To solve the initial value problem:

$y'' + 4y = 0, \quad y(0) = 1, \quad y'(0) = 0$

we follow these steps:

Step 1: Solve the Differential Equation

The equation $y'' + 4y = 0$ is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is:

$r^2 + 4 = 0$

Solving for $r$:

$r^2 = -4 \Rightarrow r = \pm 2i$

This gives us complex roots, $r = 2i$ and $r = -2i$. The general solution for such a differential equation is:

$y(x) = C_1 \cos(2x) + C_2 \sin(2x)$

Step 2: Apply Initial Conditions

1. Use $y(0) = 1$:

   $y(0) = C_1 \cos(2 \cdot 0) + C_2 \sin(2 \cdot 0) = C_1 \cdot 1 + C_2 \cdot 0 = C_1$ So, $C_1 = 1$.

2. Use $y'(0) = 0$:

   First, find $y'(x)$: $y'(x) = -2C_1 \sin(2x) + 2C_2 \cos(2x)$ Substitute $x = 0$: $y'(0) = -2C_1 \sin(0) + 2C_2 \cos(0) = 0 + 2C_2 = 0$ Thus, $C_2 = 0$.

Solution

With $C_1 = 1$ and $C_2 = 0$, the solution to the initial value problem is:

$y(x) = \cos(2x)$

Answer

The correct answer is: (d) $\cos(2x)$

Would you like further details or have any questions?

Here are some related questions to expand on this topic:

1. What is the general solution form for a second-order differential equation with complex roots?
2. How would the solution change if the initial conditions were different?
3. Can initial conditions always uniquely determine the constants in the general solution?
4. How can you verify that a solution satisfies both the differential equation and the initial conditions?
5. What types of initial value problems lead to oscillatory solutions like sine and cosine?

Tip: When solving differential equations, always solve for the roots of the characteristic equation carefully, as they determine the form of the solution.