus
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    good so you have \[\frac{dy}{dx}=\frac yx-1\] now let \(y/x=v\) \(y=vx\) \(y'=?\)

    • 8 months ago
  20. UnkleRhaukus
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    use the product rule to find y'

    • 8 months ago
  21. melmel
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    \[y'=v(1) -x\frac{dv }{ dx }\]

    • 8 months ago
  22. UnkleRhaukus
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    should be plus not minus

    • 8 months ago
  23. melmel
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    ay sorry about that

    • 8 months ago
  24. melmel
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    y′=v(1)+xdv/dx

    • 8 months ago
  25. UnkleRhaukus
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    now sub this in \[\frac{dy}{dx}=y/x-1\] becomes \[v+x\frac{dv}{dx}=v-1\]

    • 8 months ago
  26. UnkleRhaukus
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    which is now variables separable

    • 8 months ago
  27. melmel
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    \[\frac{ dv }{ dx }= \frac{ v ^{2} -v}{ x }\]

    • 8 months ago
  28. UnkleRhaukus
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    um you should have taken away the v from both sides, before dividing by x

    • 8 months ago
  29. oldrin.bataku
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    $$(x-y)\,dx+x\,dy=0$$note this is homogeneous; both our functions \(x-y,x\) are homogeneous. as @UnkleRhaukus stated, use a substitution \(y=vx\) i.e. \(dy=\left(x\dfrac{dv}{dx}+v\right)\,dx\):$$(x-vx)dx+x(x\,dv+v\,dx)=0\\(x-vx)\,dx+x^2dv+vx\,dx=0\\x\,dx+x^2\,dv=0\\x^2\,dv=-x\,dx\\dv=-\frac1x\,dx\\v=-\log x+C\\y/x=-\log x+C\\y=-x\log x+Cx$$

    • 8 months ago
  30. oldrin.bataku
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    http://en.wikipedia.org/wiki/Homogeneous_differential_equation#Homogeneous_functions

    • 8 months ago
  31. melmel
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    guys can be this one \[xdv=(v ^{2}-v) dx .....then....... (v ^{2}-v)-xdv=0\]

    • 8 months ago
  32. melmel
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    ?

    • 8 months ago
  33. melmel
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    then identify if its exact DE or not

    • 8 months ago
  34. melmel
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    ?????

    • 8 months ago
  35. UnkleRhaukus
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    \[v+x\frac{dv}{dx}=v-1\\ x\frac{dv}{dx}=-1\\dv=-xdx\\ \\∫dv=-∫dx/x\]

    • 8 months ago
  36. UnkleRhaukus
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    What is the second thing? where does the dx go?

    • 8 months ago
  37. melmel
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    v=- lnx +c is that right?

    • 8 months ago
  38. UnkleRhaukus
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    yes, now remember v=y/x

    • 8 months ago
  39. melmel
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    then sub v= y/x to v =-ln x +c

    • 8 months ago
  40. melmel
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    y = -xlnx+c ?

    • 8 months ago
  41. UnkleRhaukus
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    almost,

    • 8 months ago
  42. melmel
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    how ?

    • 8 months ago
  43. UnkleRhaukus
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    the last term,

    • 8 months ago
  44. melmel
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    cx

    • 8 months ago
  45. oldrin.bataku
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    yes

    • 8 months ago
  46. melmel
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    sorry about that

    • 8 months ago
  47. UnkleRhaukus
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    thats better \(\checkmark\)

    • 8 months ago
  48. melmel
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    tnx i other one here tech me ?

    • 8 months ago
  49. melmel
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    xdx + (y-2x )dy

    • 8 months ago
  50. UnkleRhaukus
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    is there an equals sign somewhere?

    • 8 months ago
  51. melmel
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    =0

    • 8 months ago
  52. melmel
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    \[\frac{ dy }{ dx }=\frac{ -x }{ y- }\]

    • 8 months ago
  53. melmel
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    2x

    • 8 months ago
  54. UnkleRhaukus
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    ok this one is the same technique, rearrange to dy/dx= and substitute v=y/x

    • 8 months ago
  55. UnkleRhaukus
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    divide both sides of the fraction by x, to get it in the right form \[\frac{-x}{y-2x}=\frac{-x/x}{y/x-2x/x }\]

    • 8 months ago
  56. melmel
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    always? v=y/x even to other problem?

    • 8 months ago
  57. UnkleRhaukus
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    When we can express the DE as y'(x) = f (y/x) we can call this a homogenous equation and v = y/x substitution will work if we can't get this form y'(x) = f (y/x) this technique wont work

    • 8 months ago
  58. melmel
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    ah i see

    • 8 months ago
  59. melmel
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    can you gve to me the final answer here this will serve as my basis

    • 8 months ago
  60. UnkleRhaukus
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    sometimes a 'homogenous equations" means something else

    • 8 months ago
  61. melmel
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    you gve a sense to me :)

    • 8 months ago
  62. UnkleRhaukus
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    i haven't worked out the final answer yet

    • 8 months ago
  63. melmel
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    by the way thank you :))) tnz alot

    • 8 months ago
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