The problem is to solve the second-order linear non-homogeneous differential equation $2y'' - 2y' + 25y = 5x - 1$.

1. Problem Description

The problem is to solve the second-order linear non-homogeneous differential equation $2y'' - 2y' + 25y = 5x - 1$.

2. Solution Steps

First, we find the general solution to the homogeneous equation $2y'' - 2y' + 25y = 0$.

The characteristic equation is $2r^2 - 2r + 25 = 0$.

We can solve for $r$ using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$r = \frac{2 \pm \sqrt{(-2)^2 - 4(2)(25)}}{2(2)} = \frac{2 \pm \sqrt{4 - 200}}{4} = \frac{2 \pm \sqrt{-196}}{4} = \frac{2 \pm 14i}{4} = \frac{1}{2} \pm \frac{7}{2}i$

Since the roots are complex, $r = \alpha \pm \beta i$, where $\alpha = \frac{1}{2}$ and $\beta = \frac{7}{2}$.

The general solution to the homogeneous equation is:

$y_h(x) = e^{\alpha x}(c_1\cos(\beta x) + c_2\sin(\beta x))$

$y_h(x) = e^{\frac{1}{2}x}(c_1\cos(\frac{7}{2}x) + c_2\sin(\frac{7}{2}x))$

Next, we find a particular solution $y_p(x)$ to the non-homogeneous equation.

Since the right-hand side is $5x - 1$, we assume a particular solution of the form $y_p(x) = Ax + B$.

Then, $y_p'(x) = A$ and $y_p''(x) = 0$.

Substituting these into the original equation:

$2(0) - 2(A) + 25(Ax + B) = 5x - 1$

$-2A + 25Ax + 25B = 5x - 1$

$25Ax + (25B - 2A) = 5x - 1$

Equating coefficients:

$25A = 5 \implies A = \frac{5}{25} = \frac{1}{5}$

$25B - 2A = -1 \implies 25B - 2(\frac{1}{5}) = -1 \implies 25B = -1 + \frac{2}{5} = -\frac{3}{5} \implies B = -\frac{3}{125}$

So, $y_p(x) = \frac{1}{5}x - \frac{3}{125}$

The general solution to the non-homogeneous equation is the sum of the homogeneous solution and the particular solution:

$y(x) = y_h(x) + y_p(x) = e^{\frac{1}{2}x}(c_1\cos(\frac{7}{2}x) + c_2\sin(\frac{7}{2}x)) + \frac{1}{5}x - \frac{3}{125}$

3. Final Answer

$y(x) = e^{\frac{1}{2}x}(c_1\cos(\frac{7}{2}x) + c_2\sin(\frac{7}{2}x)) + \frac{1}{5}x - \frac{3}{125}$