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?

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.