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

The given differential equation is \[ \frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0. \] **Step 1. Write the characteristic equation** Assume a solution of the form \( y = e^{rx} \). Then, we have \[ y' = re^{rx} \quad \text{and} \quad y'' = r^2e^{rx}. \] Substitute these into the differential equation: \[ r^2e^{rx} - 2re^{rx} + 2e^{rx} = 0. \] Divide through by \( e^{rx} \) (since \( e^{rx} \neq 0 \)): \[ r^2 - 2r + 2 = 0. \] **Step 2. Solve the characteristic equation** The characteristic equation is \[ r^2 - 2r + 2 = 0. \] Using the quadratic formula \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] with \( a = 1 \), \( b = -2 \), and \( c = 2 \), we have: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2}. \] Since \(\sqrt{-4} = 2i\), \[ r = \frac{2 \pm 2i}{2} = 1 \pm i. \] **Step 3. Write the general solution** For complex roots \( r = \alpha \pm \beta i \) with \(\alpha = 1\) and \(\beta = 1\), the general solution is given by \[ y(x) = e^{\alpha x} \left( C_1 \cos{(\beta x)} + C_2 \sin{(\beta x)} \right). \] Substitute \(\alpha = 1\) and \(\beta = 1\): \[ y(x) = e^{x} \left( C_1 \cos{x} + C_2 \sin{x} \right). \]