1

**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?

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

Pages: [**1**]

1

hi, is the question y''-4y'=0 instead of y''-4y''=0?

2

Find the value of b of the given equation which is exact and then solve it using the value b.

(ye^{2xy}+x) + bxe^{2xy}y^{'} = 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 + x^{2}/2 + h(y)

φ_{y} = xe^{2xy} + h^{'}(y)

φ_{y} = xe^{2xy}

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

φ = e^{2xy}/2 +x^{2}/2 = c

3

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^{-2t}y' - 2e^{-2t}y = e^{-2t}e^{2t} = e^{0} = 1

Integral both sides:

e^{-2t}y = ∫1dt

e^{-2t}y = t + c, where c is constant

y = e^{2t}t + e^{2t}c

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

2 = 0 + c

c = 2

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

Pages: [**1**]