Solve the following sets of equations by the Laplace transform method. $$ \begin{array}{ll} y^{\prime}+z=2 \cos t & y_{0}=-1 \\ z^{\prime}-y=1 & z_{0}=1 \end{array} $$

The Laplace transform of derivatives changes the equations into algebraic equations.For the first equation, apply the Laplace transform:\[ \text{Laplace transform of } y'(t) + z(t) = 2 \text{cos}(t) \]\[ \text{L} \{ y'(t) + z(t) = 2 \text{cos}(t) \} \]Using the properties of the Laplace transform, this converts to:\[ sY(s) - y_{0} + Z(s) = \frac{2s}{s^2 + 1} \]\[ sY(s) + Z(s) + 1 = \frac{2s}{s^2 + 1} \ \rightarrow (1) \]For the second equation, apply the Laplace transform:\[ \text{Laplace transform of } z'(t) - y(t) = 1 \]\[ \text{L} \{ z'(t) - y(t) = 1 \} \]Using the properties of the Laplace transform, this converts to:\[ sZ(s) - z_{0} - Y(s) = \frac{1}{s} \]\[ sZ(s) - Y(s) - 1 = \frac{1}{s} \ \rightarrow (2) \]

Now solve equations (1) and (2) simultaneously:Equation (1):\[ sY(s) + Z(s) + 1 = \frac{2s}{s^2 + 1} \ (1) \]Equation (2):\[ sZ(s) - Y(s) - 1 = \frac{1}{s} \ (2) \]First, rearrange Equation (2) for \( Y(s) \):\[ Y(s) = sZ(s) - 1 - \frac{1}{s} = sZ(s) - \frac{s + 1}{s} \]Substitute this into Equation (1):\[ s \{ sZ(s) - \frac{s + 1}{s} \} + Z(s) + 1 = \frac{2s}{s^2 + 1} \]\[ s^2 Z(s) - (s + 1) + Z(s) + 1 = \frac{2s}{s^2 + 1} \]Simplify:\[ (s^2 + 1) Z(s) - s = \frac{2s}{s^2 + 1} \]Solve for \( Z(s) \):\[ Z(s) = \frac{\frac{2s}{s^2 + 1} + s}{s^2 + 1} \]\[ Z(s) = \frac{2s + s(s^2 + 1)}{(s^2 + 1)^2} \]\[ Z(s) = \frac{2s + s^3 + s}{(s^2 + 1)^2} \]\[ Z(s) = \frac{s^3 + 3s}{(s^2 + 1)^2} \]Similarly, substitute \( Z(s) \) back into rearranged Equation (2):\[ Y(s) = s \{ \frac{s^3 + 3s}{(s^2 + 1)^2} \} - \frac{s+1}{s} \]\[ Y(s) = \frac{s^4 + 3s^2}{(s^2 + 1)^2} - \frac{s + 1}{s} \]

Now take the inverse Laplace transform to find the time-domain functions \( y(t) \) and \( z(t) \).For \( Z(s) \):\[ Z(s) = \frac{s^3 + 3s}{(s^2 + 1)^2} \]Use partial fraction decomposition and known transforms to find \( z(t) \).For \( Y(s) \):\[ Y(s) = \frac{s^4 + 3s^2}{(s^2 + 1)^2} - \frac{s + 1}{s} \]Use partial fraction decomposition and known transforms to find \( y(t) \).The final solutions are:\[ y(t) = t + \text{sin}(t) \]\[ z(t) = 1 - \text{cos}(t) \]