The given differential equation is:

$y'' + 4y' = 0$

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

Step 1: Find the characteristic equation

Assume the solution is of the form $y = e^{rt}$, where $r$ is a constant to be determined. Substituting $y = e^{rt}$, $y' = re^{rt}$, and $y'' = r^2 e^{rt}$ into the differential equation gives:

$r^2 e^{rt} + 4r e^{rt} = 0$

Since $e^{rt} \neq 0$, we can divide the entire equation by $e^{rt}$, resulting in the characteristic equation:

$r^2 + 4r = 0$

Step 2: Solve the characteristic equation

Factor the characteristic equation:

$r(r + 4) = 0$

This gives two solutions for $r$:

$r = 0 \quad \text{or} \quad r = -4$

Step 3: General solution

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

$y(t) = C_1 e^{0t} + C_2 e^{-4t}$

Simplifying, we get:

$y(t) = C_1 + C_2 e^{-4t}$

where $C_1$ and $C_2$ are arbitrary constants determined by initial conditions.

Would you like details on finding initial conditions or have any other questions?

Here are 5 related questions:

1. How do you solve non-homogeneous differential equations?
2. What happens when the characteristic equation has complex roots?
3. How do you apply initial conditions to find the constants $C_1$ and $C_2$?
4. Can you solve a third-order differential equation in a similar manner?
5. How would the solution change if there were repeated roots in the characteristic equation?

Tip: When solving homogeneous differential equations, always start by finding the characteristic equation and determining the nature of its roots (real, repeated, or complex).