Q：Find the solution of 2y''- 3y' + y = 0 with y(0) = 2 and y'(0) = 1/2
A:
Set its characteristic equation as 2r^2 - 3r + 1 = 0
Then, r1 = 1/2 , r2 = 1 , r1 ≠ r2
Thus, the solution is y(t) = C1 e^(t/2) + C2 e^t
y'(t) = (C1 /2) e^(t/2) + C2 e^t
Plug in the initial values: y(0) = 2 = C1 + C2
y'(0) = 1/2 = (C1 /2) + C2
C1 = 3 , C2 = -1
Therefore, the final solution is: y(t) = 3e^(t/2) - e^t
Clearer answer shows in the picture below
