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melmel

  • 2 years ago

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  1. melmel
    • 2 years ago
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    bernoulli's Equation

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

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

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

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

  6. melmel
    • 2 years ago
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    its right?

  7. UnkleRhaukus
    • 2 years ago
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    yeah, now move the other terms to the other side

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

  9. melmel
    • 2 years ago
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    Im correct?

  10. UnkleRhaukus
    • 2 years ago
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    good

  11. melmel
    • 2 years ago
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    then?

  12. melmel
    • 2 years ago
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    let v

  13. melmel
    • 2 years ago
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    how?

  14. UnkleRhaukus
    • 2 years ago
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    \[(y-x)/x = y/x - x/x\]

  15. melmel
    • 2 years ago
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    yes

  16. melmel
    • 2 years ago
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    after that

  17. UnkleRhaukus
    • 2 years ago
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    x/x simplifies to ...

  18. melmel
    • 2 years ago
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    1

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

  20. UnkleRhaukus
    • 2 years ago
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    use the product rule to find y'

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

  22. UnkleRhaukus
    • 2 years ago
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    should be plus not minus

  23. melmel
    • 2 years ago
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    ay sorry about that

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

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

  26. UnkleRhaukus
    • 2 years ago
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    which is now variables separable

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

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

  29. oldrin.bataku
    • 2 years ago
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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. \(dy=\left(x\dfrac{dv}{dx}+v\right)\,dx\):$$(x-vx)dx+x(x\,dv+v\,dx)=0\\(x-vx)\,dx+x^2dv+vx\,dx=0\\x\,dx+x^2\,dv=0\\x^2\,dv=-x\,dx\\dv=-\frac1x\,dx\\v=-\log x+C\\y/x=-\log x+C\\y=-x\log x+Cx$$

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

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

  32. melmel
    • 2 years ago
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    ?

  33. melmel
    • 2 years ago
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    then identify if its exact DE or not

  34. melmel
    • 2 years ago
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    ?????

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

  36. UnkleRhaukus
    • 2 years ago
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    What is the second thing? where does the dx go?

  37. melmel
    • 2 years ago
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    v=- lnx +c is that right?

  38. UnkleRhaukus
    • 2 years ago
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    yes, now remember v=y/x

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

  40. melmel
    • 2 years ago
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    y = -xlnx+c ?

  41. UnkleRhaukus
    • 2 years ago
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    almost,

  42. melmel
    • 2 years ago
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    how ?

  43. UnkleRhaukus
    • 2 years ago
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    the last term,

  44. melmel
    • 2 years ago
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    cx

  45. oldrin.bataku
    • 2 years ago
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    yes

  46. melmel
    • 2 years ago
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    sorry about that

  47. UnkleRhaukus
    • 2 years ago
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    thats better \(\checkmark\)

  48. melmel
    • 2 years ago
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    tnx i other one here tech me ?

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

  50. UnkleRhaukus
    • 2 years ago
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    is there an equals sign somewhere?

  51. melmel
    • 2 years ago
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    =0

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

  53. melmel
    • 2 years ago
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    2x

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

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

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

  57. UnkleRhaukus
    • 2 years ago
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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

  58. melmel
    • 2 years ago
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    ah i see

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

  60. UnkleRhaukus
    • 2 years ago
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    sometimes a 'homogenous equations" means something else

  61. melmel
    • 2 years ago
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    you gve a sense to me :)

  62. UnkleRhaukus
    • 2 years ago
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    i haven't worked out the final answer yet

  63. melmel
    • 2 years ago
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    by the way thank you :))) tnz alot

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