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Solving a DE via Substitution

  1. Sep 26, 2011 #1
    I need to find a 1 parameter family of solutions to:

    [tex]\frac{dy}{dt}=-\frac{1}{t^{2}} - \frac{y}{t}+y^{2}[/tex]

    By making the substitution:
    [tex]y=\frac{1}{t}+u[/tex]
    and then reducing it to a Bernoulli equation in u.

    I first took the derivative of the substitution.
    [tex]\frac{dy}{dt}=\frac{1}{t^{2}}+\frac{du}{dt}[/tex]

    Then substituted:
    [tex]\frac{1}{t^{2}}+\frac{du}{dt}=-\frac{1}{t^{2}} - \frac{\frac{1}{t}+u}{t}+(\frac{1}{t}+u)^{2}[/tex]

    After some reduction I eventually got to:
    [tex]\frac{1}{t^{2}}+\frac{du}{dt}=\frac{u+u^{2}t}{t}-1[/tex]
    I heard we were supposed to get something separable out of this, but that's not what I have here. What do I do next?
     
  2. jcsd
  3. Sep 26, 2011 #2

    LCKurtz

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    The derivative of 1/t is -1/t2. If you are careful with your algebra it should indeed simplify to a Bernoulli equation.
     
  4. Sep 26, 2011 #3
    I now have:
    [tex]u^{2}+\frac{u}{t}+\frac{1}{t^2}-\frac{du}{dt}=1[/tex]
     
  5. Sep 26, 2011 #4

    LCKurtz

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    Close, but check your algebra. I got

    [tex]u' -\frac 1 t u = u^2[/tex]

    Those 1/t2 terms should cancel out. I'm guessing it has something to do with that extra 1 you have too.
     
  6. Sep 26, 2011 #5
    The 1 comes from reducing
    [tex]\frac{y}{t}[/tex]

    [tex]\frac{\frac{1}{t}+u}{t}[/tex]
     
  7. Sep 26, 2011 #6

    LCKurtz

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    Like I said before, the non-cancelled 1/t2 and the extraneous 1 are probably related. If you are now taking differential equations, you should be able to correctly simplify this expression. You don't get a 1.
     
  8. Sep 26, 2011 #7
    I keep getting it for some reason :(

    Nevermind, I'm stupid. Lemme keep working.

    So now I have the right form. Now what? I'm reading the book and it isn't making sense to me right now.
     
    Last edited: Sep 26, 2011
  9. Sep 26, 2011 #8
    What part are you at now?

    (I solved this along with you)
     
  10. Sep 26, 2011 #9

    LCKurtz

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    If you have the above equation, multiply both sides by u-2 and make the substitution w = u-1.
     
  11. Sep 26, 2011 #10
    So I then end up with
    [tex]\frac{w}{t}-w\frac{du}{dt}=1[/tex]
     
  12. Sep 26, 2011 #11
    I think you need to check that again. Also, using the substitution, what is w'?
     
  13. Sep 26, 2011 #12
  14. Sep 26, 2011 #13
    I entered it backwards, my mistake.
    Doing the substitution:
    [tex]w=u^{-1}[/tex]
    [tex]\frac{dw}{dt}=\frac{1}{u^{2}}\frac{du}{dt}[/tex]
    [tex]u^{2}w'=\frac{du}{dt}[/tex]

    [tex]w'u-\frac{w}{t}=1[/tex]
    I'll start going through those notes now.
     
  15. Sep 26, 2011 #14
    Hmm... still not quite right.

    Starting with
    [tex]u' -\frac 1 t u = u^2[/tex]
    Let [itex]w=u^{-1}[/itex], then [itex]w'=-u^{-2} u'[/itex]. This serves as the motivation for the next step: we want to get a linear differential equation in terms of w, so only w and w' should be present. To do so, let's multiply both sides of the equation by [itex]-u^{-2}[/itex], in doing so we will get w' on the left, which is very desirable.

    [tex]-u^{-2}u' + \frac 1 t u^{-1} = -1[/tex]

    Now we can make some nice substitutions:

    [tex]w' + \frac 1 t w = -1[/tex]

    You should be able to take it from here.
     
  16. Sep 26, 2011 #15
    I really should, but I can't. I'm not having the easiest time with this class.
     
  17. Sep 26, 2011 #16
    What is the trouble you're having from the last step? Have you covered how to solve a first-order linear homogeneous differential equation?
     
  18. Sep 26, 2011 #17
    To clarify some possible confusion, at this point you would simply solve for w. This will give you an equation for w in terms of x, and then you can substitute [itex]u^{-1}=w[/itex].
     
  19. Sep 26, 2011 #18
    We have covered that. I'm trying to figure out variation of parameters, since it isn't seperable. I'll have a chance to talk to my professor tomorrow, and I'll come back if I need more help. Thank you both for all the help you've given me.
     
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