We need to solve the second-order homogeneous linear differential equation $26y'' - 2y' + y = 0$.

1. Problem Description

We need to solve the second-order homogeneous linear differential equation $26y'' - 2y' + y = 0$.

2. Solution Steps

We assume a solution of the form $y = e^{rx}$, where $r$ is a constant. Then, $y' = re^{rx}$ and $y'' = r^2e^{rx}$. Substituting these into the given differential equation, we get:

$26r^2e^{rx} - 2re^{rx} + e^{rx} = 0$

Since $e^{rx}$ is never zero, we can divide by it:

$26r^2 - 2r + 1 = 0$

This is the characteristic equation. Now, we solve for $r$ using the quadratic formula:

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

Here, $a = 26$, $b = -2$, and $c = 1$.

$r = \frac{2 \pm \sqrt{(-2)^2 - 4(26)(1)}}{2(26)}$

$r = \frac{2 \pm \sqrt{4 - 104}}{52}$

$r = \frac{2 \pm \sqrt{-100}}{52}$

$r = \frac{2 \pm 10i}{52}$

$r = \frac{1 \pm 5i}{26}$

$r = \frac{1}{26} \pm \frac{5}{26}i$

So, we have two complex conjugate roots: $r_1 = \frac{1}{26} + \frac{5}{26}i$ and $r_2 = \frac{1}{26} - \frac{5}{26}i$.

Let $\alpha = \frac{1}{26}$ and $\beta = \frac{5}{26}$.

The general solution for complex conjugate roots is given by:

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

$y(x) = e^{\frac{1}{26}x}(c_1\cos(\frac{5}{26}x) + c_2\sin(\frac{5}{26}x))$

3. Final Answer

$y(x) = e^{\frac{1}{26}x}(c_1\cos(\frac{5}{26}x) + c_2\sin(\frac{5}{26}x))$