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Laplace transforms for a nontrivial solution

  1. May 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Transform the given DE to find a nontrivial solution such that x(0)=0.


    2. Relevant equations

    3. The attempt at a solution

    Using L{f(t)}=-[itex]\frac{1}{t}[/itex]F'(s), I got


    I see that it is separable, but I do not know how to go about separating it.

    My best guess would be
    [tex]\frac{1}{X(s)}\,dx=\frac{s+1}{4}\,ds[/tex] but that seems wrong
    Last edited by a moderator: May 17, 2013
  2. jcsd
  3. May 16, 2013 #2


    Staff: Mentor

    I am assuming that the above is correct - I didn't check your work.
    For one thing, it's not separated.

    Starting with 4sX(s)+s(s+1)X'(s)=0, and abbreviating a bit, we get
    s(s + 1)X' = -4sX
    ==> X'/X = -4s/(s(s + 1))

    The equation is now separated.
    What you should probably do now is use partial fractions decomposition to break up the fraction on the right into the sum of two fractions.
  4. May 16, 2013 #3
    Oh, I see. I knew I was doing the separation wrong. Wouldn't the s cancel so I wouldn't need to do partial fractions? Also, for the next step I'm not sure what to integrate with respect to on each side. The notation is throwing me off.
    Should I do:
    ∫XdX=∫-[itex]\frac{s+1}{4}[/itex]ds ?

    Thank you
    Last edited by a moderator: May 17, 2013
  5. May 16, 2013 #4


    Staff: Mentor

    Good point - yes, the s factors cancel. I didn't notice that.

    X'/X = -4/(s + 1)
    ==> (1/X) * dX/ds = -4/(s + 1)
    ==> dX/X = -4/(s + 1) * ds
    Can you take it from here?
  6. May 16, 2013 #5
    Yes. Thanks
  7. May 16, 2013 #6


    Staff: Mentor

    When you get your solution, by all means check that it satisfies tx'' + (t - 2)x' + x = 0, and x(0) = 0.
    Was there another initial condition that you forgot to write?
  8. May 16, 2013 #7
    I got X(s)=A[itex]\frac{1}{(s+1)^{4}}[/itex], so x(t)=Bt[itex]^{3}[/itex]e[itex]^{-t}[/itex], which turned out to be correct in the back of my book. When I plugged in the derivatives of x(t) it worked out. I wasn't given any other initial conditions.
  9. May 17, 2013 #8


    Staff: Mentor

    Then how did you go from tx'' + (t -2)x' + x = 0 to 4sX(s)+s(s+1)X'(s)=0? In particular, what did you get for L(tx'')? This will involve both x(0) -- which you show -- and x'(0) -- which you didn't show.
  10. May 17, 2013 #9
    I got L{tx''}=-[itex]\frac{1}{t}[/itex][itex]\frac{d}{ds}[/itex]tL{x''}=-[itex]\frac{d}{ds}[/itex](s[itex]^{2}[/itex]X-sx(0)-x'(0))=-(2sX+s[itex]^{2}[/itex]X'). I treated x(0) and x'(0) as constants to take the derivative with respect to s
  11. May 17, 2013 #10


    Staff: Mentor

    Using the formula you wrote in post #1, it seems to me that L{tx''} would be -1/t * X'''(s).

    Maybe I'm missing something.

    In any case, it still bothers me that you're able to get a unique solution to a 2nd order diff. equation with only one initial condition.

    For example, if y'' + y = 0, then all we can say is that the solution is the family of curves y = Acos(t) + Bsin(t), where A and B are arbitrary constants.
  12. May 17, 2013 #11


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    This can't be correct. The Laplace transform of f(t) should yield a function only of ##s##; ##t## is integrated out. So ##t## shouldn't appear on the righthand side.
  13. May 17, 2013 #12


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    Of course, the formula she actually used is ##\mathcal L(tf(t))=-F'(s)## although it is poorly written in her writeup. Notice that the solution ##x(t) = Bt^3e^{-t}## satisfies ##x'(0)=0## although that wasn't given or required, and the solution isn't unique. But we do have a singular point at ##t=0## so the existence and uniqueness theorems we like so much don't necessarily apply.
  14. May 17, 2013 #13
    Sorry for the confusion, I should have written it differently. I used L{tf(t)}=-F'(s). I had the other one written down as part of the inverse L to find f(t), which was not as relevant here
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