\( y ^ { \prime \prime } + 2 y + 2 = 0 \)

To solve the differential equation \[ y'' + 2y + 2 = 0, \] we start by rewriting it in standard form: \[ y'' + 2y = -2. \] This is a second-order linear non-homogeneous differential equation. We can solve it by finding the complementary solution (solution to the homogeneous equation) and a particular solution. ### Step 1: Solve the Homogeneous Equation The homogeneous part of the equation is: \[ y'' + 2y = 0. \] To solve this, we assume a solution of the form \( y = e^{rt} \). Substituting this into the homogeneous equation gives us the characteristic equation: \[ r^2 + 2 = 0. \] Solving for \( r \): \[ r^2 = -2 \implies r = \pm i\sqrt{2}. \] Thus, the roots are purely imaginary, and we can express the general solution to the homogeneous equation as: \[ y_h(t) = C_1 \cos(\sqrt{2} t) + C_2 \sin(\sqrt{2} t), \] where \( C_1 \) and \( C_2 \) are constants determined by initial conditions. ### Step 2: Find a Particular Solution Next, we need to find a particular solution \( y_p \) to the non-homogeneous equation \( y'' + 2y = -2 \). We can try a constant solution since the right-hand side is a constant. Let \( y_p = A \), where \( A \) is a constant. Substituting \( y_p = A \) into the left-hand side of the equation gives: \[ 0 + 2A = -2. \] Solving for \( A \): \[ 2A = -2 \implies A = -1. \] Thus, a particular solution is: \[ y_p = -1. \] ### Step 3: General Solution The general solution to the original non-homogeneous equation is the sum of the complementary and particular solutions: \[ y(t) = y_h(t) + y_p = C_1 \cos(\sqrt{2} t) + C_2 \sin(\sqrt{2} t) - 1. \] ### Conclusion The general solution to the differential equation \( y'' + 2y + 2 = 0 \) is: \[ y(t) = C_1 \cos(\sqrt{2} t) + C_2 \sin(\sqrt{2} t) - 1, \] where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.