$$ \frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 2 y = 0 $$

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To solve the differential equation:

$$
\frac{d^2 y}{dx^2} - 2\frac{dy}{dx} + 2y = 0
$$

we can use the **characteristic equation** method, which is suitable for linear homogeneous differential equations with constant coefficients.

### Step 1: Form the Characteristic Equation

Assume a solution of the form $$ y = e^{rx} $$. Then:

$$
\frac{dy}{dx} = r e^{rx}, \quad \frac{d^2 y}{dx^2} = r^2 e^{rx}
$$

Substitute these into the original differential equation:

$$
r^2 e^{rx} - 2r e^{rx} + 2 e^{rx} = 0
$$

Divide through by $$ e^{rx} $$ (since $$ e^{rx} \neq 0 $$):

$$
r^2 - 2r + 2 = 0
$$

This is the **characteristic equation**:

$$
r^2 - 2r + 2 = 0
$$

### Step 2: Solve the Characteristic Equation

Use the quadratic formula:

$$
r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i
$$

Here, $$ \alpha = 1 $$ and $$ \beta = 1 $$.

### Step 3: Write the General Solution

When the roots of the characteristic equation are complex, $$ r = \alpha \pm \beta i $$, the general solution to the differential equation is:

$$
y(x) = e^{\alpha x} \left( C_1 \cos(\beta x) + C_2 \sin(\beta x) \right)
$$

Substituting $$ \alpha = 1 $$ and $$ \beta = 1 $$:

$$
y(x) = e^{x} \left( C_1 \cos(x) + C_2 \sin(x) \right)
$$

where $$ C_1 $$ and $$ C_2 $$ are arbitrary constants determined by initial conditions.

### Final Answer

After simplifying, all solutions are eˣ times (constant × cos x plus constant × sin x). Thus,

y = e^x [C₁ cos x + C₂ sin x]
The general solution to the differential equation is: $$ y = e^{x} \left( C_1 \cos x + C_2 \sin x \right) $$ where $$ C_1 $$ and $$ C_2 $$ are constants.

\( \frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + 2 y = 0 \)

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