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### Messages - Vickyyy

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##### Quiz-3 / Re: TUT0602 QUIZ3
« on: October 13, 2019, 12:57:42 AM »
hi, is the question y''-4y'=0 instead of y''-4y''=0?

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##### Quiz-2 / TUT0602 Quiz2
« on: October 07, 2019, 10:54:47 PM »
Find the value of b of the given equation which is exact and then solve it using the value b.
(ye2xy+x) + bxe2xyy' = 0
M = ye2x + x
N = bxe2xy
My = e2xy + 2xye2xy
Nx = be2xy + 2bxye2xy
Since My = Nx, b will equal to 1.
M = ye2xy + x, N = xe2xy
φx = M, φy = N
φx = ye2xy + x
φ = ∫(ye2xy+x)dx = ye2xy/2y + x2/2 + h(y)
φy = xe2xy + h'(y)
φy = xe2xy
h'(y) = 0, then h'(y) = c
φ = e2xy/2 +x2/2 = c

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##### Quiz-1 / TUT0602 Quiz1
« on: September 28, 2019, 10:50:17 PM »
Question: Find the solution of the initial value problem.
y' - 2y = e2t, y(0) = 2
Solution:
p(t) = -2, g(t) = e2t, then μ(t) = e∫-2 dt = e-2t
Multiply both sides by μ(t) and we get:
e-2ty' - 2e-2ty = e-2te2t = e0 = 1
Integral both sides:
e-2ty = ∫1dt
e-2ty = t + c, where c is constant
y = e2tt + e2tc
Substitue y(0) = 2 into the equation above:
2 = 0 + c
c = 2
Then, we obtain the solution y = e2t(t+2)

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