Author Topic: TUT0801 Quiz3  (Read 328 times)

XiaolongZhao

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TUT0801 Quiz3
« 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

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XiaolongZhao

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Re: TUT0801 Quiz3
« Reply #1 on: October 11, 2019, 02:00:56 PM »
Clearer Answer