The given differential equation is:

$y'' - 5y' + 5y = 0$

This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we follow these steps:

Step 1: Find the characteristic equation

We assume a solution of the form $y(t) = e^{rt}$, where $r$ is a constant. Substituting $y(t) = e^{rt}$ into the differential equation:

$r^2 e^{rt} - 5r e^{rt} + 5e^{rt} = 0$

Factoring out $e^{rt}$ (which is never zero):

$e^{rt}(r^2 - 5r + 5) = 0$

This simplifies to the characteristic equation:

$r^2 - 5r + 5 = 0$

Step 2: Solve the characteristic equation

We solve the quadratic equation $r^2 - 5r + 5 = 0$ using the quadratic formula:

$r = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)}$

$r = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$

So the roots are:

$r_1 = \frac{5 + \sqrt{5}}{2}, \quad r_2 = \frac{5 - \sqrt{5}}{2}$

Step 3: Write the general solution

Since we have two distinct real roots, the general solution to the differential equation is:

$y(t) = c_1 e^{r_1 t} + c_2 e^{r_2 t}$

Substituting the values of $r_1$ and $r_2$:

$y(t) = c_1 e^{\frac{5 + \sqrt{5}}{2} t} + c_2 e^{\frac{5 - \sqrt{5}}{2} t}$

This is the general solution to the differential equation.

Would you like any additional details on how this was solved or how the quadratic formula was applied?

Here are 5 related questions to further explore:

1. What happens if the characteristic equation has complex roots?
2. How would you solve a non-homogeneous second-order differential equation?
3. What is the solution method when the roots of the characteristic equation are repeated?
4. How can you solve second-order differential equations using Laplace transforms?
5. How does the solution behavior change when coefficients in the differential equation are functions of $t$?

Tip: When solving a second-order differential equation, always check if the characteristic equation has real, complex, or repeated roots, as the form of the general solution will vary accordingly.