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Math Help - Laguerre differential equation

  1. #1
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    Laguerre differential equation

    I don't really understand how to start this question, can anyone help please?

    For n \in Z with n \geq 0, the Laguerre differential equation

     <br />
t\frac{d^2f}{dt^2}+(1-t)\frac{df}{dt}+nf=0<br />

    Has the solution L_n(t) for which L_n(0)=n!

    Prove that the Laplace tranform on L_n(t) is \frac{n!(s-1)^n}{s^{n+1}}

    Hence find the Laplace transform of tL_n(t) and deduce the value of

     <br />
\int^\infty_0e^{-2t}tL_n(t)dt<br />

    Can someone please help me or even give me a push in the right direction, please, I'm really really stuck!! I have spent days trying to make a start but I really don't understand where to start!!!
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  2. #2
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    The Overall Question

    The overall question is quite a long one:

    For n \in Z with n \geq 0, the Laguerre differential equation

     <br />
t\frac{d^2f}{dt^2}+(1-t)\frac{df}{dt}+nf=0<br />

    Has the solution L_n(t) for which L_n(0)=n!

    Prove that the Laplace tranform on L_n(t) is \frac{n!(s-1)^n}{s^{n+1}}

    Hence find the Laplace transform of tL_n(t) and deduce the value of

     <br />
\int^\infty_0e^{-2t}tL_n(t)dt<br />

    if this helps???
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  3. #3
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    Quote Originally Posted by ubhik View Post
    [snip]
    Prove that the Laplace tranform of L_n(t) is \frac{n!(s-1)^n}{s^{n+1}}

    Hence find the Laplace transform of tL_n(t) and deduce the value of

     <br />
\int^\infty_0e^{-2t}tL_n(t)dt<br />

    [snip]
    Using the standard result LT[t \, f(t)] = -\frac{d}{ds}F(s) where F(s) = LT[f(t)] it follows that

    LT[t \, L_n(t)] = -\frac{d}{ds} \left[ \frac{n!(s-1)^n}{s^{n+1}} \right] = -n! \, \frac{d}{ds} \left[ \frac{(s-1)^n}{s^{n+1}} \right] = G(s)

    where you can do the differentiations to get the expression for G(s).


    Now note that by definition: LT[t \, L_n(t)] = \int^\infty_0 e^{-st}t \, L_n(t) \, dt.

    But LT[t \, L_n(t)] = G(s).

    Therefore G(s) = \int^\infty_0 e^{-st}t \, L_n(t) \, dt.

    Therefore G(2) = \int^\infty_0 e^{-2t}t \, L_n(t) \, dt ......
    Last edited by mr fantastic; May 1st 2008 at 03:51 AM. Reason: Added a line for greater clarity.
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  4. #4
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    Quote Originally Posted by ubhik View Post
    I don't really understand how to start this question, can anyone help please?

    For n \in Z with n \geq 0, the Laguerre differential equation

     <br />
t\frac{d^2f}{dt^2}+(1-t)\frac{df}{dt}+nf=0<br />

    Has the solution L_n(t) for which L_n(0)=n!

    Prove that the Laplace tranform on L_n(t) is \frac{n!(s-1)^n}{s^{n+1}}
    [snip]
    Standard results:

    LT\left[ \frac{df}{dt}\right] = s F(s) - f(0) = s F(s) - n!

    LT\left[ \frac{d^2f}{dt^2}\right] = s^2 F(s) - s f(0) - f'(0) = s^2 F(s) - n! s - f'(0)

    where F(s) = LT[f(t)]. Therefore:


    LT\left[ t \, \frac{df}{dt}\right] = -\frac{d}{ds} \left[ s F(s) - n!\right] = -F(s) - s \frac{dF}{ds}.


    LT\left[t \, \frac{d^2f}{dt^2}\right] = -\frac{d}{ds} \left[ s^2 F(s) - n! s - f'(0) \right] = -2 s F(s) - s^2 \frac{dF}{ds} + n!


    Therefore:

    LT\left[ t \, \frac{d^2 f}{d t^2} + \frac{df}{dt} - t\, \frac{df}{dt} + nf \right] = 0


    \Rightarrow \left( -2 s F(s) - s^2 \frac{dF}{ds} + n! \right) + (s F(s) - n!) - \left( -F(s) - s \frac{dF}{ds} \right) + n F(s) = 0


    \Rightarrow (s - s^2) \frac{dF}{ds} + (n + 1 - s) F(s) = 0


     \Rightarrow \int \frac{1}{F} \, dF = \int \frac{s - n - 1}{s - s^2} \, ds


    \Rightarrow F(s) = A \frac{(s - 1)^n}{s^{n+1}}.


    It's not too hard to show that f(0) = A and therefore A = n! (so I'll leave this last small detail for you to complete).
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