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MAT244--2019F => MAT244--Test & Quizzes => Quiz-3 => Topic started by: XiaolongZhao on October 11, 2019, 02:00:02 PM

Title: TUT0801 Quiz3
Post by: XiaolongZhao on October 11, 2019, 02:00:02 PM

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

Title: Re: TUT0801 Quiz3
Post by: XiaolongZhao on October 11, 2019, 02:00:56 PM

Clearer Answer