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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.

(ye^2xy+x) + bxe^2xyy^' = 0

M = ye^2x + x

N = bxe^2xy

M_y = e^2xy + 2xye^2xy

N_x = be^2xy + 2bxye^2xy

Since M_y = N_x, b will equal to 1.

M = ye^2xy + x, N = xe^2xy

φ_x = M, φ_y = N

φ_x = ye^2xy + x

φ = ∫(ye^2xy+x)dx = ye^2xy/2y + x²/2 + h(y)

φ_y = xe^2xy + h^'(y)

φ_y = xe^2xy

h'(y) = 0, then h'(y) = c

φ = e^2xy/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 = e^2t, y(0) = 2

Solution:

p(t) = -2, g(t) = e^2t, then μ(t) = e^{∫-2 dt} = e^-2t

Multiply both sides by μ(t) and we get:

e^-2ty' - 2e^-2ty = e^-2te^2t = e⁰ = 1

Integral both sides:

e^-2ty = ∫1dt

e^-2ty = t + c, where c is constant

y = e^2tt + e^2tc

Substitue y(0) = 2 into the equation above:

2 = 0 + c

c = 2

Then, we obtain the solution y = e^2t(t+2)