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Gerechte23's question

  1. May 17, 2009 #1

    HallsofIvy

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    gerechte23 added this to a thread on a different problem. I am starting a new thread for it

    Why do you think you ought to be able to resolve it? That looks to me like a very nasty non-linear equation.

    Hold on, I was assuming that the "x" in your equation was the independent variable. Did you mean it simply as "times"? (unlikely- by the time you are doing differential equations you should know better!

    IF that was what you meant, then since the independent variable does not appear explicitely, we can use "quadrature". Let v= Y'. Then Y"= V'= dV/dx= (dV/dY)(dY/dx)= V(dV/dY) so the equation becomes the separable first order equation Y^2V dV/dY= C which can be written V dV= CdY/Y^2 and integrates to [itex](1/2)V^2= -C/Y+ D[/itex] where D is the constant of integration. Now, since V= Y', we can write that as the first order equation (1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y, again a separable equation.

    [tex]\frac{YdY}{DY- C}= 2dx[/tex]
    To integrate the left side, let u= DY- C so that du= Ddy and we have
    [tex]\frac{1}{D}\frac{du}{u}= 2dx[/tex]

    That is easy to integrate.

    Of course, that only works if you meant Y2Y"= C. If your really meant xY2Y"= C with x the independent variable, I'm afraid it cannot be integrated in terms of elementary functions.
     
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  3. May 18, 2009 #2
    friend, i was working on the result and then i realized this :(1/2)(Y')^2= -C/Y+ D= (DY- C)/Y or Y'= 2(DY- C)/Y . The problem in this is Y'=[2(DY- C)/Y]^1/2. So, a little error came in. But i've tried solving it, but no. I am not good at all in differential equations. Please give me a hand again in this.
    Gerechte23
     
  4. May 24, 2009 #3
    [tex]\int\sqrt{Y}[/tex]/[tex]\int\sqrt{DY-C}[/tex] = [tex]\int\sqrt{2}dt[/tex] . When i resolve this equation using variable change i found the first part like: -[tex]\sqrt{[(DY-C)/D]}[/tex] which seems not correct when i derivate it, it doesn't give me the result i was expecting. But when i use trigonometry variable changes, it give me something with logarithm and square root. I don't know how to calculate Y in such a situation. I will post the result tomorrow.

    But please reply!
     
    Last edited: May 24, 2009
  5. May 25, 2009 #4
    Excuse me, but can you help me in this. I just can't locate where to click so that i start a new thread in here. Please help
     
  6. May 27, 2009 #5

    malawi_glenn

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    go to forums, the sub-forums then make new topic by clicking on that particular button. This is standard procedure for all forums..............
     
  7. May 29, 2009 #6
    [tex]\sqrt{\frac{Y}{DY- C}}dY = \sqrt{2}dx [/tex]

    Integrate

    [tex]\sqrt{Dy}\sqrt{Dy-C}+C\log(\sqrt{Dy}+\sqrt{Dy-C}) = t\sqrt{2D^3}+ E[/tex] :yuck:

    http://integrals.wolfram.com/index.jsp?expr=Sqrt[x/(d*x-c)]&random=false"
     
    Last edited by a moderator: Apr 24, 2017
  8. May 30, 2009 #7
    Thanks for all guys. Now my only problem is to calculate y as a function of t [Y(t)]. It seems quite complicated for me though.
    Thanks again
     
  9. May 31, 2009 #8

    HallsofIvy

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    Yes, it is! Are you required to do that?
     
  10. Jun 2, 2009 #9
    hi again, yes. I want to calculate it, but it's not a homeworh though. I've proposed myself to find it. Cause with it, i'd be able to calculate a certain acceleration depending on time. Don't you know a way that could help me solve it. If yes, please give a hand
     
  11. Mar 12, 2010 #10
    another question about that log in the result, can it be ln? (Neperian logarithm?)
     
  12. Mar 12, 2010 #11
    hi guys, maybe the way i've started resolving the equation is too difficult. Can someone help me with this please: Y''*Y^2=C, c is a real number. Please, help me with tis, i am killing myself to find that solution but failing all the time. Thanks guys for helping again* with it.
     
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