The given differential equation is:

\[
(D^2 + 2)y = e^x \cos(2x)
\]

Where \( D^2 \) denotes the second derivative with respect to \( x \). Let's solve this non-homogeneous linear differential equation.

### Step 1: Solve the Homogeneous Equation
The associated homogeneous equation is:
\[
(D^2 + 2)y = 0
\]
This simplifies to:
\[
y'' + 2y = 0
\]
The characteristic equation for this differential equation is:
\[
r^2 + 2 = 0
\]
Solving for \( r \), we get:
\[
r^2 = -2 \quad \Rightarrow \quad r = \pm i\sqrt{2}
\]
Thus, the general solution for the homogeneous equation is:
\[
y_h(x) = c_1 \cos(\sqrt{2}x) + c_2 \sin(\sqrt{2}x)
\]
where \( c_1 \) and \( c_2 \) are constants.

### Step 2: Find the Particular Solution
Now, for the particular solution, we use the method of undetermined coefficients. The non-homogeneous term is \( e^x \cos(2x) \), so we assume the particular solution to be of the form:
\[
y_p(x) = e^x(A \cos(2x) + B \sin(2x))
\]
where \( A \) and \( B \) are constants to be determined.

Taking the derivatives:
\[
y_p' = e^x[(A \cos(2x) + B \sin(2x)) + (A' \cos(2x) - 2B \cos(2x))]
\]

Substituting the derivatives and equating both, solving for constants will yield the exact solution.

Thus the general solution is

### Solution: 
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Learn how to solve the differential equation (D^2 + 2)y = e^x cos(2x) using the method of undetermined coefficients and characteristic equations. This problem involves advanced differential calculus and is suitable for college-level mathematics.

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