\( y ^ { \prime \prime } + y = x \)

To solve the differential equation: \[ y'' + y = x \] we aim to find the general solution, which consists of two parts: 1. **Complementary (Homogeneous) Solution (\( y_c \))**: Solution to the homogeneous equation \( y'' + y = 0 \). 2. **Particular Solution (\( y_p \))**: A specific solution to the non-homogeneous equation \( y'' + y = x \). ### 1. Complementary Solution (\( y_c \)) Solve the homogeneous equation: \[ y'' + y = 0 \] **Characteristic Equation:** \[ r^2 + 1 = 0 \quad \Rightarrow \quad r = \pm i \] **General Solution of Homogeneous Equation:** \[ y_c = C_1 \cos(x) + C_2 \sin(x) \] where \( C_1 \) and \( C_2 \) are arbitrary constants. ### 2. Particular Solution (\( y_p \)) Find a particular solution to: \[ y'' + y = x \] Since the right-hand side (RHS) is a polynomial of degree 1, we assume a particular solution of the form: \[ y_p = Ax + B \] where \( A \) and \( B \) are constants to be determined. **Compute Derivatives:** \[ y_p' = A \] \[ y_p'' = 0 \] **Substitute \( y_p \) into the Differential Equation:** \[ y_p'' + y_p = 0 + (Ax + B) = Ax + B = x \] **Equate Coefficients:** \[ Ax + B = x \quad \Rightarrow \quad A = 1 \quad \text{and} \quad B = 0 \] **Therefore, the Particular Solution is:** \[ y_p = x \] ### 3. General Solution Combine the complementary and particular solutions: \[ y = y_c + y_p = C_1 \cos(x) + C_2 \sin(x) + x \] where \( C_1 \) and \( C_2 \) are constants determined by initial conditions. ### Final Answer The general solution to the differential equation \( y'' + y = x \) is: \[ y(x) = C_1 \cos(x) + C_2 \sin(x) + x \] where \( C_1 \) and \( C_2 \) are arbitrary constants.