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anonymous

  • one year ago

ODE y^2=cx+1/8 c^3 ------ y=2xy'+y^2(y')^3

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  1. UsukiDoll
    • one year ago
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    ??? \[y^2=cx+\frac{1}{8}c^3..............y=2xy'+y^2(y')^3\] or \[y^2=cx+\frac{1}{8c^3}..............y=2xy'+y^2(y')^3\]

  2. UsukiDoll
    • one year ago
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    I don't know what's going on.. can you go into more detail?

  3. anonymous
    • one year ago
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    \[y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3\]

  4. anonymous
    • one year ago
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    y^2=cx+1/8 c^3 y=2xy'+y^2(y')^3 if the given left is a solution od the ODE right :))

  5. UsukiDoll
    • one year ago
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    so we are differentiating? one part of the equation with the fraction is still confusing.

  6. UsukiDoll
    • one year ago
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    so in the end we should have something like \[y=2xy'+y^2(y')^3 \]?

  7. anonymous
    • one year ago
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    yes

  8. UsukiDoll
    • one year ago
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    could you rewrite the first equation.. it's the fraction portion that is making me lost. I don't know if \[c^3y \] is in the denominator or the numerator

  9. anonymous
    • one year ago
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    its only c^3

  10. anonymous
    • one year ago
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    1/8 c^3

  11. UsukiDoll
    • one year ago
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    -_-! \[y^2=cx+\frac{1}{8}c^3. \]

  12. UsukiDoll
    • one year ago
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    omg I apologize the latex is crazy |dw:1436693748089:dw|

  13. UsukiDoll
    • one year ago
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    so we let the x and y be the variables and we treat c as a constant.

  14. anonymous
    • one year ago
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    yeah

  15. anonymous
    • one year ago
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    sorry

  16. anonymous
    • one year ago
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    @ganeshie8

  17. UsukiDoll
    • one year ago
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    well on the farthest term the one with the 1/8c^3 if c is a constant and if all numbers are constants.. we will just have 0

  18. anonymous
    • one year ago
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    i dont know how to solve it :(

  19. UsukiDoll
    • one year ago
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    do you know the derivative of y^2 ?

  20. anonymous
    • one year ago
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    2yy'

  21. UsukiDoll
    • one year ago
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    hmm... I would just leave it as 2y unless we are using implicit differentiation as well?

  22. UsukiDoll
    • one year ago
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    what about the derivative of cx?

  23. anonymous
    • one year ago
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    c(1) >

  24. UsukiDoll
    • one year ago
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    let's see... product rule is f(x)g'(x)+f'(x)g(x) so we let f(x) = c and g(x) = x what is the derivative of f'(x) and what is the derivative of g'(x) ?

  25. anonymous
    • one year ago
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    c (1) + cx ?

  26. UsukiDoll
    • one year ago
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    well .. the derivative of g(x) is 1 yes.. c(1) + that part is correct so for f'(x)g(x) f(x) = c and g(x) = x so we leave g(x) alone and take the derivative of f(x) = c. Since c is a constant, the derivative is 0 so we have c(1) +x(0) c+0 c

  27. anonymous
    • one year ago
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    \[2yy'= c+1/8 c^3\]

  28. UsukiDoll
    • one year ago
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    why is 1/8c^3 still there? a number like 1/8 is just a constant and c^3 is a constant too? I'm lost as the hills XD

  29. anonymous
    • one year ago
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    sorry

  30. anonymous
    • one year ago
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    XD

  31. anonymous
    • one year ago
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    \[3/8 c^2?\]

  32. UsukiDoll
    • one year ago
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    uh ._. I have no idea what's going on now *mind blown*

  33. anonymous
    • one year ago
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    1/8c^3 is not a constant

  34. UsukiDoll
    • one year ago
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    1/8 is a number... numbers are constants... derivatives of any number go to 0.

  35. anonymous
    • one year ago
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    what is the next step

  36. UsukiDoll
    • one year ago
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    hmmm is c also treated as a variable? I can't even...

  37. UsukiDoll
    • one year ago
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    usually x and y are variables and anything but those letters are treated as a constant...

  38. anonymous
    • one year ago
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    3/8c^2 ?

  39. anonymous
    • one year ago
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    uh ._. I have no idea what's going on now *mind blown* HAHA

  40. UsukiDoll
    • one year ago
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    in the end we are supposed to have x's and y's somehow.. so that c has to be treated as a constant.

  41. anonymous
    • one year ago
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    pass muna nahihirapan na ako e ahahaha

  42. anonymous
    • one year ago
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    i dont understand

  43. UsukiDoll
    • one year ago
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    like... maybe factor one c out to make it like \[y^2 = c(x+\frac{1}{8}c^2) \]

  44. UsukiDoll
    • one year ago
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    then differentiate...

  45. UsukiDoll
    • one year ago
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    like we know that the derivative of \[y^2 \rightarrow 2y y'\] and then product rule for c()

  46. sohailiftikhar
    • one year ago
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    what is mean of ODE ??

  47. UsukiDoll
    • one year ago
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    ODE means an ordinary differential equation...

  48. UsukiDoll
    • one year ago
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    \[y^2 dy = c(x+\frac{1}{8}c^2) dx\] if it's a separable equation, then we have to integrate both sides.

  49. anonymous
    • one year ago
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    what is the next step

  50. UsukiDoll
    • one year ago
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    integrate both sides.

  51. sohailiftikhar
    • one year ago
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    oh!! so we are given \[y=2xy' +y^{2}(y')^{3} and y ^{2}=cx+1\div8c ^{3} is \it ???\]

  52. UsukiDoll
    • one year ago
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    @sohailiftikhar supposedly we are differentiating \[y^2=cx+\frac{1}{8}c^3.\] and by doing that we get to \[y=2xy'+y^2(y')^3 \]

  53. sohailiftikhar
    • one year ago
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    just put value of \[y ^{2}\] in equation of y then take derivative of it ...

  54. sohailiftikhar
    • one year ago
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    is that ??

  55. UsukiDoll
    • one year ago
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    this isn't even my question, so .. yeah

  56. sohailiftikhar
    • one year ago
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    how it can be ...?

  57. UsukiDoll
    • one year ago
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    and with vague instructions, I can't go further... it's like hello I'm not a mind reader like what does the question want really?

  58. sohailiftikhar
    • one year ago
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    I think we are give two equations and we have to find dy/dx

  59. UsukiDoll
    • one year ago
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    this is similar to OP's previous question though.

  60. sohailiftikhar
    • one year ago
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    I can't get it what is that .....sorry .

  61. UsukiDoll
    • one year ago
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    I don't get it either.. let's get out of here.

  62. sohailiftikhar
    • one year ago
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    yes..

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