Let's solve the initial value problem step-by-step using the **Method of Undetermined Coefficients**: Given: \[ y'' + 2y' + y = t^2 + 1 - e^t, \quad y(0) = 0, \quad y'(0) = 2 \] --- ### Step 1: Solve the Homogeneous Equation The homogeneous equation is: \[ y'' + 2y' + y = 0 \] The characteristic equation: \[ r^2 + 2r + 1 = 0 \implies (r+1)^2 = 0 \implies r = -1 \text{ (double root)} \] So, the general solution to the homogeneous part: \[ y_h(t) = C_1 e^{-1t} + C_2 t e^{-1t} \] --- ### Step 2: Find a Particular Solution The nonhomogeneous part is \( t^2 + 1 - e^t \). **Guess the particular solution:** \[ y_p(t) = At^2 + Bt + C + De^t \] Substitute \( y_p \) into the ODE: First, compute derivatives: - \( y_p' = 2At + B + De^t \) - \( y_p'' = 2A + De^t \) Substitute into the original ODE: \[ y_p'' + 2y_p' + y_p = (2A + De^t) + 2(2At + B + De^t) + (At^2 + Bt + C + De^t) \] Group like terms: - \(e^t\) terms: \(De^t + 2De^t + De^t = 4De^t\) - \(t^2\) terms: \(At^2\) - \(t\) terms: \(4At + Bt\) - Constants: \(2A + 2B + C\) So, \[ y_p'' + 2y_p' + y_p = At^2 + (4A + B)t + (2A + 2B + C) + 4De^t \] Set equal to the nonhomogeneous part: \[ At^2 + (4A + B)t + (2A + 2B + C) + 4De^t = t^2 + 1 - e^t \] Match coefficients: 1. \( t^2: \quad A = 1 \) 2. \( t: \quad 4A + B = 0 \implies 4(1) + B = 0 \implies B = -4 \) 3. Constants: \( 2A + 2B + C = 1 \implies 2(1) + 2(-4) + C = 1 \implies 2 - 8 + C = 1 \implies C = 7 \) 4. \( e^t: \quad 4D = -1 \implies D = -\frac{1}{4} \) So, \[ y_p(t) = t^2 - 4t + 7 - \frac{1}{4} e^t \] --- ### Step 3: General Solution \[ y(t) = y_h(t) + y_p(t) = C_1 e^{-t} + C_2 t e^{-t} + t^2 - 4t + 7 - \frac{1}{4}e^t \] --- ### Step 4: Apply Initial Conditions Given: - \( y(0) = 0 \) - \( y'(0) = 2 \) Compute \( y(0) \): \[ y(0) = C_1 + 0 + 0 - 0 + 7 - \frac{1}{4}(1) = C_1 + 7 - \frac{1}{4} \] \[ y(0) = C_1 + \frac{28}{4} - \frac{1}{4} = C_1 + \frac{27}{4} = 0 \implies C_1 = -\frac{27}{4} \] Compute \( y'(t) \): \[ y'(t) = -C_1 e^{-t} + C_2 e^{-t} - C_2 t e^{-t} + 2t - 4 - \frac{1}{4}e^t \] At \( t = 0 \): \[ y'(0) = -C_1 + C_2 - 0 + 0 - 4 - \frac{1}{4}(1) \] \[ y'(0) = -C_1 + C_2 - 4 - \frac{1}{4} \] Set \( y'(0) = 2 \), and recall \( C_1 = -\frac{27}{4} \): \[ 2 = -(-\frac{27}{4}) + C_2 - 4 - \frac{1}{4} \] \[ 2 = \frac{27}{4} + C_2 - \frac{17}{4} \] \[ 2 = C_2 + \frac{27 - 17}{4} \] \[ 2 = C_2 + \frac{10}{4} = C_2 + \frac{5}{2} \] \[ C_2 = 2 - \frac{5}{2} = -\frac{1}{2} \] --- ### **Final Answer** \[ \boxed{ y(t) = -\frac{27}{4}e^{-t} - \frac{1}{2} t e^{-t} + t^2 - 4t + 7 - \frac{1}{4}e^t } \] --- **Summary of steps:** 1. Solve the homogeneous equation. 2. Guess and solve for a particular solution. 3. Combine for general solution. 4. Use initial conditions to solve for constants.