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Differential equations - 2nd order euler eq'n

  1. Mar 23, 2007 #1
    (a) Let alpha (a) and beta (b) be given constant. show that t^r is a solution of the Euler equation
    t^2 d^2y/dt^2 + at dy/dt + by = 0 , t>0
    if r^2 + (a-1)r + b = 0

    (b) suppose that (a-t)^2 = 4b. Show that (ln t)t^(1-a)/2 is a second solution of Euler's equation.





    please help, i have no idea how to start this.
    for (a) it says, shown that t^r is a solution, so can i just substitute in ?
     
    Last edited: Mar 23, 2007
  2. jcsd
  3. Mar 23, 2007 #2

    siddharth

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    Indeed! If y=f(t) is a solution, then it satisfies the differential equation.
     
    Last edited: Mar 23, 2007
  4. Mar 23, 2007 #3

    HallsofIvy

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    Yes, that's how you show that anything is a solution to any equation!

    However, "suppose that (a-t)^2= 4b doesn't make sense- a and b are constants while t is a variable. Surely you mean "suppose that (a-1)^2= 4b", in which case the "characteristic equation" r2+ (a-1)r+ b= 0 has (1-a)/2 as a double root. Again, just let y= (ln t)t(1-a)/2 and show that that satisfies the equation in the special case (1-a)2= 4b.
     
    Last edited: Mar 24, 2007
  5. Mar 23, 2007 #4
    Am I doing something wrong.
    I substituted y(t) = t^r in , and end up getting r^2 + (a-1)r + b = 0

    here is my work:

    y(t)=t^r
    y'(t) = rt^(r-1)
    y'(t) = r(r-1)t^(r-2)

    t^2(r)(r-1)t^(r-2) + atrt^(r-1) + bt^r = 0
    r(r-1)t^r + art^r + bt^r = 0
    r(r-1) + ar + b = 0
    r^2 - r + ar + b = 0
    r^2 + (a-1)r + b = 0

    i have no idea what to do now
     
  6. Mar 24, 2007 #5

    HallsofIvy

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    It might help to go back and read the question again! It said: ". show that t^r is a solution of the Euler equation
    t^2 d^2y/dt^2 + at dy/dt + by = 0 , t>0
    if r^2 + (a-1)r + b = 0"
    Isn't that exactly what you just did? Replacing y by rt satisfies the equation (makes it equal to 0) if and only if your last equation r2+ (a-1)r+ b= 0.

    Now do the same for tr2, remembering that you now have the additional requirement that "(a-1)2= 4b" (NOT (a-t)2!).
     
  7. Mar 24, 2007 #6
    ohh, wow i totally forgot about the if r^2+(a-1)r + b=0 part, thanks.
    but for (b), why "do the same for tr^2", why am i not subbing in the (ln t)t^1-a/2
     
  8. Mar 25, 2007 #7
    can anyone help with this?
    i have tried subbing in (ln t)t^1-a/2 but i do not get 0, so i know i am doing somethi gnwrong and it's not my work that's wrong as i've done it twice with the same answer. it keeps coming out to a^2 - 3 = 0
     
  9. Mar 25, 2007 #8
    yo

    braindead, do u by chance to queen's
     
  10. Mar 26, 2007 #9

    HallsofIvy

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    Please, read the question again! You did say that a second part of the problem was to show that (ln t)t^(1-a/2) was a solution if (a-1)2= 4b (at least I think that is your equation. I asked if you had a typo and you never answered my question).

    Good. You now recognize that you must substitute x= (ln t)t^(1-a/2) (those parentheses are necessary: t^1- a/2 is t- a/2.) using the condition that (a-1)2= 4b. I don't see how we can tell what you did wrong if you don't show exactly what you did. What are x' and x"? What equation did you get?
     
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