MAT244--2018F > Quiz-1

Q1: TUT 0201, TUT 5101 and TUT 5102

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Boyu Zheng:
Solution for TUT5101
Question: Find the solution of the given initial value problem y'-2y = e^2t , y(0)=2
let p(t)=-2 and set u=e^(integral p(t)dt) then you get u = e^(-2t)
then you multiply u on each side of the standard form equation and you get
e^(-2t)y'-2e^(-2t)y = 1
then you can find the LHS is equal to (e^(-2t)y)' = 1
integral each side you get (e^-2t)y = t +C
rearrange the equation you get y=e^2t(t+C)
y(t)=(t+C)e^2t
plug in y(0)=2 and you get C=2
y can be written as y=e^2t(t+2)

Victor Ivrii:
All "mathoperators" should be typed with escape character \: \sin \cos \ln \tan , and after it should not be  alphanumerical character (letter or digit)

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