We are given the second-order linear differential equation:

$y'' + 2y' + 2y = 2e^{-x}$

with the complementary function:

$Y_h = e^{-x}(A\cos x + B\sin x).$

We will solve this using the method of undetermined coefficients to find a particular solution.

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Step 1: Identify the Form of $Y_p$

The right-hand side of the equation is $2e^{-x}$. Normally, for an equation of the form:

$y'' + ay' + by = Ce^{rx},$

we assume a particular solution of the form:

$Y_p = Ke^{-x}.$

However, $e^{-x}$ already appears in the complementary solution, meaning that our standard assumption will lead to duplication. To account for this, we multiply by $x$ to obtain a new linearly independent function:

$Y_p = xKe^{-x}.$

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Step 2: Compute $Y_p'$ and $Y_p''$

We differentiate $Y_p = xKe^{-x}$:

$Y_p' = K e^{-x} + xK(-e^{-x}) = K e^{-x} - xK e^{-x}.$

$Y_p' = (K - xK)e^{-x}.$

Now, differentiate again to get $Y_p''$:

$Y_p'' = (-K e^{-x} - K e^{-x} + xK e^{-x}).$

$Y_p'' = (-2K + xK)e^{-x}.$

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Step 3: Substitute into the ODE

Substituting $Y_p$, $Y_p'$, and $Y_p''$ into:

$y'' + 2y' + 2y = 2e^{-x},$

we get:

$(-2K + xK)e^{-x} + 2(K - xK)e^{-x} + 2(xKe^{-x}) = 2e^{-x}.$

Factor out $e^{-x}$:

$e^{-x} \left[ (-2K + xK) + 2(K - xK) + 2xK \right] = 2e^{-x}.$

Simplify the expression inside the brackets:

$-2K + xK + 2K - 2xK + 2xK = -2K + 2K + xK - 2xK + 2xK.$

$= 0 + xK = xK.$

So we are left with:

$xK e^{-x} = 2e^{-x}.$

Since $e^{-x}$ cancels out on both sides:

$xK = 2.$

Thus,

$K = 2.$

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Step 4: Write the General Solution

Now that we have the particular solution:

$Y_p = 2x e^{-x},$

the general solution is:

$y = Y_h + Y_p.$

$y = e^{-x}(A\cos x + B\sin x) + 2x e^{-x}.$

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Final Answer:

$y = e^{-x}(A\cos x + B\sin x) + 2x e^{-x}.$

where $A$ and $B$ are arbitrary constants determined by initial conditions (if given).

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Would you like to see an example with initial conditions, or need further clarification?

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Related Questions:

1. What is the method of undetermined coefficients? 2