Find the general solution of y'' - 2y' + 2y = 0

Find the general solution of $y'' - 2y' + 2y = 0$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$: $r^2-2r+2=0.$

The discriminant is $\Delta=(-2)^2-4(1)(2)=4-8=-4<0$, so the roots are complex conjugates: \[r=\frac{2\pm\sqrt{-4}}{2}=1\pm i \implies \alpha=1,\ \ \beta=1.\]

Step 2: Fundamental system

Complex conjugate roots $r=\alpha\pm i\beta$ give the real fundamental system $e^{\alpha x}\cos\beta x,\ e^{\alpha x}\sin\beta x$: $y_1=e^{x}\cos x, \qquad y_2=e^{x}\sin x.$

Step 3: Wronskian

\[ W(y_1,y_2)=\begin{vmatrix}e^x\cos x & e^x\sin x\\ e^x(\cos x-\sin x) & e^x(\sin x+\cos x)\end{vmatrix} = e^{2x}\left(\cos^2x+\sin^2x\right) = e^{2x} \neq 0, \]

confirming $y_1,y_2$ are linearly independent for all $x$.

Step 4: General solution

The solution is recorded in the result below.