Expert: Sherman D. - 12/3/2008

Question
Hi, please could you help solve the following differential equations.

i. dy/dx = 4y / x(y - 3)

ii. dy/dx = x^2 (3 - 2y), y(0) = 1

iii. dy/dx = (1 - y) / (1 + x), y (0) = 0

Thanks very much!

Answer
i.)
dy/dx = 4y / x(y - 3)
((y - 3)/y)dy = (4/x)dx
(1 - (3/y))dy = (4/x)dx
y - 3ln(y) = 4ln(x) + c

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ii.)
dy/dx = (x^2)(3 - 2y), y(0) = 1
(1/(3 - 2y))dy = (x^2)dx
ln(3 - 2y)/(-2) = (1/3)x^3 + c
ln(3 - 2)/(-2) = (1/3)(0)^3 + c
c = 0
ln(3 - 2y)/(-2) = (1/3)x^3

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iii.)
dy/dx = (1 - y) / (1 + x), y (0) = 0
(1/(1 - y))dy = (1/(1 + x))dx
-ln(1 - y) = ln(1 + x) + c
-ln(1) = ln(1) + c
c = 0
-ln(1 - y) = ln(1 + x)

I actually had to get help with this one.

http://answers.yahoo.com/question/index;_ylt=Aof5O6fM7kLVe.yPAXngInAazKIX;_ylv=3...