The Laplace Transform of a function \( f(t) \) is defined as: \[ \mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) \, dt \] where \( s \) is a complex variable. The Laplace Transform converts a time-domain function into a complex frequency-domain representation.

The inverse Laplace Transform converts a function in the complex frequency domain back into the time domain. It is given by: \[ \mathcal{L}^{-1}\{F(s)\} = f(t) \] The inverse is typically found using partial fraction decomposition and known Laplace Transform pairs.

The Laplace Transform of \( f(t) = t^2 \) is: \[ \mathcal{L}\{t^2\} = \frac{2}{s^3} \]

The Laplace Transform of the derivative of a function \( f(t) \) is given by: \[ \mathcal{L}\{f'(t)\} = sF(s) - f(0) \] where \( F(s) \) is the Laplace Transform of \( f(t) \).

The convolution theorem states that the Laplace Transform of the convolution of two functions \( f(t) \) and \( g(t) \) is the product of their Laplace Transforms: \[ \mathcal{L}\{f(t) * g(t)\} = F(s)G(s) \] This property is useful for solving integral equations and systems of differential equations.

Taking the Laplace Transform of both sides, we get: \[ s^2Y(s) + Y(s) = \frac{1}{s^2 + 1} \] Solving for \( Y(s) \) and then taking the inverse Laplace Transform gives: \[ y(t) = 1 - \cos(t) \]

The Laplace Transform of the unit step function \( u(t - a) \) is: \[ \mathcal{L}\{u(t - a)\} = \frac{e^{-as}}{s} \]

The Laplace Transform of \( f(t) = e^{3t} \sin(2t) \) is: \[ \mathcal{L}\{e^{3t} \sin(2t)\} = \frac{2}{(s - 3)^2 + 4} \]

The shifting theorem states that if \( \mathcal{L}\{f(t)\} = F(s) \), then the Laplace Transform of \( f(t - a)u(t - a) \) is: \[ \mathcal{L}\{f(t - a)u(t - a)\} = e^{-as}F(s) \] This property is useful for dealing with time-delayed signals.

Using partial fraction decomposition, we get: \[ F(s) = \frac{1}{2s} - \frac{1}{2(s + 2)} \] Taking the inverse Laplace Transform, we find: \[ f(t) = \frac{1}{2} - \frac{1}{2}e^{-2t} \]