1. ## 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

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

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

$
\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!!!

2. ## The Overall Question

The overall question is quite a long one:

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

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

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

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

if this helps???

3. Originally Posted by ubhik
[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

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

[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 $G(2) = \int^\infty_0 e^{-2t}t \, L_n(t) \, dt$ ......

4. Originally Posted by ubhik
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

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

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).