Author Topic: TUT0301  (Read 470 times)

sunyife2

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TUT0301
« on: September 27, 2019, 02:22:41 PM »
y’ - y = 2te^(2t)   y(0)=1

p(t)= -1 u=e^(∫-1dt)= e^(-t)
y’u-yu=2te^(2t)*u
y’e^(-t)-ye^(-t)=e^(-t)*2te^(2t)
d/dt(y*e^(-t))=2te^t
y*e^(-t) = ∫2te^t dt
           = 2e^t*(t-1)+c
y = (2e^t*(t-1)+c)*e^(t)
  = 2 e^(2t)*(t-1)+c*e^(t)
t = 0, y = 1
1 = 2*1*-1?+c
c = 3

sol: y=2 e^(2t)*(t-1)+3*e^(t)