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!