We are given the second-order linear ordinary differential equation:

$y'' + 2y' + 2y = 2e^{-x}$ and the complementary (homogeneous) solution: $Y_h = e^{-x}(A \cos x + B \sin x)$ where $A$ and $B$ are constants to be determined from initial conditions.

Step 1: Find the particular solution using the method of undetermined coefficients.

Since the right-hand side of the equation is $2e^{-x}$, a function of the form $e^{-x}$, we propose a particular solution of the form:

$y_p = e^{-x}(C)$ where $C$ is an undetermined constant to be determined.

Step 2: Differentiate $y_p$.

First, find the first and second derivatives of $y_p$:

$y_p' = \frac{d}{dx}\left(e^{-x}C\right) = -Ce^{-x}$ $y_p'' = \frac{d}{dx}\left(-Ce^{-x}\right) = Ce^{-x}$

Step 3: Substitute into the original equation.

Now, substitute $y_p$, $y_p'$, and $y_p''$ into the left-hand side of the ODE:

$y_p'' + 2y_p' + 2y_p = Ce^{-x} + 2(-Ce^{-x}) + 2(Ce^{-x})$ $= Ce^{-x} - 2Ce^{-x} + 2Ce^{-x}$ $= Ce^{-x}$

Step 4: Equate to the right-hand side of the equation.

The right-hand side of the equation is $2e^{-x}$, so we equate:

$Ce^{-x} = 2e^{-x}$

From this, we conclude that:

$C = 2$

Step 5: Write the full solution.

The general solution to the differential equation is the sum of the complementary function and the particular solution:

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

Thus, the general solution is:

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

Summary:

The particular solution is found by using undetermined coefficients, and the general solution to the differential equation is:

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

Would you like any further details or clarifications on any steps?

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Related Questions:

1. How would the solution change if the right-hand side were a polynomial, such as $2x^2$?
2. What would happen if the right-hand side were $2e^{x}$?
3. How can the complementary function change if the characteristic equation has complex roots?
4. What happens if we have repeated roots in the characteristic equation?
5. Can this method of undetermined coefficients be applied to non-linear ODEs?

Tip:

When using the method of undetermined coefficients, ensure that the form of the particular solution accounts for any overlaps with the complementary function. For example, if $e^{-x}$ is part of $Y_h$, then $y_p$ should involve an extra factor of $x$ to avoid duplication.