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nikita2

  • 4 years ago

Could you help me with this equation y'' = 1/y^2

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  1. JamesJ
    • 4 years ago
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    This is a tricky equation. I'll come back in a minute and help you with it.

  2. anonymous
    • 4 years ago
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    are u asking for y?

  3. Shayaan_Mustafa
    • 4 years ago
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    it is a differential equation.

  4. Shayaan_Mustafa
    • 4 years ago
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    ask me what are you trying to ask?

  5. nikita2
    • 4 years ago
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    yes it is differential eq.

  6. Shayaan_Mustafa
    • 4 years ago
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    so what are asking for? finding general solution or anything else?

  7. nikita2
    • 4 years ago
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    general

  8. Shayaan_Mustafa
    • 4 years ago
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    ok you need a general solution. There are two solution. first is complementary solution and other is particular solution. do you know this?

  9. Shayaan_Mustafa
    • 4 years ago
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    you just get relax nikita2. I am here for help. take time and post a correct and complete question.

  10. nikita2
    • 4 years ago
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    ok, I will )

  11. Shayaan_Mustafa
    • 4 years ago
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    ok

  12. JamesJ
    • 4 years ago
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    @shayaan, do you know how to find the particular solution? This is quite non-standard.

  13. Shayaan_Mustafa
    • 4 years ago
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    yes i know. there are several methods. but good one is variation of parameters .

  14. JamesJ
    • 4 years ago
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    No, that's not correct. This is non-linear equation.

  15. nikita2
    • 4 years ago
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    \[ (D _{2}^{t}u)^{2} (D _{2}^{x}D _{2}^{t}u) = 1\] And i should to solve Cauchy problem

  16. Shayaan_Mustafa
    • 4 years ago
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    is it non-linear?

  17. nikita2
    • 4 years ago
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    yes(

  18. Shayaan_Mustafa
    • 4 years ago
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    hmmm... then fine.

  19. JamesJ
    • 4 years ago
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    y = y(t), y'' - 1/y^2 = 0. Not linear, unfortunately.

  20. Shayaan_Mustafa
    • 4 years ago
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    do you know how to start with cauchy euler method?? you get start then we are here to volunteerily

  21. nikita2
    • 4 years ago
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    yes i'll try it now

  22. JamesJ
    • 4 years ago
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    ok ... multiply first both sides by y' dt \[ y'' y' dt = \frac{y'}{y^2} dt \] Writing \( d(y') = y' dt \) and \( dy = y' dt \) we have \[ (y')' d(y') = \frac{1}{y^2} dy \] Now integrate both sides and we have ....

  23. Shayaan_Mustafa
    • 4 years ago
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    @JamesJ give him reason why we did so? So that he could get clear concepts. Thanks.

  24. nikita2
    • 4 years ago
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    Now integrate both sides and we have .... \[(y')^2/2 = -1/3*1/y^3\]

  25. nikita2
    • 4 years ago
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    + Const

  26. JamesJ
    • 4 years ago
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    \[ \frac{1}{2} (y')^2 = -\frac{1}{y} + C \]

  27. nikita2
    • 4 years ago
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    Oh! i've done mistake here!

  28. JamesJ
    • 4 years ago
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    Now.... \[ \frac{dy}{dt} = \pm \sqrt{ \frac{ 2(Cy-1)}{y}} \] This is separable and it isn't too hard to find an equation for t, t = t(y).

  29. nikita2
    • 4 years ago
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    i see...

  30. JamesJ
    • 4 years ago
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    Interesting, eh? I only figured this out recently myself, working it out in this problem: http://openstudy.com/study#/updates/4f0de661e4b084a815fcff56

  31. nikita2
    • 4 years ago
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    Oh! I've not notise that it is forse of atraktion!

  32. nikita2
    • 4 years ago
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    But anyway if i'll try to do next. Solve \[D _{2}^{t}u(t,x) = y(t,x)\] where y we found

  33. JamesJ
    • 4 years ago
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    Tricky, but probably not impossible analytically. What's the original motivation for this problem?

  34. nikita2
    • 4 years ago
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    I solving the Cauchy problem \[(D _{2}^{t}u(t,x))^2*(D _{2}^{x}D _{2}^{t}u(t,x)) =\] the initial condition is on {t=0}

  35. nikita2
    • 4 years ago
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    =1

  36. JamesJ
    • 4 years ago
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    right. btw, you might find this site helpful: math.stackexchange.com

  37. JamesJ
    • 4 years ago
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    but in the meantime, it would be great if you wanted to help out here answering some elementary questions now and again!

  38. nikita2
    • 4 years ago
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    I did it and I will )

  39. nikita2
    • 4 years ago
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    Thank you very much!

  40. JamesJ
    • 4 years ago
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    Happy to help. This was a fun problem.

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