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Thread: 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 $\displaystyle n \in Z$ with $\displaystyle n \geq 0$, the Laguerre differential equation

    $\displaystyle
    t\frac{d^2f}{dt^2}+(1-t)\frac{df}{dt}+nf=0
    $

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

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

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

    $\displaystyle
    \int^\infty_0e^{-2t}tL_n(t)dt
    $

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

    $\displaystyle
    t\frac{d^2f}{dt^2}+(1-t)\frac{df}{dt}+nf=0
    $

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

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

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

    $\displaystyle
    \int^\infty_0e^{-2t}tL_n(t)dt
    $

    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 $\displaystyle L_n(t)$ is $\displaystyle \frac{n!(s-1)^n}{s^{n+1}}$

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

    $\displaystyle
    \int^\infty_0e^{-2t}tL_n(t)dt
    $

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

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

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


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

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

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

    Therefore $\displaystyle G(2) = \int^\infty_0 e^{-2t}t \, L_n(t) \, dt$ ......
    Last edited by mr fantastic; May 1st 2008 at 02:51 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 $\displaystyle n \in Z$ with $\displaystyle n \geq 0$, the Laguerre differential equation

    $\displaystyle
    t\frac{d^2f}{dt^2}+(1-t)\frac{df}{dt}+nf=0
    $

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

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

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

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

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


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


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


    Therefore:

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


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


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


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


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


    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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