TUT0801 QUIZ3

[taojinwe]

Question: Find the solution of 2y''- 3y' + y = 0 with y(0) = 2 and y'(0) = 1/2

Answer:
Convert the original equation into
                             2r² - 3r + 1 = 0

Then solve the equation as
                             (r-0.5) (r-1) = 0

Therefore,  r₁ = 0.5 , r₂ = 1

Since r1 ≠ r2

We use the formula y(t) = C₁ e^r1t + C₂ e^r2t
                                   = C₁ e^0.5t + C₂ e^t
               
                             y'(t) = (C₁ /2) e^0.5t + C₂ e^t

Then we put the initial values: y(0) = 2 = C₁ + C₂
             
                                             y'(0) = 0.5 = (C₁ /2) + C₂

                                             C1 = 3 , C2 = -1
Thus, the final solution is: y(t) = 3e^0.5t - e^t