Third Order Differential Equation using the reduction of order method.
Solution 1:
You are correct, $$y'''-2y''+y'-2y=(\underbrace{y''-2y'}_{z'})'+\underbrace{y'-2y}_{z}=0,$$ so $z''+z=0.$ So $z=c_1\cos(t)+c_2\sin(t),$ and then you can use the assignment $z = y'-2y$ to see you need only solve $y'-2y = c_1\cos(t)+c_2\sin(t).$ The solutions to $z=y'-2y$ provides the solutions to $z'=y''-2y',$ so we don't need to put any extra effort there.