The initial value problem given is:

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

Step 1: Solve the homogeneous differential equation

The differential equation is a second-order linear homogeneous differential equation with constant coefficients. The general form is:

$y'' + y = 0$

To solve this, we start by finding the characteristic equation associated with the differential equation. Assuming a solution of the form $y = e^{rt}$, we substitute this into the differential equation:

$r^2 e^{rt} + e^{rt} = 0$

Factor out $e^{rt}$:

$e^{rt} (r^2 + 1) = 0$

Since $e^{rt}$ is never zero, we have:

$r^2 + 1 = 0$

Solve for $r$:

$r^2 = -1$ $r = \pm i$

So the general solution to the differential equation is:

$y(t) = C_1 \cos(t) + C_2 \sin(t)$

Step 2: Apply the initial conditions

Given the initial conditions $y(0) = 0$ and $y'(0) = 0$, we will apply these to find the constants $C_1$ and $C_2$.

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

   $y(0) = C_1 \cos(0) + C_2 \sin(0) = C_1 \cdot 1 + C_2 \cdot 0 = C_1$

   Therefore:

   $C_1 = 0$

   So the solution simplifies to:

   $y(t) = C_2 \sin(t)$

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

   First, compute the derivative of $y(t)$:

   $y'(t) = C_2 \cos(t)$

   Now apply the initial condition:

   $y'(0) = C_2 \cos(0) = C_2 \cdot 1 = C_2$

   Therefore:

   $C_2 = 0$

Final Solution

Since both $C_1$ and $C_2$ are zero, the solution to the initial value problem is:

$y(t) = 0$

This means that the function $y(t)$ is identically zero for all $t$.

Do you want further details or have any questions?

Here are some related questions:

1. What is the significance of the characteristic equation in solving differential equations?
2. Can you solve a similar initial value problem for a different differential equation, such as $y'' - y = 0$?
3. What would happen if the initial conditions were different?
4. How would the solution change if the differential equation was non-homogeneous, for example, $y'' + y = \cos(t)$?
5. Can you explain the physical interpretation of this differential equation in a mechanical or electrical system?

Tip: When solving second-order differential equations, the initial conditions are crucial to determining the specific solution to the problem. Always double-check them to ensure accuracy.