Author Topic: TUT0801 QUIZ3  (Read 247 times)

taojinwe

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TUT0801 QUIZ3
« on: October 11, 2019, 03:38:48 PM »
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 er1t + C2 er2t
                                   = C1 e0.5t + C2 et
               
                             y'(t) = (C1 /2) e0.5t + C2 et

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) = 3e0.5t - et