We are given the 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 the 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

The characteristic equation for the differential equation $y'' + 4y = 0$ is $r^2 + 4 = 0$.

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

Since the roots are complex conjugates, the general solution is of the form

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

We are given the boundary condition $y(0) = 1$.

Substituting $x = 0$ 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 $c_1 = 1$.

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

We are given the boundary condition $y(\frac{\pi}{4}) = 2$.

Substituting $x = \frac{\pi}{4}$ into the solution, we get

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

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

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

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

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

We want to find $y'(0)$.

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

3. Final Answer

$y'(0) = 4$