Register to reply 
ODE y`+2y=y^2by asdf1
Tags: None 
Share this thread: 
#1
Oct1605, 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=[1ce^(x)]/2 so y=1/u=2/[1ce^(x)] but the correct answer should be y=x/[1+2cx^(2x)] does anybody know where my calculations went wrong? 


#2
Oct1605, 03:47 AM

HW Helper
P: 876

so du/(1+2u)=dx (1/2)ln(1+2u) = x +lnC ln(1+2u) = 2x + lnC² ln{(1+2u)/C²} = 2x 1+ 2u = C²e^(2x) =>1+2u=ce^(2x) 


#3
Oct1605, 04:50 AM

P: 736

opps~ thanks!!! :)



Register to reply 