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We are given the general solution of the differential equation and need to find the specific solution that satisfies the initial value problem. The given differential equation and its general solution are:

\[
y = C_1 e^{4x} + C_2 e^{-x}
\]
with the initial conditions:
\[
y(0) = 1, \quad y'(0) = 1
\]

### Step 1: Find the first derivative \(y'(x)\)
The derivative of \(y(x)\) is:

\[
y'(x) = C_1 \cdot 4 e^{4x} - C_2 e^{-x}
\]

### Step 2: Apply the initial conditions
First, use the initial condition \(y(0) = 1\):

\[
y(0) = C_1 e^{4(0)} + C_2 e^{-0} = C_1 + C_2 = 1
\]
This gives the equation:
\[
C_1 + C_2 = 1 \quad \text{(Equation 1)}
\]

Next, apply the second initial condition \(y'(0) = 1\):

\[
y'(0) = C_1 \cdot 4 e^{4(0)} - C_2 e^{-(0)} = 4C_1 - C_2 = 1
\]
This gives the equation:
\[
4C_1 - C_2 = 1 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of equations
We now solve the system of equations:
1. \(C_1 + C_2 = 1\)
2. \(4C_1 - C_2 = 1\)

Add the two equations together to eliminate \(C_2\):

\[
(4C_1 - C_2) + (C_1 + C_2) = 1 + 1
\]
\[
5C_1 = 2
\]
\[
C_1 = \frac{2}{5}
\]

Substitute \(C_1 = \frac{2}{5}\) into Equation 1:

\[
\frac{2}{5} + C_2 = 1
\]
\[
C_2 = 1 - \frac{2}{5} = \frac{3}{5}
\]

### Step 4: Write the particular solution
Now that we have \(C_1 = \frac{2}{5}\) and \(C_2 = \frac{3}{5}\), the particular solution is:

\[
y = \frac{2}{5} e^{4x} + \frac{3}{5} e^{-x}
\]

This is the solution to the initial-value problem.

Would you like more details or further clarifications? Here are some related questions:

1. How is the first derivative calculated for exponential functions?
2. What method was used to solve the system of equations?
3. How do you verify that the solution satisfies the original differential equation?
4. Why do we need two initial conditions for this second-order differential equation?
5. How can the solution behavior change based on the values of \(C_1\) and \(C_2\)?

**Tip:** Always double-check initial conditions after solving to ensure the particular solution is correct!

The given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem: y = C_1 e^{4x} + C_2 e^{-x}, (-∞, ∞); y'' - 3y' - 4y = 0, y(0) = 1, y'(0) = 1.

Learn how to solve the initial-value problem for the differential equation y'' - 3y' - 4y = 0, with general solutions involving exponential functions. This problem involves solving a second-order linear differential equation with initial conditions y(0) = 1 and y'(0) = 1, and applying key mathematical concepts such as solving systems of equations.

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