The given differential equation is:

$y'' = k^2 y$

with initial conditions:

$y(0) = A, \quad y'(0) = B.$

Step 1: Solve the homogeneous equation

The general solution of the second-order linear homogeneous differential equation $y'' = k^2 y$ depends on the sign of $k^2$. Here, $k^2 > 0$ implies real $k$. The characteristic equation is:

$r^2 - k^2 = 0 \quad \Rightarrow \quad r = \pm k.$

Thus, the general solution to the differential equation is:

$y(x) = C_1 e^{kx} + C_2 e^{-kx}.$

Step 2: Apply initial conditions

1. Use $y(0) = A$: $y(0) = C_1 e^{0} + C_2 e^{0} = C_1 + C_2 = A.$ Therefore, we get the first equation: C_1 + C_2 = A. \tag{1}

2. Use $y'(0) = B$: First, differentiate the general solution: $y'(x) = C_1 k e^{kx} - C_2 k e^{-kx}.$ At $x = 0$, this becomes: $y'(0) = C_1 k - C_2 k = B.$ Therefore, we get the second equation: k(C_1 - C_2) = B. \tag{2}

Step 3: Solve for $C_1$ and $C_2$

From equation (2), solve for $C_1 - C_2$: $C_1 - C_2 = \frac{B}{k}.$ Now, solve the system of equations:

1. $C_1 + C_2 = A$
2. $C_1 - C_2 = \frac{B}{k}$

Adding these two equations: $2C_1 = A + \frac{B}{k} \quad \Rightarrow \quad C_1 = \frac{A + \frac{B}{k}}{2}.$

Subtracting these two equations: $2C_2 = A - \frac{B}{k} \quad \Rightarrow \quad C_2 = \frac{A - \frac{B}{k}}{2}.$

Step 4: Write the final solution

Substitute $C_1$ and $C_2$ back into the general solution:

$y(x) = \frac{A + \frac{B}{k}}{2} e^{kx} + \frac{A - \frac{B}{k}}{2} e^{-kx}.$

This is the general solution to the given differential equation with the specified initial conditions.

Final Answer:

$y(x) = \frac{A + \frac{B}{k}}{2} e^{kx} + \frac{A - \frac{B}{k}}{2} e^{-kx}.$

Let me know if you'd like further details or clarifications! Here are five related questions to deepen your understanding:

1. What happens to the solution when $B = 0$?
2. How does the solution change if $k$ is complex?
3. How can you apply this solution to boundary value problems instead of initial conditions?
4. Can this method be used for non-homogeneous differential equations?
5. How would the solution change if $k^2 < 0$?

Tip: Always check the behavior of your solution at boundary points to confirm that it satisfies the initial or boundary conditions!