We are asked to solve the second-order linear non-homogeneous differential equation: $y'' + y' = 4x + 1$.

1. Problem Description

We are asked to solve the second-order linear non-homogeneous differential equation:

$y'' + y' = 4x + 1$.

2. Solution Steps

First, we solve the homogeneous equation:

$y'' + y' = 0$

The characteristic equation is:

$r^2 + r = 0$

$r(r+1) = 0$

So, $r_1 = 0$ and $r_2 = -1$.

The homogeneous solution is:

$y_h = c_1e^{0x} + c_2e^{-x} = c_1 + c_2e^{-x}$

Now, we find the particular solution. Since the right-hand side is $4x+1$, we would normally assume a solution of the form $Ax+B$. However, since the homogeneous solution contains a constant term, we multiply our assumed solution by $x$. Thus, we assume a particular solution of the form:

$y_p = x(Ax+B) = Ax^2 + Bx$

Now we find the first and second derivatives:

$y_p' = 2Ax + B$

$y_p'' = 2A$

Plugging into the original differential equation:

$2A + 2Ax + B = 4x + 1$

Comparing coefficients, we have:

$2A = 4 \implies A = 2$

$2A + B = 1 \implies 2(2) + B = 1 \implies 4 + B = 1 \implies B = -3$

Therefore, the particular solution is:

$y_p = 2x^2 - 3x$

The general solution is the sum of the homogeneous and particular solutions:

$y = y_h + y_p$

$y = c_1 + c_2e^{-x} + 2x^2 - 3x$

3. Final Answer

$y = c_1 + c_2e^{-x} + 2x^2 - 3x$