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February 13th, 2017, 11:28 AM   #1
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Struggling to understand this integration and logs..

Hi all,

I have a question which I've started. I'm a bit stumped.


Solve x=0 y=1/2

1/y + 1/(1-y) dy = 1 dx

lny - ln(1-y) = x +c

c=0

This is where I'm unsure how to proceed.

Is it log law?

ln y/(1-y) = x

then take e's to both side?

y/(1-y) =e^x


Thanks for any advice.

Last edited by skipjack; February 15th, 2017 at 12:34 PM.
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February 13th, 2017, 02:02 PM   #2
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Solve what expression? Solve for what?
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February 13th, 2017, 02:04 PM   #3
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Quote:
y/(1-y) =e^x
yes ... now solve for $y$ in terms of $x$
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February 15th, 2017, 09:03 AM   #4
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Quote:
Originally Posted by mathman View Post
Solve what expression? Solve for what?
I presume that nicevans1 problem was "solve the differential equation
(1/y+ 1/(1-y)) dy = 1 dx with the initial condition y(0)= 1/2".
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