The Laplace transform of \(y(t)\) is \(Y(s), y(0)=2\) and \(y^{\prime}(0)=-1\). Find the Laplace transform of the following expressions: (a) \(y^{\prime}\) (b) \(y^{\prime \prime}\) (c) \(2 y^{\prime \prime}-y^{\prime}+3 y\) (d) \(-y^{\prime \prime}+4 y^{\prime}-6 y\) (e) \(\frac{y^{\prime \prime}}{2}+3 y^{\prime}-y\)

Answer: \(L\{y'(t)\} = sY(s) - 2\) (b) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of y''(t)? Answer: \(L\{y''(t)\} = s^2Y(s) - 2s + 1\) (c) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of \(2y'' - y' + 3y\)? Answer: \(L\{2y'' - y' + 3y\} = 2[s^2Y(s) - 2s + 1] - [sY(s) - 2] + 3Y(s)\) (d) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of \(-y'' + 4y' - 6y\)? Answer: \(L\{-y'' + 4y' - 6y\} = -[s^2Y(s) - 2s + 1] + 4[sY(s) - 2] - 6Y(s)\) (e) Given the Laplace transform of y(t), y(0), and y'(0), what is the Laplace transform of \(\frac{y''}{2} + 3y' - y\)? Answer: \(L\{\frac{y''}{2} + 3y' - y\} = \frac{1}{2}[s^2Y(s) - 2s + 1] + 3[sY(s) - 2] - Y(s)\)