Laplace Transform L[x(t)] given, find L[tx(t)]

[Color_of_Cyan]

Homework Statement

If L[x(t)] = (s + 4)/(s² + 1), find L[tx(t)]

Homework Equations

Laplace transform:

F(s) = 0∫ f(t)e^-stdtLaplace table

The Attempt at a Solution

Clearly it's not just asking for a Laplace transform. Not sure what it's specifically asking to be honest.

t multiplied by whatever is inside the equation definitely isn't the answer.

 

[Mark44]

Homework Statement

If L[x(t)] = (s + 4)/(s² + 1), find L[tx(t)]

Homework Equations

Laplace transform:

F(s) = 0∫ f(t)e^-stdtLaplace table

The Attempt at a Solution

Clearly it's not just asking for a Laplace transform.

Clearly it is. There are a couple of approaches you could take, but I'd like to see what you have tried before I share them with you.

Not sure what it's specifically asking to be honest.

t multiplied by whatever is inside the equation definitely isn't the answer.

I don't know what this means...

 

[PeterDonis]

Not sure what it's specifically asking to be honest.

Is it asking for the Laplace transform of the function $t x(t)$ (i.e., the function obtained by multiplying $x(t)$ by $t$), given the Laplace transform of $x(t)$? That's what it looks like to me, but I'd like to make sure you have transcribed the problem statement correctly.

Also, please show us explicitly your attempt at a solution(i.e., with equations showing the steps of the calculation you attempted). Just describing it in words isn't enough.

 

[Color_of_Cyan]

Okay, there seems to be one property that sticks out for this:

L[tf(t)] = -dF(s)/ds

L[x(t)] = (s+4)/(s² + 1)

L[f(t)] = (s+4)/(s² + 1). Then find L[tf(t)]

$$\frac{d} {ds} [\frac {s+4} {s^2 + 1}] $$

Just this derivative?

= (d/ds)[(s+4)/(s² + 1)^-1]

=
$$ \frac {(1)(s^2 + 1) - (2s)(s+4)} {{(s^2 + 1)}^2} $$

Would I need only simplify the rest of this to get L[tx(t)] ?

 

[Mark44]

Okay, there seems to be one property that sticks out for this:

L[tf(t)] = -dF(s)/ds

L[x(t)] = (s+4)/(s² + 1)

L[f(t)] = (s+4)/(s² + 1). Then find L[tf(t)]

$$\frac{d} {ds} [\frac {s+4} {s^2 + 1}] $$

Just this derivative?

= (d/ds)[(s+4)/(s² + 1)^-1]

=
$$ \frac {(1)(s^2 + 1) - (2s)(s+4)} {{(s^2 + 1)}^2} $$

Would I need only simplify the rest of this to get L[tx(t)] ?

Yes

 

[Color_of_Cyan]

Okay, thanks! :)

Edit: Oh wait I forgot that there was a negative sign by the derivative, so

(s² + 8s - 1)/(s² + 1)²