Let's solve the differential equation: \[ y'' - 2y' + y = e^x \sec^2 x \] We'll use the **method of undetermined coefficients**. --- ### 1. **Solve the homogeneous equation** \[ y'' - 2y' + y = 0 \] The characteristic equation is: \[ r^2 - 2r + 1 = 0 \implies (r-1)^2 = 0 \implies r = 1 \text{ (double root)} \] So the **general solution of the homogeneous part** is: \[ y_h(x) = (A + Bx)e^{x} \] --- ### 2. **Find a particular solution \( y_p \)** We need to find a particular solution to: \[ y'' - 2y' + y = e^{x} \sec^2 x \] Let's use the **variation of parameters** method (since the right-hand side is not a simple polynomial, exponential, sine, or cosine). The homogeneous solutions are: - \( y_1 = e^x \) - \( y_2 = x e^x \) The general formula for a particular solution is: \[ y_p = u_1(x) y_1(x) + u_2(x) y_2(x) \] where \( u_1' = -\frac{y_2(x) f(x)}{W(y_1, y_2)} \), \( u_2' = \frac{y_1(x) f(x)}{W(y_1, y_2)} \). **First, compute the Wronskian:** \[ W(y_1, y_2) = \begin{vmatrix} e^x & x e^x \\ e^x & e^x + x e^x \end{vmatrix} = e^x (e^x + x e^x) - x e^x e^x = e^{2x} + x e^{2x} - x e^{2x} = e^{2x} \] So \( W = e^{2x} \). --- #### Compute \( u_1' \): \[ u_1' = -\frac{y_2(x) f(x)}{W} = -\frac{x e^x \cdot e^x \sec^2 x}{e^{2x}} = -x \sec^2 x \] \[ u_1 = -\int x \sec^2 x \, dx \] #### Compute \( u_2' \): \[ u_2' = \frac{y_1(x) f(x)}{W} = \frac{e^x \cdot e^x \sec^2 x}{e^{2x}} = \sec^2 x \] \[ u_2 = \int \sec^2 x \, dx = \tan x \] --- So, \[ y_p = u_1(x) y_1(x) + u_2(x) y_2(x) = u_1(x) e^x + u_2(x) x e^x \] Now, let's compute \( u_1(x) \): \[ u_1 = -\int x \sec^2 x \, dx \] **Let's solve this integral:** Use integration by parts: Let \( u = x \), \( dv = \sec^2 x dx \) - \( du = dx \) - \( v = \tan x \) So, \[ \int x \sec^2 x dx = x \tan x - \int \tan x dx = x \tan x + \ln|\cos x| \] Therefore, \[ u_1 = -\left( x \tan x + \ln|\cos x| \right) \] --- ### **Now plug back in:** \[ y_p = u_1(x) e^x + u_2(x) x e^x = \left[ -\left( x \tan x + \ln|\cos x| \right) e^x \right] + \left( \tan x \cdot x e^x \right) \] \[ = -x \tan x e^x - \ln|\cos x| e^x + x \tan x e^x \] Notice that \( -x \tan x e^x + x \tan x e^x = 0 \). So, \[ y_p = -\ln|\cos x| \, e^x \] --- ## **General Solution** \[ \boxed{ y(x) = (A + Bx) e^x - e^x \ln|\cos x| } \] where \( A, B \) are arbitrary constants.