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

$y'' + 6y' + 13y = 0$

We also know the general solution to the corresponding homogeneous equation, where $y_1 = e^{-3x} \cos 2x$ and $y_2 = e^{-3x} \sin 2x$ are two linearly independent solutions. The general solution to the differential equation can be written as:

$y(x) = c_1 y_1 + c_2 y_2 = c_1 e^{-3x} \cos 2x + c_2 e^{-3x} \sin 2x$

where $c_1$ and $c_2$ are constants that we need to determine using the initial conditions. The initial conditions provided are:

1. $y(0) = 2$
2. $y'(0) = 0$

Step 1: Apply the initial condition $y(0) = 2$

Substitute $x = 0$ into the general solution:

$y(0) = c_1 e^0 \cos 0 + c_2 e^0 \sin 0 = c_1 (1) (1) + c_2 (1) (0) = c_1$

So, $c_1 = 2$.

Step 2: Differentiate the general solution to find $y'(x)$

Now, differentiate $y(x)$ with respect to $x$:

$y'(x) = \frac{d}{dx} \left( c_1 e^{-3x} \cos 2x + c_2 e^{-3x} \sin 2x \right)$

Using the product rule and chain rule:

$y'(x) = c_1 \left( -3e^{-3x} \cos 2x - 2e^{-3x} \sin 2x \right) + c_2 \left( -3e^{-3x} \sin 2x + 2e^{-3x} \cos 2x \right)$

Simplifying:

$y'(x) = e^{-3x} \left[ c_1 (-3 \cos 2x - 2 \sin 2x) + c_2 (-3 \sin 2x + 2 \cos 2x) \right]$

Step 3: Apply the initial condition $y'(0) = 0$

Substitute $x = 0$ into $y'(x)$:

$y'(0) = e^0 \left[ c_1 (-3 \cos 0 - 2 \sin 0) + c_2 (-3 \sin 0 + 2 \cos 0) \right]$

$y'(0) = \left[ c_1 (-3 \cdot 1 - 2 \cdot 0) + c_2 (-3 \cdot 0 + 2 \cdot 1) \right]$

$0 = -3c_1 + 2c_2$

Substitute $c_1 = 2$ from the earlier step:

$0 = -3(2) + 2c_2$

$0 = -6 + 2c_2$

$2c_2 = 6 \quad \Rightarrow \quad c_2 = 3$

Final Solution

Thus, the particular solution to the differential equation is:

$y(x) = 2e^{-3x} \cos 2x + 3e^{-3x} \sin 2x$

This is the solution that satisfies the given initial conditions.

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Would you like further details or have any questions? Here are five related problems for you to explore:

1. Solve $y'' + 4y' + 5y = 0$ with initial conditions $y(0) = 1$ and $y'(0) = 0$.
2. Find the general solution to $y'' - 2y' + 5y = 0$.
3. Use the method of undetermined coefficients to solve $y'' + 6y' + 13y = 10e^{-3x}$.
4. Solve $y'' + 9y = 0$ and find the solution if $y(0) = 0$ and $y'(0) = 1$.
5. Show that $y_1 = e^{-3x} \cos 2x$ and $y_2 = e^{-3x} \sin 2x$ are linearly independent.

Tip: To determine whether two solutions are linearly independent, you can check their Wronskian determinant.