Register to reply

ODE y`+2y=y^2

by asdf1
Tags: None
Share this thread:
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?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
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!!! :)


Register to reply