The problem asks you to solve the boundary value problem:

$y'' + 2y' + y = 3t^2 e^{-t}$ with boundary conditions: $y(0) = 4 \quad \text{and} \quad y'(0) = -2.$

To solve this, we will follow these steps:

Step 1: Solve the corresponding homogeneous equation

The homogeneous part of the differential equation is: $y'' + 2y' + y = 0.$ Solve this characteristic equation: $r^2 + 2r + 1 = 0.$ Factorizing: $(r + 1)^2 = 0,$ so the roots are $r = -1$ with multiplicity 2. Therefore, the general solution to the homogeneous equation is: $y_h(t) = (C_1 + C_2 t) e^{-t}.$

Step 2: Find a particular solution

For the non-homogeneous equation, we need to find a particular solution. We will use the method of undetermined coefficients.

The right-hand side is $3t^2 e^{-t}$, so we assume the form of the particular solution as: $y_p(t) = (At^2 + Bt + C) e^{-t}.$ Differentiate $y_p(t)$ twice and substitute into the original differential equation to find the coefficients $A$, $B$, and $C$.

Step 3: Apply the boundary conditions

Once you have the general solution, which is the sum of the homogeneous and particular solutions: $y(t) = y_h(t) + y_p(t),$ you can use the boundary conditions $y(0) = 4$ and $y'(0) = -2$ to solve for the constants $C_1$ and $C_2$.

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