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 need 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 need to find the value of $y'(0)$.

2. Solution Steps

The given differential equation is $y'' + 4y = 0$.

The characteristic equation is $r^2 + 4 = 0$.

Solving for $r$, we have $r^2 = -4$, so $r = \pm 2i$.

Thus, the general solution is $y(x) = c_1 \cos(2x) + c_2 \sin(2x)$.

Now, we apply the boundary conditions.

First, $y(0) = 1$. Plugging in $x=0$ into the general solution, we get:

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

Since $y(0) = 1$, we have $c_1 = 1$.

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

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

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

Since $y(\frac{\pi}{4}) = 2$, we have $c_2 = 2$.

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

Now, we need to find $y'(x)$.

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

Finally, we need to find $y'(0)$.

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

3. Final Answer

$y'(0) = 4$