To solve for the complementary function of the given differential equation:

$(D^2 - 4D + 5)y = 4e^x,$

we proceed as follows:

Step 1: Auxiliary Equation

The complementary function is derived from solving the homogeneous part of the equation:

$D^2 - 4D + 5 = 0.$

This is a quadratic equation in $D$. Solving for $D$ using the quadratic formula:

$D = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = -4$, and $c = 5$:

$D = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)}.$

Simplify step by step:

$D = \frac{4 \pm \sqrt{16 - 20}}{2}.$ $D = \frac{4 \pm \sqrt{-4}}{2}.$

Since $\sqrt{-4} = 2i$, we get:

$D = \frac{4 \pm 2i}{2}.$

Simplify further:

$D = 2 \pm i.$

Step 2: General Solution of the Homogeneous Equation

The roots $D = 2 + i$ and $D = 2 - i$ are complex. For such roots, the complementary function takes the form:

$y_c = e^{\alpha x} \left( A \cos(\beta x) + B \sin(\beta x) \right),$ where $\alpha$ is the real part of the root and $\beta$ is the imaginary part.

Here, $\alpha = 2$ and $\beta = 1$. Therefore:

$y_c = e^{2x} \left( A \cos x + B \sin x \right).$

Final Answer:

The correct option is (D):

$e^{2x} \left( A \cos x + B \sin x \right).$

Do you want further clarification or a step-by-step breakdown?

Here are 5 related questions for practice:

1. Solve for the complementary function of $D^2 + 4 = 0$.
2. Find the general solution of $D^2 - 6D + 9 = 0$.
3. Derive the complementary function when the auxiliary equation has repeated roots.
4. Solve for the auxiliary equation of $D^2 + 2D + 5 = 0$.
5. Explain the role of $e^{\alpha x}$ and trigonometric functions in the solution.

Tip: For complex roots, always split them into their real and imaginary parts for constructing the complementary solution.