The given second-order linear differential equation is:

$y'' + 5y' = 0$

with the initial conditions:

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

Step 1: Solve the homogeneous differential equation.

The characteristic equation for the differential equation is:

$r^2 + 5r = 0$

Factorizing the equation:

$r(r + 5) = 0$

So, the roots of the characteristic equation are:

$r = 0 \quad \text{or} \quad r = -5$

Step 2: Write the general solution.

The general solution to the differential equation is given by:

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

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

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

Step 3: Apply the initial conditions.

1. Use $y(0) = -4$:

Substitute $t = 0$ into the general solution:

$y(0) = C_1 + C_2 e^{-5(0)} = C_1 + C_2 = -4$

So, we get the first equation:

$C_1 + C_2 = -4 \quad \text{(Equation 1)}$

2. Use $y'(0) = -5$:

First, find $y'(t)$:

$y'(t) = -5C_2 e^{-5t}$

Now substitute $t = 0$ into $y'(t)$:

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

This gives the second equation:

$-5C_2 = -5 \quad \Rightarrow \quad C_2 = 1$

Step 4: Solve for $C_1$.

Substitute $C_2 = 1$ into Equation 1:

$C_1 + 1 = -4 \quad \Rightarrow \quad C_1 = -5$

Step 5: Write the solution.

The solution to the initial value problem is:

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

Final Answer:

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

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Would you like further details on this solution, or is there anything unclear? Here are some related questions to consider:

1. How do you determine the type of solution for different types of differential equations?
2. What would happen if the right-hand side of the equation was not zero, such as $y'' + 5y' = f(t)$?
3. How do initial conditions affect the constants in the general solution?
4. What are the physical interpretations of the constants in solutions to second-order differential equations?
5. How would the solution change if there were complex roots in the characteristic equation?

Tip: When solving homogeneous linear differential equations with constant coefficients, always start by finding the roots of the characteristic equation!