How to find inverse Laplace transform?

[matematikuvol]

Homework Statement

Find inverse Laplace transform
\mathcal {L}^{-1}[\frac{1}{(s^2+a^2)^2}]

Homework Equations

The Attempt at a Solution

I try with theorem
\mathcal{L}[f(t)*g(t)]=F(s)G(s)
So this is some multiple of
\mathcal{L}[\sin at*\sin at]
So \mathcal {L}^{-1}[\frac{1}{(s^2+a^2)^2}]=\propto \sin at*\sin at
Or
\mathcal {L}^{-1}[\frac{1}{(s^2+a^2)^2}]=\propto \int^t_0\sin aq \sin(at-aq)dq
Is there some easier way?
Tnx for answer.

 

[Ray Vickson]

Homework Statement

Find inverse Laplace transform
\mathcal {L}^{-1}[\frac{1}{(s^2+a^2)^2}]

Homework Equations

The Attempt at a Solution

I try with theorem
\mathcal{L}[f(t)*g(t)]=F(s)G(s)
So this is some multiple of
\mathcal{L}[\sin at*\sin at]
So \mathcal {L}^{-1}[\frac{1}{(s^2+a^2)^2}]=\propto \sin at*\sin at
Or
\mathcal {L}^{-1}[\frac{1}{(s^2+a^2)^2}]=\propto \int^t_0\sin aq \sin(at-aq)dq
Is there some easier way?
Tnx for answer.

(1) Don't just say "some multiple"; give the exact factor---it matters a lot!
(2) Just do the convolution integral; it is about as easy a way as any (other than using tables or a computer algebra package).

 

[matematikuvol]

\sin at*\sin at=\int^t_0\sin aq\sin (at-aq)dq=\int^t_0\sin aq(\sin at\cos aq-\sin aq\cos at)dq
So we have to solve to different integrals
\sin at\int^t_0 \sin aq \cos aqdq=\frac{1}{2}\sin^3 at
and
\cos at\int^t_0 \sin^2 aqdq=\cos at\int^t_0\frac{1-\cos 2aq}{2}dq=\frac{1}{2}t\cos at-\frac{1}{4a}\sin 2at
So
\sin at*\sin at=\frac{1}{2}\sin^3 at+\frac{1}{2}t\cos at-\frac{1}{4a}\sin 2at
Laplace transform od $\sin at$ is $\frac{a}{s^2+a^2}$.
So
\mathcal{L}[\sin at*\sin at]=\frac{a^2}{(s^2+a^2)^2}
So
\mathcal{L}^{-1}[\frac{1}{(s^2+a^2)^2}]=\frac{1}{a^2}(\frac{1}{2}\sin^3 at+\frac{1}{2}t\cos at-\frac{1}{4a}\sin 2at)
Is that correct? Is there some easier way to do it? Tnx for the answer.

 

[Ray Vickson]

\sin at*\sin at=\int^t_0\sin aq\sin (at-aq)dq=\int^t_0\sin aq(\sin at\cos aq-\sin aq\cos at)dq
So we have to solve to different integrals
\sin at\int^t_0 \sin aq \cos aqdq=\frac{1}{2}\sin^3 at
and
\cos at\int^t_0 \sin^2 aqdq=\cos at\int^t_0\frac{1-\cos 2aq}{2}dq=\frac{1}{2}t\cos at-\frac{1}{4a}\sin 2at
So
\sin at*\sin at=\frac{1}{2}\sin^3 at+\frac{1}{2}t\cos at-\frac{1}{4a}\sin 2at
Laplace transform od $\sin at$ is $\frac{a}{s^2+a^2}$.
So
\mathcal{L}[\sin at*\sin at]=\frac{a^2}{(s^2+a^2)^2}
So
\mathcal{L}^{-1}[\frac{1}{(s^2+a^2)^2}]=\frac{1}{a^2}(\frac{1}{2}\sin^3 at+\frac{1}{2}t\cos at-\frac{1}{4a}\sin 2at)
Is that correct? Is there some easier way to do it? Tnx for the answer.

\frac{1}{s^2+a^2} \leftrightarrow \frac{1}{a} \sin(a t)
so
\frac{1}{(s^2+a^2)} \leftrightarrow \frac{1}{a^2} \int_0^t \sin(ay) \sin(a(t-y)) \, dy<br /> = \frac{1}{2a^3} \sin(at) -\frac{1}{2a^2} t \cos(a t).
I took the lazy way out and just used the computer package Maple 11. You could also use Wolfram Alpha, which is free for use.