The given second-order linear differential equation is:

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

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

Step 1: Form the characteristic equation

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

$r^2 e^{rt} + 5r e^{rt} - 6e^{rt} = 0$

Dividing by $e^{rt}$ (which is never zero), we get the characteristic equation:

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

Step 2: Solve the characteristic equation

Solve the quadratic equation $r^2 + 5r - 6 = 0$ using the quadratic formula:

$r = \frac{-5 \pm \sqrt{5^2 - 4(1)(-6)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 24}}{2} = \frac{-5 \pm \sqrt{49}}{2}$

$r = \frac{-5 \pm 7}{2}$

Thus, the two roots are:

$r_1 = \frac{-5 + 7}{2} = 1, \quad r_2 = \frac{-5 - 7}{2} = -6$

Step 3: Write the general solution

Since the characteristic equation has two distinct real roots, the general solution to the differential equation is:

$y(t) = c_2 e^{r_1 t} + c_3 e^{r_2 t}$

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

$y(t) = c_2 e^{t} + c_3 e^{-6t}$

Final Answer:

The general solution to the differential equation $y'' + 5y' - 6y = 0$ is:

$y(t) = c_2 e^{t} + c_3 e^{-6t}$

Do you need further details or have any questions?

Here are five related questions:

1. How do we solve a homogeneous linear differential equation with repeated roots?
2. What is the significance of the characteristic equation in solving differential equations?
3. How do you determine the form of the solution when complex roots occur in the characteristic equation?
4. Could this method be extended to non-homogeneous differential equations, and how?
5. How do initial conditions affect the form of the general solution?

Tip: When solving differential equations, always start by finding the characteristic equation for homogeneous equations with constant coefficients.