We are given the second-order linear differential equation:

$y'' + 5y' = 0$ with initial conditions:

$y(0) = -4, \quad y'(0) = -5.$

Step 1: Solve the homogeneous equation

The given differential equation is:

$y'' + 5y' = 0.$

This is a linear homogeneous differential equation with constant coefficients. The characteristic equation associated with this differential equation is:

$r^2 + 5r = 0.$

We can factor this equation:

$r(r + 5) = 0.$

So, the roots are:

$r = 0, \quad r = -5.$

Thus, the general solution to the differential equation is:

$y(t) = C_1 e^{0t} + C_2 e^{-5t}.$

Since $e^{0t} = 1$, we simplify this to:

$y(t) = C_1 + C_2 e^{-5t}.$

Step 2: Use the initial conditions to determine $C_1$ and $C_2$

1. Apply the first initial condition $y(0) = -4$:

Substitute $t = 0$ into the solution:

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

Thus, we get the equation:

C_1 + C_2 = -4. \tag{1}

2. Apply the second initial condition $y'(0) = -5$:

First, find $y'(t)$:

$y'(t) = \frac{d}{dt} \left( C_1 + C_2 e^{-5t} \right) = -5C_2 e^{-5t}.$

Substitute $t = 0$:

$y'(0) = -5C_2 e^{0} = -5C_2 = -5.$

So, we get:

C_2 = 1. \tag{2}

Step 3: Solve for $C_1$

Substitute $C_2 = 1$ into equation (1):

$C_1 + 1 = -4,$ which gives:

$C_1 = -5.$

Step 4: Write the final solution

Now that we have $C_1 = -5$ and $C_2 = 1$, the solution to the differential equation is:

$y(t) = -5 + e^{-5t}.$

Final Answer:

$y(t) = -5 + e^{-5t}.$

Would you like further explanation on any part of the solution? Or do you have any other questions?

Related Questions:

1. What is the behavior of the solution as $t \to \infty$?
2. How would the solution change if the equation had a different constant in front of $y'$?
3. Can this method be applied to higher-order linear differential equations?
4. How do you solve non-homogeneous differential equations with similar methods?
5. What is the significance of the initial conditions in determining the constants $C_1$ and $C_2$?

Tip: Always check if the roots of the characteristic equation are real or complex, as this determines the form of the solution (exponential vs. sinusoidal).