Author Topic: tut0401 quiz 3  (Read 285 times)

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tut0401 quiz 3
« on: October 11, 2019, 02:15:17 PM »
hi everyone here is the quiz question find the general solution of the given differential equation
2y'' -3y' +y = 0
therefore 2r^2 -3r + 1 = 0
therefore (2r-1)(r-1) = 0
R1 = 1/2 R2 = 1
therefore y = C1e^((1/2)t)+C2e^t