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anonymous

  • one year ago

find D.E x^2/a^2+y^2/b^2=1 ans. xyy''+xy'^2=yy'

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  1. anonymous
    • one year ago
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    diff. twice and eliminate a^2 and b^2 from three equations.

  2. anonymous
    • one year ago
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    how @surjithayer

  3. anonymous
    • one year ago
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    help please

  4. ganeshie8
    • one year ago
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    Have you used the surjithayer's hint and differentiated twice ?

  5. anonymous
    • one year ago
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    yeah

  6. ganeshie8
    • one year ago
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    show you work please

  7. anonymous
    • one year ago
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    wait

  8. anonymous
    • one year ago
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  9. anonymous
    • one year ago
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  10. anonymous
    • one year ago
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    number 14

  11. ganeshie8
    • one year ago
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    whats the derivative of \(\dfrac{x^2}{a^2}\) with respect to \(x\) ?

  12. anonymous
    • one year ago
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    but i think , may solution is not right

  13. anonymous
    • one year ago
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    yes

  14. ganeshie8
    • one year ago
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    Keep in mind, \(a\) and \(b\) are constants here

  15. ganeshie8
    • one year ago
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    \[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\] differentiating implicitly with respect to \(x\) gives \[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0\tag{2}\] yes ?

  16. anonymous
    • one year ago
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    okay :)

  17. anonymous
    • one year ago
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    but , i did it first , but my answer was not right

  18. ganeshie8
    • one year ago
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    your work in that picture is wrong you're not treating a^2 as constant, instead trying to differentiate it and writing 2a.. which is wrong

  19. anonymous
    • one year ago
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    can you help please

  20. ganeshie8
    • one year ago
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    \[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\] differentiating implicitly with respect to \(x\) gives \[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'} \] Plug this in first equation and get \[\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1 \] \[\implies -xyy' +y^2 = b^2 \tag{2}\] differentiate this again and you're done!

  21. anonymous
    • one year ago
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    @ganeshie8 gets :)) thank you very much :))

  22. ganeshie8
    • one year ago
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    yw, are you able to differentiate it again and arrive at the answer ?

  23. anonymous
    • one year ago
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    @ganeshie8 im very confused

  24. ganeshie8
    • one year ago
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    remember product rule ?

  25. anonymous
    • one year ago
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    yeah i remember

  26. anonymous
    • one year ago
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    how about the b^2 ?

  27. ganeshie8
    • one year ago
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    b^2 is just a constant, like, 3^2 or 5^2

  28. ganeshie8
    • one year ago
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    it doesn't change, so the derivative is 0

  29. anonymous
    • one year ago
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    \[-x y y'' . -1 + x y^2.y'=b^2\]

  30. ganeshie8
    • one year ago
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    \[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\tag{1}\] differentiating implicitly with respect to \(x\) gives \[\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0 \implies a^2 = -\dfrac{b^2x}{yy'} \] Plug this in first equation and get \[\dfrac{x^2}{-b^2x/(yy')}+\dfrac{y^2}{b^2}=1 \] \[\implies -xyy' +y^2 = b^2 \tag{2}\] differentiating w.r.t \(x\) again \[-(yy' +y'xy' + y''xy) + 2yy'= 0\] \[yy' = xyy'' + x{y'}^2\]

  31. ganeshie8
    • one year ago
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    Recall below product rules \[\left(f(x)*g(x)\right)' = f'(x)g(x) + f(x)g'(x)\] \[\left(f(x)*g(x)*h(x)\right)' = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)\]

  32. anonymous
    • one year ago
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    @ganeshie8 i got the answer :))))

  33. anonymous
    • one year ago
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    how about x^2+axy+bx+c=0

  34. ganeshie8
    • one year ago
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    Give it a try first

  35. anonymous
    • one year ago
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    i have a answer

  36. anonymous
    • one year ago
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    @ganeshie8 number 11

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  37. ganeshie8
    • one year ago
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    Looks you forgot to use product rule here ? |dw:1436688898948:dw|

  38. UsukiDoll
    • one year ago
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    isn't that we have to find the second and third derivatives and then plug them back into the equation to check if the equation we are given (that ans. thing) holds true?

  39. anonymous
    • one year ago
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    @ganeshie8

  40. UsukiDoll
    • one year ago
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    x and y's are variables treat a b c as constants which when the derivative is taken it goes to 0

  41. anonymous
    • one year ago
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    ai derivative of x is 1 = a (1) ?

  42. ganeshie8
    • one year ago
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    @UsukiDoll this is like constructing the differential equation from the given general solution by eliminating the arbitrary constants

  43. UsukiDoll
    • one year ago
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    oh I see so we're not verifying... we're finding a DE

  44. ganeshie8
    • one year ago
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    yeah we can think of the equations of given ellipses/parabolas as general solutions

  45. anonymous
    • one year ago
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    yeah , were finding D.E

  46. ganeshie8
    • one year ago
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    @jacalneaila did you find ur mistake in #11 ?

  47. anonymous
    • one year ago
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    yeah, i try to solve ut again , but may answer is different now :(

  48. anonymous
    • one year ago
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    *it

  49. ganeshie8
    • one year ago
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    show ur work

  50. anonymous
    • one year ago
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    wait , i dont have a cellphone

  51. ganeshie8
    • one year ago
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    #11 should be fairly simple compared to the earlier one

  52. anonymous
    • one year ago
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    \[ax+yy'+a(1)? \]

  53. anonymous
    • one year ago
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    @ganeshie8

  54. ganeshie8
    • one year ago
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    |dw:1436690038639:dw|

  55. anonymous
    • one year ago
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    ugh

  56. anonymous
    • one year ago
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    im still doing it , and still different

  57. anonymous
    • one year ago
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    @ganeshie8 Thank you very much , its very helpful :)) .

  58. anonymous
    • one year ago
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    @ganeshie8 Thank you again ^_^

  59. UsukiDoll
    • one year ago
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    so for this part axy to make it less confusing I would put a () like this a(xy) and then take the product rule of xy a(xy'+y) then I have to take the product rule for xy' a(xy''+y'+y') and then another product rule for xy'' a(xy'''+y''+y''+y'') so that leaves (xy'''+3y'') I think I'm missing a step.. that a is supposed to go away x.x

  60. ganeshie8
    • one year ago
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    the missing part is \(=0\) in every line :)

  61. UsukiDoll
    • one year ago
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    oh boy xD! I was doing this while dealing with a migraine.

  62. ganeshie8
    • one year ago
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    From personal experience, regular diet+good sleep are the best medicine for migraine

  63. UsukiDoll
    • one year ago
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    yeah...that's what I lot of people say...but I'm such a nighthawk and it runs in the family.

  64. ganeshie8
    • one year ago
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    Also there are some ellergetic foods which you need to avoid eating

  65. UsukiDoll
    • one year ago
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    if it's chocolate, then I can't =(

  66. ganeshie8
    • one year ago
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    For me they are potato and some other vegetable whose name idk in english

  67. ganeshie8
    • one year ago
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    they depend on the person i think..

  68. ganeshie8
    • one year ago
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    this one http://www.mommyscuisine.com/wp-content/uploads/chikkudukayatomato1-300x225.jpg

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