$$I(s)=\frac{6}{Ls^2 + Rs + \frac{1}{C}}$$
Having fixed $R$, $C$ and $L$ (for example $R=6$, $C=1/5$, $L=1$), we have:
$$I(s) = \frac{6}{s^2+6s+5}$$
First of all, find the zeros of denominator:
$$s^2 + 4s + 1= 0 \Rightarrow s = -5 \vee s = -1$$
Then $s^2+4s+1= (s+5)(s+1)$.
We can write
$$I(s) = \frac{a}{s+5} + \frac{b}{s+1} = \frac{a(s+1)+b(s+5)}{(s+1)(s+5)} = \frac{s(a+b) + (a + 5b)}{s^2 + 6s + 5} = \frac{6}{s^2+6s+5}$$
Then:
$$\left\{\begin{array}{lcl}a+b & = & 0\\a + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -b\\-b + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -\frac{3}{2}\\b & = & \frac{3}{2} \end{array}\right.$$
Then:
$$I(s) = -\frac{\frac{3}{2}}{s+5} + \frac{\frac{3}{2}}{s+1}$$
The antitransformation yield to:
$$i(t) = -\frac{3}{2}e^{-5t} + \frac{3}{2}e^{-t}$$
I can remove the derivative for you, but my expression still involves $ \mathcal L(y(t)^2) $. Maybe you can work it as a convolution?
Anyways, check my work, but using integration by parts, with $ u = e^{-st} $ and $ v = (y'(t))^2 $, and the fact that $\int (y(t))^2 dt = y'(t)y(t) + \int y'(t)y(t)dt $, and $\int y'(t)y(t)dt = y(t)^2/2 $
$$\int _0^\infty e^{-st} (y'(t))^2 dt = e^{-st}y(t)(y'(t)-y(t)/2)|^\infty_0 + s\int_0^\infty e^{-st}y(t)(y'(t)-y(t)/2) dt $$
Since
$$\int_0^\infty e^{-st}y(t)y'(t)dt = e^{-st}(y(t))^2/2|^\infty_0 + s\int_0^\infty e^{-st}(y(t))^2/2dt$$
Then, under condition that the following limit applies for arbitrary constants A and B (they should for polynomial t-space) $$\lim_{t\to \infty} e^{-st}(A*y(t)y'(t) + B*(y(t))^2) = 0$$
$$\int _0^\infty e^{-st} (y'(t))^2 dt = y(0)y'(0) - \frac{(s+1)y(0)^2}{2} + \frac{s(s-1)}{2}(\int_0^\infty e^{-st}(y(t))^2 dt = \mathcal L(y(t)^2)) $$
Best Answer
Hint:
$$\mathcal{L}(t^2*f(t))=+\dfrac{{d}^2F(s)}{{ds}^2}$$
And you can rewrite the equation to the following: $${cos}^2t=\dfrac{1}{2}(cos(2t)+1)$$
Applying this gives:
$$\mathcal{L}(t^2*{cos}^2t)=\mathcal{L}(t^2*\dfrac{1}{2}(cos2t+1))$$ $$=\dfrac{1}{2}\mathcal{L}(t^2cos2t)+\dfrac{1}{2}\mathcal{L}(t^2)$$ This gives: $$\dfrac{1}{2}*\dfrac{d^2(\frac{s}{s^2+4})}{{ds}^2}+\dfrac{1}{s^3}=\dfrac{s(s^2-12)}{(s^2+4)^3}+\dfrac{1}{s^3}$$