In given question,
The formula is,
L(c^at)=\(\frac{1}{s-a log( c )}\)
L(3^2t)=\(\frac{1}{s-2 log?(3)}\).

In given question,
L(sint)=\(\frac{1}{s^2+1}\)
By effect of division by t,
\(L(\frac{sin?t}{t})=\int_{s}^{\infty}\frac{1}{s^2+1} ds\)
=\(\frac{\pi}{2}-tan^{-1}s\)
=cot^-1s
By effect of derivative of Laplace Transform,
\(L\left (\frac{d}{dt}\left (\frac{sin?t}{t}\right )\right )=s�cot^{-1}s-1\).

It is not possible to solve an equation if input and initial conditions are zero becase Y(s) becomes zero where Y(s) is Laplace Transform of y(t) function.

\(g(t)=t^{5/2}=\frac{5}{2} \int_0^t t^{\frac{3}{2}} dt=\frac{15}{4} \int_0^t \int_0^t ?t dt dt\)
let f(t)=?t, hence, F(s)=\(\frac{\sqrt{?}}{2s^{\frac{3}{2}}}\)
hence, G(s)=\(\frac{15}{4} \,\frac{1}{s^2} \,F(s)=\frac{15}{8} \frac{??}{s^{7/2}}\)

The Laplace Transform of a functions is given by
\(L\{f(t)\}=F(s)=\int_0^{\infty}f(t)e^{-st}dt\)
Put f(t) = cos(at)
On solving above integral, we get s ⁄ s²+a².

We know that,
Given, Y(s)=\(\frac{1}{s(s-1)(s^2+1)}\)
Let, G(s)=\(\frac{1}{(s-1)(s^2+1)}=\frac{1}{2(s^2-1)}-\frac{s+1}{2(s^2+1)}=\frac{1}{2*(s-1)}-\frac{s}{2(s^2+1)}-\frac{1}{2(s^2+1)}\)
Now, g(t)=\(\frac{1}{2}e^t-\frac{1}{2}cos(t)-\frac{1}{2}cos(t)\)
Now, Y(s)=\(\frac{1}{2}G(s)=>y(t)=\int_0^t g(t)dt=\frac{1}{2}e^t+\frac{1}{2}sin(t)-\frac{1}{2}cos(t)\)

Given, 5 ^d3⁄_dt³ y(t) + 10 ^d⁄_dt y(t) – 5y(t) = x(t) + ?₀^t x(t)dt,
By laplace transform, 5s³ Y(s)+10sY(s)-5Y(s)=X(s)+1/s X(s)
Hence,
\(\frac{Y(s)}{X(s)} =\frac{[1+1/s]}{5s^3+10s-5}=\frac{[s+1]}{s(5s^3+10s-5)}\)
Poles of \(\frac{Y(s)}{X(s)}\) are, 0 ,0.455,-0.226�1.467
Hence system is Unstable.

In given question,
\(L^{-1} \left (\frac{s+1}{(s-1)(s+2)^2}\right )\)
Using properties of partial fractions-
s+1=A(s+2)²+B(s-1)(s+2)+C(s-1)
At s=1, A=\(\frac{2}{9}\)
At s=2, C=\(\frac{1}{3}\)
At s=0, B=\(\frac{-2}{9}\)
Re substituting all se values in original fraction,
=\(L^{-1} \left (\frac{2}{9(s-1)} + \frac{-2}{9(s+2)} + \frac{1}{3(s+2)^2}\right)\)
=\(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}�t\).

The Laplace transform of a function is given by
\(\{f(t)\}=F(s)=\int_0^{\infty} f(t)e^{-st}dt\)
put f(t) = 1 ⁄ √?t
The solution for above question is obtained by solving above integral.

We know that,
F(s)=\(\int_{-\infty}^\infty sin?(at)u(t) e^{-st} dt=\int_0^? sin?(at)e^{-st} dt\)
=\(\left [\frac{e^{-st}}{a^2+s^2}[-ssin(at)-acos?(at)]\right ]_?^0\)
=\(\frac{a}{a^2+s^2}\)