To solve the given second-order linear homogeneous differential equation with constant coefficients, we first find the characteristic equation. The roots of this equation will help us determine the general solution. Then, we use the initial conditions to find the specific solution.
We start with the second-order linear homogeneous differential equation given by:
\[
y^{\prime \prime} + 5y^{\prime} + 4y = 0
\]
The characteristic equation corresponding to the differential equation is obtained by substituting \( y = e^{rt} \):
\[
r^2 + 5r + 4 = 0
\]
Factoring the characteristic equation, we find:
\[
(r + 1)(r + 4) = 0
\]
This gives us the roots:
\[
r_1 = -1, \quad r_2 = -4
\]
The general solution of the differential equation is:
\[
y(t) = C_1 e^{-t} + C_2 e^{-4t}
\]
We apply the initial conditions \( y(0) = 4 \) and \( y^{\prime}(0) = -19 \) to find the constants \( C_1 \) and \( C_2 \).
- From \( y(0) = 4 \):
\[
C_1 + C_2 = 4 \quad \text{(1)}
\]
- To find \( y^{\prime}(t) \):
\[
y^{\prime}(t) = -C_1 e^{-t} - 4C_2 e^{-4t}
\]
Applying the second initial condition \( y^{\prime}(0) = -19 \):
\[
-C_1 - 4C_2 = -19 \quad \text{(2)}
\]
We now solve the system of equations (1) and (2):
From equation (1):
\[
C_2 = 4 - C_1
\]
Substituting into equation (2):
\[
-C_1 - 4(4 - C_1) = -19
\]
This simplifies to:
\[
-C_1 - 16 + 4C_1 = -19
\]
Combining like terms:
\[
3C_1 - 16 = -19
\]
Solving for \( C_1 \):
\[
3C_1 = -3 \implies C_1 = -1
\]
Substituting back to find \( C_2 \):
\[
C_2 = 4 - (-1) = 5
\]
Substituting \( C_1 \) and \( C_2 \) back into the general solution gives us the specific solution:
\[
y(t) = -1 e^{-t} + 5 e^{-4t}
\]
The specific solution to the differential equation is:
\[
\boxed{y(t) = -e^{-t} + 5e^{-4t}}
\]
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