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Let's solve Q1 from the image:

Given:
\[
y'' + 4y' - 12y = \alpha x + \cos(3x)
\]
with \(\alpha = 1\).

Substituting \(\alpha = 1\), the equation becomes:
\[
y'' + 4y' - 12y = x + \cos(3x)
\]

This is a second-order non-homogeneous differential equation. To solve it, we can break it down into two parts:
1. **Solve the corresponding homogeneous equation**.
2. **Find a particular solution for the non-homogeneous part**.

### Step 1: Solve the Homogeneous Equation
The homogeneous equation is:
\[
y'' + 4y' - 12y = 0
\]

To solve it, we assume a solution of the form \(y = e^{\lambda x}\), leading to the characteristic equation:
\[
\lambda^2 + 4\lambda - 12 = 0
\]

Solving this quadratic equation:
\[
\lambda = \frac{-4 \pm \sqrt{4^2 + 4 \cdot 12}}{2} = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm \sqrt{64}}{2} = \frac{-4 \pm 8}{2}
\]

Thus, we get:
\[
\lambda = 2 \quad \text{and} \quad \lambda = -6
\]

So, the general solution to the homogeneous equation is:
\[
y_h = C_1 e^{2x} + C_2 e^{-6x}
\]

### Step 2: Find a Particular Solution
We need to find a particular solution \(y_p\) for the non-homogeneous term \(x + \cos(3x)\).

#### Particular Solution for \(x\)
For the \(x\) term, we try a particular solution of the form \(y_p = Ax + B\).

Differentiating, we get:
\[
y_p' = A
\]
\[
y_p'' = 0
\]

Substitute \(y_p = Ax + B\) into the differential equation:
\[
0 + 4A - 12(Ax + B) = x
\]
This

Learn how to solve the second-order non-homogeneous differential equation y'' + 4y' - 12y = x + cos(3x) by finding both the homogeneous solution and a particular solution. This problem covers core concepts in differential equations, including the characteristic equation, particular solutions for polynomial and trigonometric terms, and the superposition principle. Suitable for college-level students.

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