The equation you've provided is a general solution to a second-order homogeneous linear differential equation with constant coefficients. The general form of such a solution is typically:

$y = c_1 e^{\alpha_1 x} + c_2 e^{\alpha_2 x}$

In your case, the solution is:

$y = c_1 e^x + c_2 e^{5x}$

Here, $c_1$ and $c_2$ are arbitrary constants, and the exponents $x$ and $5x$ suggest that the characteristic equation associated with the differential equation has roots $1$ and $5$.

Explanation:

This function likely solves a differential equation of the form:

$y'' - 6y' + 5y = 0$

Deriving the differential equation:

1. The characteristic equation for the differential equation can be written as:

$r^2 - 6r + 5 = 0$

2. Solving for $r$ gives the roots $r = 1$ and $r = 5$.

3. Therefore, the general solution to the differential equation is:

$y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$

Where $r_1 = 1$ and $r_2 = 5$, leading to:

$y = c_1 e^x + c_2 e^{5x}$

Do you want more details or have any questions?

Here are 5 related questions:

1. How do you find the roots of the characteristic equation?
2. What are some initial conditions that could determine the values of $c_1$ and $c_2$?
3. How can this general solution be applied to a real-world problem?
4. What happens if the roots of the characteristic equation are complex?
5. How do you solve a non-homogeneous differential equation?

Tip: For a non-homogeneous differential equation, the general solution is the sum of the homogeneous solution and a particular solution.