Inverse Laplace Transform(s/((s^2)+1)^2

[fysiikka111]

Homework Statement

Find inverse Laplace transform of:
s/((s^2)+1)^2

Homework Equations

The Attempt at a Solution

The answer is:
(t/2)sin t
First shifting theorem or partial fractions don't work, I think. From a Laplace transforms table, sin at = a/((s^2)+(a^2)), which is almost the form that the problem is in.
Thanks

 

[hunt_mat]

Here is something that may help, take the sin transform and differentiate it w.r.t s and what do you get?

Mat

 

[fysiikka111]

Thanks for replying.
d/ds of sin transform = -2sa/(s^2+a^2)^2
This would give an inverse of the problem of:
(d/ds)-sin(t)/2
I'm not sure I did that right.

 

[hunt_mat]

Set a=-1 and what do you get?

 

[hunt_mat]

Not quite, we have:
$$\frac{1}{s^{2}+a^{2}}=\int_{0}^{\infty}e^{-st}\sin atdt$$
Differentiate this w.r.t and tell me what you get?

 

[fysiikka111]

I'm getting:
1/(a-s)
with integration-by-parts twice.

 

[Mark44]

No, hunt_mat said to differentiate, not integrate.

 

[fysiikka111]

Oh, right, sorry.
(d/ds)1/(s^2+a^2)=-2s/(s^2+a^2)^2

 

[hunt_mat]

and on the RHS?

 

[fysiikka111]

RHS:
integral from 0 to infinity(-s(e^-st)sin atdt

 

[hunt_mat]

set a=-1 and what have you got?

 

[fysiikka111]

If a=-1:
-2s/(s^2+a^2)^2 = -s/(-s-1)

 

[fysiikka111]

Sorry:
-2s/(s^2+1)^2=-s/(-s-1)

 

[hunt_mat]

No, if a=-1:
$$\frac{s}{(s^{2}+1)^{2}}=\int_{0}^{\infty}\frac{t}{2}e^{-st}\sin tdt$$
What does that tell you?

 

[fysiikka111]

I know then that:
L{tsin(at)/2}=s/(s^2+a^2)?
I'm not sure.

 

[hunt_mat]

Let me re-arrange the RHS:
$$\int_{0}^{\infty}\frac{t\sin t}{2}e^{-st}dt$$
Does it make sense now?

 

[fysiikka111]

No, I'm sorry, I'm just not seeing it. Thanks for all your help, though.

 

[hunt_mat]

Think of it like this, you know that:
$$\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots$$
Now differentiate both sides to show that:
$$\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+4x^{3}+\cdots$$
and now you know the power series for $$(1-x)^{2}$$ is just by differentiating the series.

 

[fysiikka111]

So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
Thanks

 

[hunt_mat]

when you differentiate the integral you take in differentiation under the integral sign:
$$\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt$$
Differentiating the integrand:
$$\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t$$
As now we regard s as variable and t is fixed. So the above can be written as:
$$(-t\sin t)e^{-st}$$
and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?

 

[fysiikka111]

I think I got it now.
After differentiating both sides and dividing by -2:
s/(s^2+a^2)^2=L{tsin at/2}
Hence,
L-1{s/(s^2+1)^2}=tsint/2
However, how do we know that "a" isn't -1, since (-1)^2=1.

 

[hunt_mat]

well done, got there in the end.