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Y'' + ty' + ty

  1. May 1, 2015 #1
    I know [itex]y=e^{t-t^2/2} [/itex] is a solution but how do I reach it? I tried substituting y=uv, making u' terms disappear then I noticed that the resulting u''=(__)u could be solved by a gaussian but this seems like luck to me
     
  2. jcsd
  3. May 1, 2015 #2
    Have you tried substituting y=ekx?
     
  4. May 1, 2015 #3
    [itex]e^{kx}[/itex] cannot be a solution but I decided to try [itex]e^{k(x)}[/itex] which gave me a ricatti equation and after I substituted [itex]k=q\frac{p'}{p}[/itex] I got the original equation

    edit: this is the solution
    [itex]c_1e^{\frac{-x^2}{4}+x}+c_2e^{\frac{-x^2}{4}+x} \int ^x e^{\frac{-s^2}{2}+2s}ds[/itex]
     
    Last edited: May 1, 2015
  5. May 1, 2015 #4

    Mark44

    Staff: Mentor

    A little nitpicky, but y'' + ty' + ty is not an equation (differential or otherwise), so it doesn't make any sense to talk about a solution. Is the full equation y'' + ty' + ty = 0?
     
  6. May 1, 2015 #5
    yes
     
  7. May 2, 2015 #6

    Svein

    User Avatar
    Science Advisor

    More nitpicking: Is y supposed to be a function of t or a function of x with t as a parameter?
     
  8. May 2, 2015 #7
    you don't have to answer the question if you don't want to
     
  9. May 2, 2015 #8
    I actually solved this qustion after much work, however I do not like m solution. It would be great if someone could link me somewhere that derives error function laplace transforms

    6hujRto.jpg
     

    Attached Files:

  10. May 2, 2015 #9

    Mark44

    Staff: Mentor

    Of course that's true, but if you're asking for help with a problem, it's best to provide as much information about the problem as you can.
     
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