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).