Solve the initial value problem  y"+y=0  y(0)=0=y'(0)

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.

Explore the solution to the initial value problem y'' + y = 0 with y(0) = 0 and y'(0) = 0. Learn how to solve second-order homogeneous differential equations and apply initial conditions effectively. Suitable for undergraduate students studying differential equations.

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