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To solve the given differential equation:

\[
\frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} + 2y = 0
\]

with initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \), let's follow these steps.

### Step 1: Formulate the Characteristic Equation
The equation is a second-order linear homogeneous differential equation with constant coefficients. We can solve it by finding the characteristic equation.

1. Write the differential equation in terms of \( D = \frac{d}{dx} \):
   \[
   D^2 y + 2 D y + 2 y = 0.
   \]

2. The characteristic polynomial is:
   \[
   D^2 + 2D + 2 = 0.
   \]

### Step 2: Solve the Characteristic Equation
The characteristic equation is:
\[
r^2 + 2r + 2 = 0.
\]

Solving for \( r \) using the quadratic formula:
\[
r = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2}.
\]

This simplifies to:
\[
r = \frac{-2 \pm 2i}{2} = -1 \pm i.
\]

So the roots are \( r = -1 + i \) and \( r = -1 - i \).

### Step 3: Write the General Solution
Since the roots are complex, the general solution is:
\[
y(x) = e^{-x} \left( C_1 \cos(x) + C_2 \sin(x) \right).
\]

### Step 4: Apply Initial Conditions
Now we use the initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \) to solve for \( C_1 \) and \( C_2 \).

1. **Apply \( y(0) = 0 \):**
   \[
   y(0) = e^{0} \left( C_1 \cos(0) + C_2 \sin(0) \right) = C_1.
   \]
   Since \( y(0) = 0 \), we have \( C_1 = 0 \).

2. **Apply \( y'(0) = 1 \):**
   First, calculate \( y'(x) \):
   \[
   y(x) = e^{-x} \left( C_1 \cos(x) + C_2 \sin(x) \right).
   \]
   Differentiating using the product rule:
   \[
   y'(x) = -e^{-x} \left( C_1 \cos(x) + C_2 \sin(x) \right) + e^{-x} \left( -C_1 \sin(x) + C_2 \cos(x) \right).
   \]
   Substitute \( C_1 = 0 \):
   \[
   y'(x) = -C_2 e^{-x} \sin(x) + C_2 e^{-x} \cos(x) = C_2 e^{-x} (\cos(x) - \sin(x)).
   \]
   Now, substitute \( x = 0 \) and \( y'(0) = 1 \):
   \[
   y'(0) = C_2 e^{0} (\cos(0) - \sin(0)) = C_2 \cdot 1 = C_2.
   \]
   Since \( y'(0) = 1 \), we have \( C_2 = 1 \).

### Final Solution
The solution to the differential equation is:
\[
y(x) = e^{-x} \sin(x).
\]

Would you like further details or have any questions?

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### Related Questions

1. How can we solve a differential equation with non-constant coefficients?
2. What method can be used to solve higher-order linear homogeneous differential equations?
3. How does the method change for a non-homogeneous differential equation?
4. What are some examples of physical systems that use this type of differential equation?
5. How can initial conditions change the behavior of the solution in a differential equation?

### Tip
When solving second-order differential equations, always check if the characteristic roots are real, repeated, or complex, as this changes the form of the general solution.

Solve: d²y/dx² + 2 dy/dx + 2y = 0; y(0) = 0, y'(0) = 1

Learn how to solve the second-order homogeneous differential equation d²y/dx² + 2 dy/dx + 2y = 0 with initial conditions y(0) = 0 and y'(0) = 1. This solution covers characteristic equations, complex roots, and the application of initial conditions for an accurate solution. Suitable for undergraduate math students.

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