Author Topic: y' - 2y = e2t, y(0) = 2  (Read 333 times)

Victorwoshinidie

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y' - 2y = e2t, y(0) = 2
« on: September 29, 2019, 12:12:05 AM »
 μ(t) = e^∫-2 dt = e^(-2t)
Multiply both sides by μ(t):
e^(-2t)y' - 2e^(-2)ty = e^(-2t)e^(2t)
d/dt (e^(-2t)y) = I
e^(-2t)y = t+c(By integrating both sides)
y=(t+c)/(e^(-2t))
y = (t+c)e^2t
Substitue y(0) = 2 into the equation above, then:
2 = (0 + c)e^0 = C*1
 The solution to the IVP is y = (t+2)e^(2t)