We are given a second-order linear homogeneous differential equation $y'' + 4y = 0$ with boundary conditions $y(0) = 1$ and $y(\frac{\pi}{4}) = 2$. We are asked to find the value of $y'(0)$.

1. Problem Description

We are given a second-order linear homogeneous differential equation $y'' + 4y = 0$ with boundary conditions $y(0) = 1$ and $y(\frac{\pi}{4}) = 2$. We are asked to find the value of $y'(0)$.

2. Solution Steps

First, we solve the differential equation $y'' + 4y = 0$. The characteristic equation is $r^2 + 4 = 0$, which has roots $r = \pm 2i$. Thus, the general solution is

$y(x) = c_1 \cos(2x) + c_2 \sin(2x)$.

Now, we use the boundary condition $y(0) = 1$.

$y(0) = c_1 \cos(0) + c_2 \sin(0) = c_1(1) + c_2(0) = c_1 = 1$.

So, $c_1 = 1$, and $y(x) = \cos(2x) + c_2 \sin(2x)$.

Next, we use the boundary condition $y(\frac{\pi}{4}) = 2$.

$y(\frac{\pi}{4}) = \cos(2 \cdot \frac{\pi}{4}) + c_2 \sin(2 \cdot \frac{\pi}{4}) = \cos(\frac{\pi}{2}) + c_2 \sin(\frac{\pi}{2}) = 0 + c_2(1) = c_2 = 2$.

So, $c_2 = 2$, and the solution is $y(x) = \cos(2x) + 2 \sin(2x)$.

Now, we find the derivative of $y(x)$:

$y'(x) = -2 \sin(2x) + 4 \cos(2x)$.

Finally, we evaluate $y'(0)$:

$y'(0) = -2 \sin(0) + 4 \cos(0) = -2(0) + 4(1) = 4$.

3. Final Answer

$y'(0) = 4$