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The given differential equation is:

\[
\frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + 4y = 4x^2 + 8
\]

This is a second-order linear non-homogeneous differential equation. To solve this, we can follow these steps:

### Step 1: Solve the homogeneous equation
First, solve the homogeneous part of the equation:

\[
\frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + 4y = 0
\]

The characteristic equation for this is:

\[
r^2 + 2r + 4 = 0
\]

Solving this quadratic equation using the quadratic formula:

\[
r = \frac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2}
\]

\[
r = \frac{-2 \pm 2i\sqrt{3}}{2} = -1 \pm i\sqrt{3}
\]

Thus, the roots are complex:

\[
r = -1 \pm i\sqrt{3}
\]

The general solution to the homogeneous equation is then:

\[
y_h(x) = e^{-x} \left( C_1 \cos(\sqrt{3}x) + C_2 \sin(\sqrt{3}x) \right)
\]

### Step 2: Find a particular solution
Next, we find a particular solution to the non-homogeneous equation. Since the right-hand side is \(4x^2 + 8\), we can try a particular solution of the form:

\[
y_p(x) = Ax^2 + Bx + C
\]

Taking derivatives:

\[
\frac{dy_p}{dx} = 2Ax + B
\]
\[
\frac{d^2 y_p}{dx^2} = 2A
\]

Substituting these into the non-homogeneous equation:

\[
2A + 2(2Ax + B) + 4(Ax^2 + Bx + C) = 4x^2 + 8
\]

Simplifying:

\[
2A + 4Ax + 2B + 4Ax^2 + 4Bx + 4C = 4x^2 + 8
\]

Now, collect like terms:

\[
(4A)x^2 + (4A + 4B)x + (2A + 2B + 4C) = 4x^2 + 8
\]

Equating the coefficients:

For \(x^2\): \(4A = 4 \Rightarrow A = 1\)

For \(x\): \(4A + 4B = 0 \Rightarrow 4 + 4B = 0 \Rightarrow B = -1\)

For the constant: \(2A + 2B + 4C = 8 \Rightarrow 2(1) + 2(-1) + 4C = 8 \Rightarrow 4C = 8 \Rightarrow C = 2\)

Thus, the particular solution is:

\[
y_p(x) = x^2 - x + 2
\]

### Step 3: General solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(x) = y_h(x) + y_p(x)
\]

\[
y(x) = e^{-x} \left( C_1 \cos(\sqrt{3}x) + C_2 \sin(\sqrt{3}x) \right) + x^2 - x + 2
\]

This is the general solution to the given differential equation.

Would you like more details or have any questions?

Here are 5 questions to further explore the topic:
1. How do we solve a characteristic equation with complex roots?
2. What is the significance of the homogeneous solution in differential equations?
3. Why did we choose a quadratic polynomial for the particular solution?
4. How would the solution change if the non-homogeneous term was different?
5. What are other methods for solving non-homogeneous differential equations?

**Tip:** When solving non-homogeneous differential equations, always check if the particular solution you propose matches the form of the non-homogeneous term (right-hand side).

Find the general solution of the differential equation: d^2y/dx^2 + 2dy/dx + 4y = 4x^2 + 8.

Learn how to solve the second-order linear differential equation d^2y/dx^2 + 2dy/dx + 4y = 4x^2 + 8. This guide covers solving the homogeneous and non-homogeneous parts, finding characteristic equations, and applying the method of undetermined coefficients. Suitable for undergraduate students studying differential equations.

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