Inverse Laplace transform of $\frac{6s^2-2}{\left(s^2+1\right)^3}$

Question

I would like to determine the inverse Laplace transform of $$\frac{6s^2-2}{\left(s^2+1\right)^3}.$$

What's the easiest way to do this? (Easy in the sense that it follows easily from some Laplace transform formula or is less computationally heavy.)

The official solution uses multiplication by the $t^n$ rule, but this works because they knew what function works when you take its derivative. I'm not sure how to "guess" the correct function or how to solve this question.

I have a test tomorrow and this question has been bugging me all day. I would really appreciate the help!

Answer 1

Observe that

$$\frac{6s^2-2}{(s^2+1)^3}=\left[6\frac{s^2-1}{(s^2+1)^2}+\frac4{(s^2+1)^2}\right]\cdot\frac1{s^2+1}$$

and thus we have that

$$\mathscr{L}^{-1}\left(\frac{6s^2-2}{(s^2+1)^3}\right)=f(t)\implies F(s):=\mathscr L\{f(t)\}(s)=\left[6\frac{s^2-1}{(s^2+1)^2}+\frac4{(s^2+1)^2}\right]\cdot\frac1{s^2+1}$$

and we can see the rightmost expression as the product of the transform of two functions, making $\;f\;$ a convolution:

$$\mathscr L^{-1}\left(6\frac{s^2-1}{(s^2+1)^2}\right)=6t\cos t+\mathscr L^{-1}\left(\frac4{(s^2+1)^2}\right)=6t\cos t+\mathscr L^{-1}\left(2\frac{2\cdot1^3}{(s^2+1)^2}\right)=$$

$$=6t\cos t+2\sin t-2t\cos t=2\sin t +4t\cos t\,,\;\;\text{and of course}\;\;\mathscr L^{-1}\left(\frac1{s^2+1}\right)=\sin t$$

and we thus get that

$$f(t)=\sin t*\left(2\sin t+4t\cos t\right)=2\sin t*\sin t+4\sin t*t\cos t$$

and now:

$$2\sin t*\sin t=2\int_0^t\sin(t-y)\sin y\,dy=\int_0^t\left(\cos(t-2y)-\cos t\right) dy=$$

$$=\left.-\frac12\sin(t-2y)\right|_0^t-t\cos t=-\frac12\left(\sin(-t)-\sin t\right)+t\cos t=\sin t-t\cos t$$

whereas

$$4\sin t*t\cos t=4\int_0^t\sin(t-y) y\cos y\,dy=4\int_0^ty\,\frac12\left(\sin t+\sin(t-2y)\right)dy=$$

$$=\left.\left[t^2\sin t+\int_0^ty\sin (t-2y)dy\right]\stackrel{\text{by parts}}=t^2\sin t+y\cos(t-2y)\right|_0^t-\int_0^t\cos(t-2y)dy=$$

$$=t^2\sin t+t\cos t+\left.\frac12\sin(t-2y)\right|_0^t=t^2\sin t+t\cos t+\frac12\left(\sin(-t)-\sin t\right)=$$

$$=t^2\sin t+t\cos t-\sin t$$

and finally:

$$=f(t)=\sin t-t\cos t+t^2\sin t+t\cos t-\sin t=t^2\sin t$$

Answer 2

DonAntonio's answer covers the partial fraction/convolution approach. One could fully expand into partial fractions to get

$$F(s) = \frac6{(s^2+1)^2} - \frac8{(s^2+1)^3}$$

The inverse transform of the first term will involve convolving the inverse of $\frac1{s^2+1}$ with itself, and the second term can be evaluated by convolving that result with $\frac1{s^2+1}$. But that's rather cumbersome.

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As another option, recall the inverse transform of the second derivative,

$$\mathscr L^{-1}_t\left\{F''(s)\right\} = t^2 f(t)$$

Integrating twice, we have

$$\int \left(\int \frac{6\sigma^2-2}{(\sigma^2+1)^3}\,d\sigma\right) \, ds = c_1 - \int \frac{2s}{(s^2+1)^2} \, ds = c_2 + c_1s + \frac1{1+s^2}$$

Now,

$$F(s) = c_2 + c_1 s + \frac1{1+s^2} \implies f(t) = c_2 \delta(t) + c_1\delta'(t) + \sin(t)$$

which effectively reduces to $f(t)=\sin(t)$ when $t>0$, and hence

$$\mathscr L^{-1}_t\left\{\frac{6s^2-2}{(s^2+1)^3}\right\} = t^2 f(t) = \boxed{t^2\sin(t)}$$

Answer 3

A solution with almost no calculation: since partial fractions would include $\frac{1}{(s-i)^3}$, which is both $\mathscr{L}\frac12t^2e^{it}$ and $\frac{s^3-3s+i(3s^2-1)}{(s^2+1)^3}$, and since $6s^2-2=0(s^3-3s)+2(3s^2-1)$, the solution is $2\Im\frac12t^2e^{it}=t^2\sin t$.