\( \frac { d ^ { 2 } y } { d x ^ { 2 } } - 4 \frac { d y } { d x } + 5 y = c \)

1. **Identify the differential equation:** We are given \[ \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = c, \] where \( c \) is a constant. 2. **Solve the homogeneous part:** The homogeneous equation is \[ \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = 0. \] - Form the characteristic equation: \[ r^2 - 4r + 5 = 0. \] - Solve for \( r \) using the quadratic formula: \[ r = \frac{4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 5}}{2} = \frac{4 \pm \sqrt{16-20}}{2} = \frac{4 \pm \sqrt{-4}}{2}. \] Since \(\sqrt{-4} = 2i\), the roots are \[ r = 2 \pm i. \] - Therefore, the general solution to the homogeneous equation is given by \[ y_h = e^{2x}\left(C_1 \cos x + C_2 \sin x\right), \] where \( C_1 \) and \( C_2 \) are arbitrary constants. 3. **Find a particular solution:** Since the nonhomogeneous term is the constant \( c \), we assume a particular solution of the form \[ y_p = A, \] where \( A \) is a constant. - Substitute \( y_p = A \) into the differential equation. Note that \[ \frac{dy_p}{dx} = 0 \quad \text{and} \quad \frac{d^2 y_p}{dx^2} = 0. \] - Substitute into the equation: \[ 0 - 4(0) + 5A = c \quad \Longrightarrow \quad 5A = c. \] - Solve for \( A \): \[ A = \frac{c}{5}. \] Hence, the particular solution is \[ y_p = \frac{c}{5}. \] 4. **Combine the solutions:** The general solution to the nonhomogeneous differential equation is the sum of the homogeneous and particular solutions: \[ y = y_h + y_p = e^{2x}\left(C_1 \cos x + C_2 \sin x\right) + \frac{c}{5}. \]