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ODE y`+2y=y^2

by asdf1
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asdf1
#1
Oct16-05, 03:32 AM
P: 736
for the following question:
y`+2y=y^2

my problem:
suppose u=1/y
so u`=-[y^(-2)]*y`=-1+2u
so du/(1+2u)=-dx
=>1+2u=ce^(-x)
=>u=[1-ce^(-x)]/2
so y=1/u=2/[1-ce^(-x)]

but the correct answer should be y=x/[1+2cx^(2x)]
does anybody know where my calculations went wrong?
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Fermat
#2
Oct16-05, 03:47 AM
HW Helper
P: 876
Quote Quote by asdf1
for the following question:
y`+2y=y^2
my problem:
suppose u=1/y
so u`=-[y^(-2)]*y`=-1+2u
so du/(1+2u)=-dx
=>1+2u=ce^(-x)

=>u=[1-ce^(-x)]/2
so y=1/u=2/[1-ce^(-x)]
but the correct answer should be y=x/[1+2cx^(2x)]
does anybody know where my calculations went wrong?
missed out dividing by 2 when integrating.

so du/(1+2u)=-dx
(1/2)ln(1+2u) = -x +lnC
ln(1+2u) = -2x + lnC
ln{(1+2u)/C} = -2x
1+ 2u = Ce^(-2x)
=>1+2u=ce^(-2x)
asdf1
#3
Oct16-05, 04:50 AM
P: 736
opps~ thanks!!! :)


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